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# Machine Learning - A Probabilistic Perspective Exercises - Chapter 4

This is a continuation of the exercises in "Machine learning - a probabilistic perspective" by Kevin Murphy. Chapter 4 is on "Gaussian Models". Let's get started!

4.1 Uncorrelated does not imply independent

Let $$X \sim U(-1,1)$$ and $$Y = X^2$$. Clearly Y is dependent on X, show $$\rho(X,Y)=0$$.

$$\rho(X,Y)$$ is just a normalised version of the covariance, so we just need to show the covariance is zero, i.e.:

$$\text{Cov}(X,Y) = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]$$

Clearly $$\mathbb{E}[X] = 0$$ and so we just need to calculate $$\mathbb{E}[XY]$$ and show this is zero. We can write:

$$\mathbb{E}[XY] = \int_{-1}^1 dx \int_0^1 dy \ xy p(x,y)$$

Then we say $$p(x,y) = p(y|x) p(x)$$, but $$p(y|x) = \delta(y - x^2)$$, i.e. a dirac-delta function, and $$p(x)=1/2$$, i.e. just a constant. This means we can evaluate the integral over y to get:

$$\mathbb{E}[XY] = 1/2 \int_{-1}^1 x^3$$

This is the integral of an odd function and so is clearly equal to zero.

4.2 Uncorrelated and Gaussian does not imply independent, unless jointly Gaussian

Let $$X \sim \mathcal{N}(0,1)$$ and $$Y=WX$$, where W takes values $$\pm 1$$ with equal probability. Clearly X and Y are not independent, as Y is a function of X.

(a) Show$$Y \sim \mathcal{N}(0,1)$$

This is kind of obvious from symmetry because $$\mathcal{N}(0,1)$$ is symmetric, i.e. $$\mathcal{N}(x|0,1) = \mathcal{N}(-x|0,1)$$. This means we can write:

$$P(Y=y) = P(W=1)P(X=y) + P(W=-1)P(X=-y) = P(X=y) = \mathcal{N}(0,1)$$

(b) Show covariance between X and Y is zero

We know that $$\mathbb{E}[X] = \mathbb{E}[Y] = 0$$, so we just need to evaluate $$\mathbb{E}[XY]$$:

$$\mathbb{E}[XY] = \int \int dx \ dy \ xy p(x,y)$$

But again $$p(x,y) = p(y|x)p(x)$$, and we can write $$p(y|x) = 0.5 \delta(y-x) + 0.5 \delta(y+x)$$. This means we are left with:

$$\mathbb{E}[XY] = \int_{-\infty}^{\infty} x \mathcal{N}(x|0,1)(0.5(x-x)) dx = 0$$

which proves the result.

4.3 Prove $$-1 \le \rho(X,Y) \le 1$$

$$\rho(X,Y) = \frac{\text{Cov}(X,Y)}{\sqrt{\text{Var}(X) \text{Var}(Y)}}$$

$$\text{Cov}(X,Y) = \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])]$$

$$\text{Var}(X) = \mathbb{E}[(X-\mathbb{E}[X])^2]$$

Let us write $$\mu_X = \mathbb{E}[X]$$ and $$\mu_Y = \mathbb{E}[Y]$$, for notational convenience. If now for any constants a and b we consider:

$$\mathbb{E}[(a(X-\mu_X) + b(Y-\mu_Y))^2]$$

which is clearly greater than or equal to zero. Multiplying out, this inequality gives:

$$a^2 \mathbb{E}[(X-\mu_X)^2] + b^2 \mathbb{E}[(Y-\mu_Y)^2] + 2ab \mathbb{E}[(X-\mu_X)(Y-\mu_Y)] \ge 0$$

Which we can re-write as:

$$2ab \text{Cov}(X,Y) \ge -a^2 \text{Var}(X) - b^2 \text{Var}(Y)$$

Now let us substitute in $$a^2 = \text{Var}(Y)$$ and $$b^2 = \text{Var}(X)$$:

$$2 \sqrt{\text{Var}(X) \text{Var}(Y)} \text{Cov}(X,Y) \ge -2 \text{Var}(X) \text{Var}(Y)$$

$$\implies \frac{\text{Cov}(X,Y)}{\sqrt{\text{Var}(X) \text{Var}(Y)}} = \rho(X,Y) \ge -1$$

If we do the same thing, but instead now consider $$\mathbb{E}[(a(X-\mu_X) - b(Y-\mu_Y))^2]$$, with the same definitions of a and b, it's easy to show that $$\rho(X,Y) \le 1$$ as well.

4.4 Correlation coefficient for linearly related variables

If $$Y=aX + b$$, then if $$a > 0$$ show that $$\rho(X,Y)=1$$, and if $$a < 0$$ that $$\rho(X,Y) = -1$$.

Let's say $$\mathbb{E}[X] = \mu_X$$ and $$\text{Var}(X) = \sigma_X^2$$. It follows that:

$$\mathbb{E}[Y] = a \mu_X + b$$ and $$\text{Var}(Y) = a^2 \sigma_X^2$$.

Now, to evaluate the correlation we need $$\mathbb{E}[XY] = \mathbb{E}[aX^2 + bX] = a \mathbb{E}[X^2] + b \mu_X$$

This means that the covariance is:

$$\text{Cov}(X,Y) = a \mathbb{E}[X^2] + b \mu_X - \mu_X(a \mu_X + b) = a \sigma_X^2$$

This allows us to get the correlation:

$$\rho(X,Y) = \frac{ \text{Cov}(X,Y)}{ \sqrt{\sigma_X^2 \sigma_Y^2}} = \frac{a \sigma_X^2}{\sqrt{a^2 \sigma_X^4}} = \frac{a \sigma_X^2}{|a| \sigma_X^2} = sgn(a)$$

Which is all we were asked to show!

4.5 Normalization constant for MV Gaussian

Prove that: $$(2 \pi)^{d/2} | \mathbf{\Sigma}|^{1/2} = \int \exp(-\frac{1}{2} (\mathbf{x-\mu}^T \mathbf{\Sigma}^{-1} (\mathbf{x-\mu})) d \mathbf{x}$$

We are told to diagonalize the covariance matrix, which can always be done since it is symmetric. That is, we can write:

$$D = P^{-1} \Sigma P$$

Where D is a diagonal matrix where the entries are the eigenvalues of $$\Sigma$$ and the columns of P are the eigenvectors. In fact, since $$\Sigma$$ is symmetric the eigenvectors can form an orthogonal basis, and it is possible to make P an orthogonal matrix, such that $$P^{-1} = P^T$$. This allows us to say:

$$D^{-1} = P^T \Sigma^{-1} P \implies \Sigma^{-1} = P D^{-1} P^T$$

As such, we can write the integral as:

$$\int \exp(-\frac{1}{2}(x-\mu)^T P D^{-1} P^T(x-\mu)) dx = \int \exp(-\frac{1}{2} (P(x-\mu))^T \begin{bmatrix} \frac{1}{\lambda_1} & & \\ & \ddots & \\ & & \frac{1}{\lambda_d} \end{bmatrix} (P(x-\mu))) dx$$

Now let us define $$y = P(x-\mu)$$. Because P is an orthogonal matrix (which has determinant 1), the Jacobian is 1 and we can replace $$dx$$ with $$dy$$. The term inside the exponential is then:

$$\sum_{ij} y_i \delta_{ij} \frac{1}{\lambda_i} y_j = \sum_i \frac{y_i^2}{\lambda_i}$$. Effectively by transforming to the eigenbasis we have decoupled the components of y, so we can write:

$$= \int_{-\infty}^{\infty} dy_1 e^{-\frac{y_1^2}{2 \lambda_1}} \dots \int_{-\infty}^{\infty} dy_d e^{-\frac{y_d^2}{2 \lambda_d}}$$

i.e. just the product of many one-dimensional Gaussians. This is equal to:

$$\sqrt{2 \pi \lambda_1} \sqrt{2 \pi \lambda_2} \dots \sqrt{2 \pi \lambda_d} = (2 \pi)^{d/2} \sqrt{\lambda_1 \dots \lambda_d}$$

We then use that $$det(\Sigma) = \prod_{i=1}^d \lambda_i$$, which gives us the final answer we want!

# Continuous Blackjack

Recently I had an interview where as an extra "bonus" question at the end I was asked an interesting maths problem. With a couple of hints from the interviewer, I was able to sketch out a rough solution, however afterwards I wanted to look up a proper solution just to verify it. Interestingly, I wasn't able to find one (I'm sure it's out there, I just need to look harder). Anyway, I thought it was a nice little problem and so I thought it was worth posting what I believe to be the correct solution.

The problem is this - consider that you are playing a game with one other person, and you are going to be going first. The rules are you pick a number randomly from the Uniform(0,1) distribution (i.e. a random number between 0 and 1). You then decide either to stick with this total, or you play on and choose another such random number. You can do this as many times as you like, however if the sum total of the numbers you pick goes over 1, you go bust and automatically lose. If you decide to stick with a number less than 1, the other player has a go and plays by the same rules. The person who sticks at the higher number (or doesn't go bust when their opponent does) is the winner. Clearly it is an advantage to go second, and the optimal strategy for player 2 is extremely simple - keep playing until you get a higher total than your opponent or until you go bust. The question is, given that you are player 1 what is the best strategy you can adopt?

The first thing to realise is that our optimal strategy will be divided at some number t, which I shall call the "decision boundary", and where if we have a sum less than t we will draw a new number, and if we have a sum greater than t we will stick. We can then think about what the probability of winning is, given that we stick at a particular value t. This is equal to 1 minus the probability that we lose - and the probability that we lose is the probability that the second player is able to land their sum within the interval $$[t,1]$$, given that they play on until they either reach this interval or go bust. To go about calculating this, let us define $$P_t[x]$$ to be the probability that we are able to land in the interval $$[t,1]$$, given that we are currently at x and definitely going to play on if we have not yet reached t. We can write down an equation for this as follows:

$$P_t[x] = (1-t) + \int_x^t P_t[y]dy$$

where the first term is the probability that we reach the interval in the next turn, and the second term is the integral of the probability of reaching y < t (=1 as we are drawing uniform(0,1) random variables) multiplied by the probability that we reach the interval $$[t,1]$$ starting from y. We can convert this from an integral equation into an ODE:

$$\frac{dP_t[x]}{dx} = -P_t[x] \ \ \ \implies P_t[x] = A e^{-x}$$

We can obtain the constant A by noting that $$P_t[t] = 1-t$$, and hence that $$A = (1-t)e^t$$. This means that:

$$P_t[x] = (1-t)e^{t-x}$$. Now, player 2 starts from $$x=0$$, and so the probability of losing given that we stuck at a value t is simply $$P_t[0] = (1-t)e^t$$. The probability that we win is $$1-P_t[0] = 1-(1-t)e^t$$.

The final step we need to solve this is to say that the threshold at which our strategy changes should be the following point: where the probability of winning given that we stick at t is exactly the same as the probability that we win given that we choose one more number. We can write this condition as:

$$1-(1-t)e^t = \int_t^1 \left[ 1-(1-t')e^{t'}\right]dt' = (1-t) - e^t(t-2) - e$$

This gives a non-linear equation for the optimal decision boundary t, which cannot be re-arranged nicely but numerically we can solve to find that $$t \approx 0.57$$. That is, if our sum is less than approximately 0.57 we should pick another number, and if it's more we should stick!

# Machine Learning - A Probabilistic Perspective Exercises - Chapter 3

This is a continuation of the exercises in "Machine learning - a probabilistic perspective" by Kevin Murphy. Chapter 3 is on "Generative Models for Discrete Data".

3.1 MLE for the Bernoulli/ binomial model

We start off with a nice simple one. If we have data D consisting of N1 heads out of a total number N trials, then the likelihood is $$P(D|\theta) = \theta^{N_1} (1-\theta)^{N-N_1}$$. It's a lot easier to work with the log likelihood here:

$$\log(P(D|\theta)) = N_1 \log(\theta) + (N-N_1) \log(1-\theta)$$

Taking the derivative:

$$\frac{d}{d \theta} = \frac{N_1}{\theta} - \frac{N-N_1}{1-\theta} = 0 \ \ \implies \frac{N_1}{\theta} = \frac{N-N_1}{1-\theta}$$

Rearrange this and we find $$\theta = \frac{N_1}{N}$$

3.2 Marginal likelihood for the Beta-Bernoulli model.

This question is looking at deriving the marginal likelihood, $$P(D) = \int P(D|\theta) P(\theta) d\theta$$. We are told to use the chain rule of probability: $$P(x_{1:N}) = p(x_1) p(x_2 | x_1) p(x_3: x_{1:2})\dots$$

and reminded that in the chapter we derived the posterior predictive distribution:

$$P(X=k | D_{1:N}) = \frac{N_k + \alpha_k}{\sum_i N_i + \alpha_i}$$

We are the given an example - suppose D = H,T,T,H,H (or D=1,0,0,1,1). It follows that:

$$P(D) = \frac{\alpha_1}{\alpha} \frac{\alpha_0}{\alpha + 1} \frac{\alpha_0+1}{\alpha+2} \frac{\alpha_1 + 1}{\alpha + 3} \frac{\alpha_1+2}{\alpha+4}$$

where we have just applied the chain rule, using the posterior predictive distribution after each data point has been collected. It's clear that if we do this more generally (for any collection of data), that we will be left with:

$$P(D) = \frac{\left[ \alpha_0 \dots (\alpha_0 + N_0 - 1) \right] \left[ \alpha_1 \dots (\alpha_1 + N_1 - 1) \right]}{\alpha \dots (\alpha + N-1)}$$

We then note that this can be re-written using factorials as follows:

$$P(D) = \frac{(\alpha_0+N_0-1)! (\alpha_1 + N_1 -1)! (\alpha-1)!}{(\alpha_0-1)! (\alpha_1-1)! (\alpha + N -1)!}$$

Remembering that $$\Gamma(N) = (N-1)!$$, and that $$\alpha = \alpha_0 + \alpha_1$$, we get the result which is given in the question:

$$P(D) = \frac{ \Gamma(\alpha_0 + N_0) \Gamma(\alpha_1 + N_1) \Gamma(\alpha_1 + \alpha_0) }{\Gamma(\alpha_1 + \alpha_0 + N) \Gamma(\alpha_1) \Gamma(\alpha_0) }$$

3.3 Posterior predictive for Beta-Binomial model

In the text the posterior predictive distribution for the Beta-Binomial model was derived for the case of predicting the outcome of multiple future trials given the data:

$$P(x|n,D) = \frac{B(x+\alpha_1', n-x+\alpha_0')}{B(\alpha_1', \alpha_0')} \binom{n}{x}$$

where $$\alpha_1'$$ and $$\alpha_0'$$ involve the prior parameters and the data. The question simply asks to show that when $$n=1$$ that we have: $$P(x=1|D) = \frac{\alpha_1'}{\alpha_0' + \alpha_1'}$$.

We need to remember that by definition, $$B(a,b) = \frac{\Gamma(a) \Gamma(b)}{\Gamma(a+b)}$$, hence:

$$\frac{B(1+\alpha_1', \alpha_0')}{B(\alpha_1', \alpha_0')} = \frac{\Gamma(1+\alpha_1') \Gamma(\alpha_0')}{\Gamma(1 + \alpha_0' + \alpha_1')} \frac{\Gamma(\alpha_0' + \alpha_1')}{\Gamma(\alpha_0') \Gamma(\alpha_1')}$$

But then we simply note the following: $$\Gamma(1+\alpha_1') = \alpha_1'! = \alpha_1' (\alpha_1-1)! = \alpha_1' \Gamma(\alpha_1')$$. Using this and simplifying clearly leaves us with the desired result.

3.4 Beta updating from censored likelihood

Suppose we toss a coin $$n=5$$ times. Let X be the number of heads. We observe there are fewer than 3 heads, but we don't know how many precisely. Prior we use is $$P(\theta) = \text{Beta}(\theta|1,1)$$. Compute posterior, $$P(\theta | X < 3)$$.

Now $$\text{Beta}(\theta|1,1)$$ is an uninformative prior, so this is just a constant. So the posterior, $$P(\theta | X<3) \propto P(X < 3|\theta) P(\theta) \propto P(X<3 | \theta)$$. So we need to consider the likelihood, $$P(X<3|\theta)$$. This is straightforward to calculate as it is the sum of the probability of no heads, one head and two heads, i.e.:

$$P(X<3 | \theta) = (1-\theta)^5 + \binom{5}{4}(1-\theta)^4 \theta + \binom{5}{3} (1-\theta)^3 \theta^2 = Bin(0|\theta, 5) + Bin(1|\theta,5) + Bin(2|\theta,5)$$

It follows that:

$$P(X<3 | \theta) \propto \text{Beta}(1,6) + \text{Beta}(2,5) + \text{Beta}(3,4)$$

which is a mixture distribution.

3.5 Uninformative prior for log-odds ratio

Let $$\phi = \text{logit}(\theta) = \log(\frac{\theta}{1-\theta})$$. If $$p(\phi) \propto 1$$, show $$p(\theta) \propto \text{Beta}(\theta| 0,0)$$.

If we just apply the change of variables formula from chapter 2:

$$p(\theta) = \bigg| \frac{d\phi}{d\theta} \bigg| p(\phi)$$

but $$p(\phi)$$ is a constant, and $$\phi = \log(\theta) - \log(1-\theta)$$, so:

$$\frac{d\phi}{d\theta} = \frac{1}{\theta} + \frac{1}{1-\theta} = \frac{1}{\theta(1-\theta)}$$

Remembering the definition for the Beta distribution: $$\text{Beta}(x|a,b) = \frac{1}{B(a,b)} x^{a-1}(1-x)^{b-1}$$, and so clearly $$p(\theta) \propto \text{Beta}(\theta|0,0)$$.

3.6 MLE for Poisson distribution

Definition: $$\text{Poi}(x|\lambda) = \frac{\lambda^x}{x!} e^{-\lambda}$$

So the likelihood of a set of data $$\{x_i\}$$ is:

$$L(\lambda ; \{x_i\}) =\prod_i \frac{\lambda^{x_i}}{x_i!} e^{-\lambda}$$

Unsurprisingly, it's easier to work with the log-likelihood:

$$l(\lambda; \{x_i\}) = \sum_i \left[ -\lambda + x_i \log(\lambda) - \log(x_i !) \right]$$

If we ignore the term that doesn't depend on $$\lambda$$ then we are left with $$-\lambda N + \log(\lambda \left( \sum_i x_i \right)$$. Differentiating w.r.t $$\lambda$$ and setting equal to zero:

$$-N + \frac{1}{\lambda} \sum_i x_i = 0 \ \ \ \implies \hat{\lambda} = \frac{1}{N} \sum_i x_i$$

3.7 Bayesian analysis of the Poisson distribution

(a) Derive the posterior assuming a Gamma prior

The prior is: $$P(\lambda) = Ga(\lambda |a,b) \propto \lambda^{a-1} e^{-\lambda b}$$.

and the posterior is proportional to the likelihood times the prior, i.e. $$P(\lambda | D) \propto P(D | \lambda) P(\lambda)$$.

We already looked at the likelihood for the Poisson distribution in the previous section, so:

$$P(\lambda | D) \propto \prod_{i=1}^N \frac{e^{-\lambda} \lambda^{x_i}}{x_1!} \lambda^{a-1} e^{-\lambda b} \propto e^{-\lambda(N+b)}\lambda^{a-1+\sum_i x_i} = Ga(\lambda | a + \sum_i x_i, N+b)$$

So we see that the posterior is also a Gamma distribution, making the Gamma distribution a conjugate prior for the Poisson distribution.

(b) Posterior mean as a->0, b->0

We use that the fact that we derived the mean of a Gamma distribution in the text, finding that it is equal to a/b. So clearly it's just:

$$\frac{1}{N} \sum_{i=1}^N x_i$$, which is the MLE we found in the previous section.

3.8 MLE for the uniform distribution

Consider a uniform distribution centered on 0 with width 2a. $$p(x) = \frac{1}{2a} I(x \in [-a,a])$$ is the density function.

(a) Given a data set x1, ..., xn, what is the MLE estimate of a?

The key point here is that $$P(D|a) = 0$$ for any a which is less than the data point with the largest magnitude, and equal to $$\frac{1}{(2a)^n}$$ for any a larger than this. This is clearly minimised when a is made as small as possible, i.e. $$\hat{a} = max|x_i|$$.

(b)What probability would the model assign to a new data point x_n+1 using the MLE estimate for a?

Clearly $$\frac{1}{2\hat{a}}$$ if $$|x_{n+1}| \le \hat{a}$$ and 0 otherwise.

(c) Do you see a problem with this approach?

Clearly there is an issue here as any value with an absolute value larger than $$max |x_i|$$ is assigned zero probability. For relatively small data sets this will be a big issue, but even for larger data sets it seems far from ideal.

3.9 Bayesian analysis of the uniform distribution

This is very much a continuation of the previous question, although here we are told to consider a $$Unif(0,\theta)$$ distribution. The MLE is now max(D). Overall I'm fairly sure the question is either extremely poorly worded or has some mistakes, so I'm just going to go through it in the way that makes sense to me.

We are told to use a Pareto prior - the Pareto distribution is defined as:

$$p(x | k,m) = k m^k x^{-(k+1)} I(x \ge m)$$

where I is an indicator function. So a $$\text{Pareto}(\theta | b, K)$$ prior is:

$$P(\theta) = K b^K \theta^{-(K+1)} I(\theta \ge b)$$

We more or less established the likelihood in the previous section is given by:

$$P(D | \theta) = \frac{1}{\theta^N} I(\theta \ge max(D))$$

This means that the joint distribution, $$P(D,\theta)$$ is given by $$P(D,\theta) = P(D | \theta) P(\theta) = \frac{K b^K}{\theta^{N+K+1}} I(\theta \ge m)$$ where $$m = \text{max}(b, D)$$.

We can use this to write the marginal likelihood:

$$P(D) = \int P(D,\theta) d\theta = \int_m^{\infty} \frac{K b^K}{\theta^{N+K+1}} d\theta = \frac{K b^K}{(N+K) m^{N+K}}$$

Now the posterior is given by:

$$P(\theta | D) = \frac{P(\theta, D)}{P(D)} = \frac{(N+K) m^{N+K}}{\theta^{N+K+1}} I(\theta \ge m) = \text{Pareto}(\theta | N+K, m=\text{max}(D,b))$$

3.11 Bayesian analysis of the exponential distribution

$$p(x|\theta) = \theta e^{-\theta x}$$ for $$x \ge 0, \theta > 0$$ defines the exponential distribution.

(a) Derive the MLE

We can write the likelihood function as:

$$L(\mathbf{x};\theta) = \theta^N e^{-\theta \sum_{i=1}^N x_i}$$

Clearly working with the log-likelihood will be better here:

$$l(\mathbf{x}; \theta) = N \log(\theta) - \theta \sum_i x_i$$

Taking the derivative wrt theta and setting it equal to zero:

$$0 = \frac{N}{\theta} - \sum_i x_i$$

and so clearly the MLE is: $$\hat{\theta} = \frac{1}{\frac{1}{N} \sum_i x_i} = \frac{1}{\bar{x}}$$

(b) Suppose we observe X1=5, X2=6, X3=4. What is the MLE?

The mean is 5, so $$\hat{\theta} = 1/5$$.

(c) Assume a prior $$p(\theta) = Expon(\theta | \lambda)$$. What value should lambda take to give $$\mathbb{E} [\theta] = 1/3$$?

The exponential distribution is a special case of the Gamma distribution where $$a=1$$ and $$b=\lambda$$. We derived that the mean of a Gamma distribution is just given by a/b, and so we want:

$$1/3 = \frac{1}{\hat{\lambda}}$$ and hence $$\hat{\lambda}=3$$.

(d) What is the posterior?

$$P(\theta | D) \propto \theta^N e^{-N \theta \bar{x}} \lambda e^{-\lambda \theta} \propto \theta^N e^{-\theta(N \bar{x} + \lambda)} = Ga(\theta | N+1, N\bar{x}+\lambda)$$

(e) Is exponential prior conjugate to exponential likelihood?

Kind of, in the sense that both the prior and the posterior are Gamma distributions. But the posterior is not also an Exponential distribution.

(f) What is the posterior mean?

Again, mean of Gamma is a/b so we have $$\frac{N+1}{N \bar{x} + \lambda}$$. If $$\lambda = 0$$ and as $$N \to \infty$$ we recover the MLE.

(g) Why do they differ?

Bit of a stupid question - because we calculated them in a different manner. We should expect the posterior mean to be less prone to overfitting though, and generally be a bit better.

3.12 MAP estimation for Bernoulli with non-conjugate priors

In the book we discussed Bayesian inference of a Bernoulli rate parameter when we used a $$\text{Beta}(\theta|\alpha, \beta)$$ prior. In this case, the MAP estimate was given by:

$$\theta_{MAP} = \frac{N_1 + \alpha - 1}{N + \alpha + \beta - 2}$$

(a) Now consider the following prior:

$$p(\theta) = 0.5$$ if $$\theta = 0.5$$, $$p(\theta) = 0.5$$ if $$\theta = 0.4$$ and 0 otherwise. Clearly this is a weird prior to use, but I guess it's just for an exercise to make a point so let's go with it. The question is to derive the MAP estimate as a function of N1 and N.

We can write the posterior as:

$$P(\theta | D) \propto \theta^{N_1} (1-\theta)^{N-N_1} I( \theta \in \{0.4, 0.5 \})$$

So the MAP is simply: $$\text{max}(0.5^{N_1} 0.5^{N-N_1}, 0.4^{N_1} 0.6^{N-N_1}) = \text{max}(0.5^{N}, 0.4^{N_1} 0.6^{N-N_1})$$.

With some algebraic manipulations we can show that the MAP is 0.5 if: $$N_1 > \frac{\log(6/5)}{\log(6/4)} N \approx 0.45 N$$. I was kind of surprised that it wasn't exactly 0.45N, but I guess it has something with it being constrained to be between 0 and 1. I'm actually not sure though.

(b) Suppose the true theta is 0.41. Which prior leads to the better estimate?

When N is small, we would expect this second approach to work better (as it is choosing only between 0.4 and 0.5), however as N becomes larger eventually the other prior, where $$\theta$$ can take any value, will give a better estimate.

3.13 Posterior predictive for a batch of data using the dirichlet-multinomial model

Derive an expression for $$P(\tilde{D} | D, \alpha)$$, i.e. use the posterior to predict the results for a whole batch of data. Now, the definition of the Dirichlet distribution is:

$$\text{Dir}(\mathbf{\theta} | \mathbf{\alpha}) = \frac{1}{B(\mathbf{\alpha})} \prod_{k=1}^K \theta_k^{\alpha_k-1} I(\mathbf{\theta} \in S_k)$$ (the identity function is ensuring the components of $$\theta$$ sum to one.)

where $$B(\mathbf{\alpha}) = \frac{\prod_k \Gamma(\alpha_k)}{\Gamma[\sum_k \alpha_k]}$$.

Using a $$\text{Dir}(\mathbf{\theta} | \mathbf{\alpha})$$ prior for $$\theta$$ we showed in the book that the posterior was given by:

$$P(\mathbf{\theta} | D) = \text{Dir}(\mathbf{\theta} | \alpha_1 + N_1, \dots, \alpha_K + N_K)$$

This means that we can write:

$$P(\tilde{D} | D, \alpha) = \int P(\tilde{D} | \theta) P(\theta | D) d\theta = \int \prod_{k=1}^K \theta_k^{x_k} \text{Dir}(\theta | \alpha_1 + N_1, \dots, \alpha_K + N_K) d\theta$$

where $$\mathbf{x} = \{x_k\}$$ is the numbers of each class in the data we are predicting. This is equal to:

$$\int \prod_k \theta_k^{x_k} \frac{1}{B(\mathbf{\alpha + N})} \prod_k \theta_k^{\alpha_k + N_k} I(\theta \in S_k) d\theta$$

$$= \int \frac{B(\mathbf{\alpha + N + X})}{B(\mathbf{\alpha+N})} \text{Dir}(\theta | \alpha_1 + N_1 + x_1, \dots, \alpha_K + N_K + x_K) d\theta = \frac{B(\mathbf{\alpha + N + X})}{B(\mathbf{\alpha+N})}$$

where we just converted the parameters in the Dirichlet distribution and introduced the correct normalisation parameter. Since it is a probability distribution, it then of course integrates to 1 (the final step).

3.14 Posterior predictive for the Dirichlet-Multinomial

(a) Suppose we compute the empirical distribution over letters of the Roman alphabet plus the space character (27 values), from 2000 samples. Suppose we see the letter "e" 260 times. What is $$P(x_{2001} = e | D)$$, assuming a Dirichlet prior with alpha_k = 10 for all k?

We showed in the text that the posterior predictive is given by:

$$P(X=j | D) = \frac{\alpha_j + N_j}{\sum_k (\alpha_k + N_k)} = \frac{10 + 260}{270 + 2000} \simeq 0.119$$

(b) Suppose actually we saw "e" 260 times, "a" 100 times and "p" 87 times. What is $$P(x_{2001}=p, x_{2002} = a | D)$$ under the same assumptions?

We basically just derived what we need for this in the previous question. We are looking for the probability of the data vector $$\mathbf{X} = (1,0,\dots, 1, 0, \dots, 0)$$, where the non-zero components are at indices 1 ("a") and 16 ("p"). We showed:

$$P(X | D) = \frac{B(\mathbf{\alpha + N + X})}{B(\mathbf{\alpha+N})}$$

This is equal to:

$$\frac{\prod_k \Gamma(\alpha_k + N_k + x_k) \Gamma(\sum_k \alpha_k + N_k)}{\Gamma(\sum_k \alpha_k + N_k + x_k) \prod_k \Gamma(\alpha_k + N_k)}$$

Now in the product terms, everything cancels except the components corresponding to p and a, where we pick up factors of $$\frac{\Gamma(98)}{\Gamma(97)}$$ and $$\frac{\Gamma(111)}{\Gamma(110)}$$ respectively. Overall, we are left with:

$$P(X|D) = \frac{ \Gamma(111) \Gamma(98) \Gamma(2270)}{\Gamma(110) \Gamma(97) \Gamma(2272)} = \frac{(110!)(97!)(2269!)}{(109!)(96!)(2271!)} = \frac{110*97}{2270*2271} \simeq 0.002$$

3.17 Marginal likelihood for beta-binomial under uniform prior

Suppose we toss a coin N times and observe N1 heads. Let $$N_1 \sim Bin(N,\theta)$$ and $$\theta \sim Beta(1,1)$$. Show that the marginal likelihood is given by: $$P(N_1 | N) = \frac{1}{N+1}$$.

We can write:

$$P(N_1 | N) = \int P(N_1 | N, \theta) P(\theta) d\theta$$

But a $$Beta(1,1)$$ distribution is just uniform, so we don't need to take this into account. So we can say:

$$P(N_1 | N) = \int_0^1 \text{Bin}(N_1 | N, \theta) d\theta = \int_0^1 \frac{N!}{N_1! (N-N_1)!} \theta^{N_1} (1-\theta)^{N-N_1} d\theta$$

It helps to re-write the factorials in terms of Gamma functions ($$\Gamma(n+1) = n!$$):

$$P(N_1 | N) = \int_0^1 \frac{\Gamma(N+1)}{\Gamma(N_1+1) \Gamma(N-N_1 + 1)} \theta^{N_1} (1-\theta)^{N-N_1} d\theta$$

Now, by definition $$\text{Beta}(\theta | N_1 + 1, N-N_1+1) = \frac{\Gamma(N+2)}{\Gamma(N_1+1) \Gamma(N-N_1+1)} \theta^{N_1} (1-\theta)^{N-N_1}$$. This means that we can say:

$$P(N_1 | N) = \int_0^1 \frac{\Gamma(N+1)}{\Gamma(N+2)} \text{Beta}(\theta | N_1+1, N-N_1+1) d\theta = \frac{N!}{(N+1)!} = \frac{1}{N+1}$$

since a probability distribution must integrate to 1. This result kind of surprised me to be honest - I guess it kind of makes sense although intuitively I would have expected a uniform prior over $$\theta$$ to lead to it being most likely to have $$N_1 = N/2$$, rather than being completely uniform!

3.18 Bayes factor for coin tossing

Suppose we toss a coin $$N=10$$ times and observe $$N_1 = 9$$ heads. Let the null hypothesis be that the coin is fair, and the alternative be that the coin can have any bias - $$p(\theta) = Unif(0,1)$$. Derive the Baye's factor in favour of the biased coin hypothesis. What if $$N_1=90$$ and $$N=100$$.

I think this just means we need to look at the ratios of the likelihoods under each assumption. Under the fair assumption:

$$P(N_1 | \theta = 1/2) = \frac{N!}{N_1!(N-N_1)!} \frac{1}{2}^N$$

Under the biased assumption, we just calculated this in the previous exercise:

$$P(N_1 | \theta \sim Unif(0,1)) = \frac{1}{N+1}$$

So $$BF = \frac{N_1! (N-N_1)! 2^N}{(N+1)!}$$ which for $$N_1 = 9$$ and $$N=10$$ I find $$BF \simeq 9.31$$. For $$N_1 = 90$$, $$N=100$$ I find: $$BF \simeq 7.25 \times 10^{14}$$ - clearly the coin is amazingly unlikely to be fair in the second case.

3.19 Irrelevant features with Naive Bayes

Let $$x_{iw}=1$$ if word w occurs in document i, and be 0 otherwise. Let $$\theta_{cw}$$ be the estimated probability that word w occurs in documents of class c. The log-likelihood that document x belongs to class c is:

$$\log(p(\mathbf{x_i}|c,\theta)) = \log \prod_w \theta_{cw}^{x_{iw}}(1-\theta_{cw})^{1-x_{iw}}$$

$$= \sum_w x_{iw} \log(\frac{\theta_{cw}}{1-\theta_{cw}}) + \sum_w \log(1-\theta_{cw})$$

This can be written more succinctly as $$\log(p(\mathbf{x_i}) = \mathbf{\phi(x_i)}^T \mathbf{\beta_c}$$, where $$\mathbf{\phi(x_i)} = (\mathbf{x_i},1)$$ and:

$$\mathbf{\beta_c} = (\log \frac{\theta_{c,1}}{1-\theta_{c,1}}, \dots, \log \frac{\theta_{c,W}}{1-\theta_{c,W}} , \sum_w \log(1-\theta_{cw}))^T$$

i.e. a linear classifier, as the class-conditional density is a linear function of the params $$\mathbf{\beta_c}$$.

(a) Assuming P(C=1) = P(C=2) = 0.5, write an expression for the log posterior odds ratio in terms of the features and the parameters.

We just use Bayes's theorem: $$P(C=1 | \mathbf{x_i}) = P(\mathbf{x_i} | C=1) P(C) / P(\mathbf{x_i})$$, and likewise for C=2. However, as $$P(C=1) = P(C=2)$$ we get a cancellation such that:

$$\log \frac{P(C=1 | \mathbf{x_i})}{P(C=2 | \mathbf{x_i})} = \mathbf{\phi(x_i)}^T(\mathbf{\beta_1 - \beta_2})$$

(b) Consider a particular word w. State the conditions on $$\theta_{1,w}$$ and $$\theta_{2,w}$$ under which the presence or absence of the word will have no effect on the class posterior.

For this, we want to poster odds ratio to be 1, and hence for the logarithm of this to be zero. This means that $$\beta_{1,w} = \beta_{2,w}$$.

(c) The posterior mean estimate of theta, using a Beta(1,1) prior, is given by:

$$\hat{\theta_{cw}} = \frac{1+ \sum_{i \in c} x_{iw}}{2 + n_c}$$

where the sum is over the nc documents in class c. Consider a word w, and suppose it occurs in every document, regardless of class. Let there be n1 documents of class 1, and n2 of class 2, with n1 not equal to n2. Will this word be ignored by our classifier?

Clearly not, as we are told that $$\hat{\theta_{1,w}} = \frac{1+n_1}{2+n_c}$$ and $$\hat{\theta_{2,w}} = \frac{1+n_2}{2+n_c}$$, which are not equal as $$n_1 \neq n_2$$, and hence the necessary condition derived in (b) does not hold.

(d) What other ways can you think of to encourage "irrelevant" words to be ignored?

I guess things like preprocessing and feature selection would be good for this.

3.21 Mutual information for Naive Baye's with binary features

The result was stated in the chapter, here we are asked to derive it. We are looking for the mutual information between feature j and the class label Y, i.e. $$I(X_j, Y)$$. By definition, this is equal to:

$$I(X_j;Y) = \sum_{x_j \in \{0,1\}} \sum_{y \in C} P(x_j, y) \log \left( \frac{P(x_j, y)}{p(x_j) p(y)} \right)$$

To get the joint values, we can say $$P(x_j = 1, y=c) = P(x_j=1 | y=c) P(y=c) = \theta_{jc} \pi_c$$

and then $$P(x_j=0, y=c) = P(x_j=0 | y=c) P(y=c) = (1-\theta_{jc}) \pi_c$$.

By definition, $$P(y=c) = \pi_c$$, and then we can say:

$$P(x_j=1) = \sum_{c'} P(x_j=1, y=c') = \sum_{c'} \theta_{jc'} \pi_{c'} \equiv \theta_j$$

$$P(x_j=0) = \sum_{c'} P(x_j=0, y=c') = \sum_{c'} (1-\theta_{jc'}) \pi_{c'} = 1-\theta_j$$

Putting this together we get the desired result:

$$I(X_j, Y) = \sum_c \left[ \theta_{jc} \pi_c \log \left( \frac{\theta_{jc}}{\theta_j} \right) + (1-\theta_{jc})\pi_c \log \left( \frac{1-\theta_{jc}}{1-\theta_j} \right) \right]$$

# Machine Learning - A Probabilistic Perspective Exercises - Chapter 2

Recently I've been becoming more and more interested in machine learning, and so far my attention has been primarily focused on reinforcement learning. I had a lot of fun working on the Big 2 AI, but I feel like I really need to invest more time to studying the fundamentals of machine learning. I've got myself a copy of "Machine Learning - A Probabilistic Perspective", which seems like a great text book, and so I'm going to work my way through it. I've decided to make a decent attempt at doing as many of the exercises as possible, and I feel like actually writing up an explanation for them is quite useful for me in making sure I actually understand what's going on. Potentially it might also be useful for other people too, so I thought I would post my answers as I go! I may skip some exercises if I think they're boring (or too "wordy"), or of course if I'm unable to do them (although I will probably mention them in this case)!

2.1 My neighbor has two children. Assuming that the gender of a child is like a coin flip, it is most likely, a priori, that my neighbour has one boy and one girl, with probability 1/2. The other possibilities—two boys or two girls—have probabilities 1/4 and 1/4.
(a) Suppose I ask him whether he has any boys, and he says yes. What is the probability that one child is a girl?

This is a standard probability puzzle. The key here is in the wording of the question we ask him - whether he has any boys? A priori, there are four possible options for child 1 and child 2: BB (boy boy), BG, GB and GG, each with equal probability. If he says yes to our question, this is compatible with three of the initial four options: BB, BG and GB. That is, we can only exclude the possibility GG. In two out of these three remaining possibilities he has one girl, and so the probability is 2/3.

(b) Suppose instead that I happen to see one of his children run by, and it is a boy. What is the probability that the other child is a girl

In this case the question is more specific with its wording - we have seen one specific child, but this tells us nothing about the other child, and so we get the more intuitive answer of 1/2.

2.3 Show Var[X+Y] = Var[X] + Var[Y] + 2 Cov[X,Y] for any two random variables X and Y

This one is relatively straightforward and just involves starting from a basic relationship for the variance:

$$Var[X+Y] = \mathbb{E}[(X+Y)^2] - \mathbb{E}[X+Y]^2 = \mathbb{E}[(X+Y)^2] - (\mathbb{E}[X] + \mathbb{E}[Y])^2$$

(using the linearity of expectation). Expanding this out:

$$\mathbb{E}[X]^2 + \mathbb{E}[Y]^2 + 2 \mathbb{E}[XY] - \mathbb{E}[X]^2 - 2\mathbb{E}[X] \mathbb{E}[Y] - \mathbb{E}[Y]^2$$

which clearly gives the required result (as $$Cov[X,Y] = \mathbb{E}[XY]-\mathbb{E}[X]\mathbb{E}[Y]$$ ).

2.4 After your yearly checkup, the doctor has bad news and good news. The bad news is that you tested positive for a serious disease, and that the test is 99% accurate (i.e., the probability of testing positive given that you have the disease is 0.99, as is the probability of testing negative given that you don’t have the disease). The good news is that this is a rare disease, striking only one in 10,000 people. What are the chances that you actually have the disease?

This is another famous example of the use of Baye's theorem. If we define two events as follows:

A - you have the disease

B - the test returns a positive result

If you go and take the test, and you get a positive result saying that you have the disease, then what you want to calculate is $$p(A|B)$$. That is, the probability that you have the disease given that you tested positive. By Baye's theorem, this is equivalent to:

$$p(A|B) = \frac{p(B|A) p(A)}{p(B)}$$

Now, since the test is 99% accurate, $$P(B|A) = 0.99$$, i.e. if you have the disease you will test positive 99% of the time. But p(A) = 1/10000, i.e. the probability of an "average" person in the population having the disease (with no other information about them). Finally, we can calculate p(B) as follows:

$$P(B) = 0.99 \frac{1}{10000} + 0.01 \frac{9999}{10000}$$

where the first term is the probability of being positive if you have the disease*prob of actually having the disease and the second term is prob of testing positive if you don't have the disease (false positive) * probability of not having the disease.

Putting the numbers in you find that $$P(B|A) \approx 0.98 \%$$, which makes it extremely unlikely that you have the disease despite testing positive! This is one example of why it would be a good idea if doctors learned some stats!

2.5 Monty Hall problem - On a game show, a contestant is told the rules as follows: There are three doors, labelled 1, 2, 3. A single prize has been hidden behind one of them. You get to select one door. Initially your chosen door will not be opened. Instead, the gameshow host will open one of the other two doors, and he will do so in such a way as not to reveal the prize. For example, if you first choose door 1, he will then open one of doors 2 and 3, and it is guaranteed that he will choose which one to open so that the prize will not be revealed. At this point, you will be given a fresh choice of door: you can either stick with your first choice, or you can switch to the other closed door. All the doors will then be opened and you will receive whatever is behind your final choice of door. Imagine that the contestant chooses door 1 first; then the gameshow host opens door 3, revealing nothing behind the door, as promised. Should the contestant (a) stick with door 1, or (b) switch to door 2, or (c) does it make no diﬀerence? You may assume that initially, the prize is equally likely to be behind any of the 3 doors.

This is another famous problem which has quite an interesting history, apparently leading to many genuine mathematicians complaining about the correct solution which was given in a newspaper column (although this was a little while ago, I still find this extremely surprising!) It also comes up online every now and then with people arguing/trying to explain why you should switch, so I'll add my best attempt at explaining it here I guess!

For me, the crucial detail that helps in understanding it is that the host has information about where the prize is, which means that he is not choosing a door at random to open, since he will never open the door with the prize behind. If you take the starting point as the prize having equal probability of being behind each door, then you only need to consider two possibilities - 1) the initial door you chose does not have the prize behind it (which happens 2/3 of the time), and 2) the initial door does have the prize behind it. In case 1), you have selected one of the doors without the prize behind it - which means the host has no choice but to open the only other door that doesn't have the prize behind it. This means in this case, if you switch, you win the prize - and remember this happens 2/3 of the time. Obviously if you did initially pick the winning door and you switch then you lose, but this only happens 1/3 of the time. So it's better to switch!

2.6 Conditional Independence

(a) Let $$H \in \{1,\dots,K\}$$ be a discrete RV, and let e1 and e2 be the observed values of two other RVs E1 and E2. Suppose we wish to calculate the vector:

$$\vec{P}(H|e_1, e_2) = (P(H=1|e_1, e_2), \dots, P(H=K|e_1, e_2))$$

Which of the following are sufficient for the calculation?

(i) P(e1,e2), P(H), P(e1|H), P(e2|H)

(ii) P(e1,e2), P(H), P(e1,e2|H)

(iii) P(e1|H), P(e2|H), P(H)

We can use Baye's theorem to write:

$$P(H| E_1, E_2) = \frac{P(E_1, E_2 | H) P(H)}{ P(E_1, E_2)}$$

so clearly (ii) is sufficient. The others are not in general sufficient.

(b) Now suppose we assume $$E_1 \bot E_2 | H$$. Now which are sufficient?

Well, clearly (ii) still is. But conditional independence of E1 and E2 given H means that we can write:

$$P(E_1, E_2 | H) = P(E_1 | H) P(E_2 | H)$$

which means that (i) is also sufficient now.

However, we can also write:

$$P(e_1, e_2) = \sum_h p(e_1, e_2, h) = \sum_h p(e_1, e_2 | h) p(h)$$

which means that (iii) is also sufficient too!

2.7 Show that pairwise independence does not imply mutual independence

The best way to do this is to show an example where we have pairwise independence, but not mutual independence. Consider rolling a fair, four-sided dice. If we define 3 events, A = {1,2}, B={1,3} and C={1,4}, then clearly P(A) = P(B) = P(C) = 1/2. But also, P(A,B) = P(A,C) = P(B,C) = P({1}) = 1/4 = P(A)P(B) = P(A)P(C) = P(B)P(C). This means we have pairwise independence.

However, P(A,B,C) = P({1}) also, which is not equal to P(A)P(B)P(C) = 1/8, and so we do not have mutual independence.

2.9 Conditional Independence. Are the following properties true?

(a) $$(X \bot W|Z,Y) \ AND \ (X \bot Y|Z) \implies (X \bot Y , W|Z)$$

OK initially this looks pretty confusing, but once you break it down it's not too bad.

The first condition tells us that:

$$P(X,W | Z,Y) = P(X | Z,Y) P(W| Z,Y)$$

and the second tells us:

$$P(X,Y | Z) = P(X | Z) P(Y | Z)$$

The condition we need to show is:

$$P(X,Y,W | Z) = P(X | Z) P(Y, W | Z)$$

Completely generally, we can use the chain rule of probability to say:

$$P(X,Y,W | Z) = P(W| X,Y,Z) P(Y| X,Z) P(X | Z)$$

However, since X and W are conditionally independent given Z,Y, we can say that $$P(W| X,Y,Z) = P(W | Y,Z)$$ (this is because generally $$P(X,W | Y,Z) = P(W | X, Y, Z) P(X | Y, Z)$$, so if conditional dependence is true, then so must be the previous identity. Similarly, P(Y|X,Z) = P(Y|Z) since X and Y are conditionally independent given Z. This means that:

$$P(X,Y,W | Z) = P(W|Y,Z)P(Y|Z) P(X|Z)$$

Then we can say that $$P(W|Y,Z)P(Y|Z) = P(W,Y|Z)$$, which gives us exactly what we need.

I couldn't be bothered to do part (b), but I expect it's kind of similar.

2.10 Deriving the inverse Gamma density

Let $$X \sim Ga(a,b)$$, such that:

$$p(x|a,b) = \frac{b^a}{\Gamma(a)} x^{a-1} e^{-xb}$$.

If $$Y = 1/X$$, show that $$Y \sim IG(a,b)$$, where:

$$IG(x |a,b) = \frac{b^a}{\Gamma(a)} x^{-(a+1)}e^{-b/x}$$

Let's start from the CDF of Y:

$$P(Y \le y) = P(\frac{1}{X} \le y) = P(X \ge \frac{1}{y}) = 1 - P(X \le \frac{1}{y})$$

If we let $$u = 1/y$$, then:

$$p(y) = \frac{d}{dy} P(Y \le y) = \frac{du}{dy} \frac{d}{du}(1-P(X \le u)) = \frac{1}{y^2} Ga(u | a, b)$$

Then we substitute y back in to find:

$$p(y) = \frac{b^a}{\Gamma(a)} y^{-(a+1)} e^{-b/y}$$

which is the result we require.

2.11 Basically asking to evaluate: $$\int_{-\infty}^{\infty} e^{-\frac{x^2}{2 \sigma^2}} dx$$

I remember seeing how to do this before - the trick is to consider the squared value and then convert to circular polar coordinates, so that you're just left with an integral from $$r=0 \to \infty$$ and a constant integral over $$\theta$$. Since we change from $$dx dy$$ to $$r dr d\theta$$ this allows the integral in r to be evaluated. I can't really be bothered to fill in the details but it should be easy to find elsewhere.

2.12 Expressing mutual information in terms of entropies: show that $$I(X;Y) = H(X) - H(X|Y) = H(Y) - H(Y|X)$$

If we start from the definition in the textbook of the MI being equal to the KL divergence between $$P(X,Y)$$ and $$P(X)P(Y)$$ then we have:

$$I(X;Y) = \sum_{x,y} P(x,y) \log(\frac{P(x,y)}{P(x)P(y)}) = \sum_{x,y} P(x,y) [ \log(P(x,y)) - \log(P(x)) - \log(P(y)) ]$$

If we write $$P(x,y) = P(x | y) p(y)$$ then we have:

$$I(X;Y) = \sum_{x,y} P(x|y)P(y) [ \log(P(x|y) - \log(P(x))]$$

We can then use that the conditional entropy $$H(X|Y) = -\sum_y p(y) \sum_x p(x|y) \log(p(x|y))$$, which is the negative of the first term we have. The second term, $$-\sum_{x,y} P(x,y) \log(P(x)) = H(X)$$, clearly, and so we have found that $$I(X;Y) = H(X) - H(X|Y)$$. It is easy to show the second identity, replacing $$P(x,y) = P(y|x)p(x)$$ instead.

2.13 Mutual information for correlated normals. Find the mutual information I(X1; X2) where X has a bivariate normal distribution:

$$\begin{bmatrix} X_1 \\ X_2 \end{bmatrix} \sim \mathcal{N} \left( \mathbf{0}, \begin{bmatrix} \sigma^2 & \rho \sigma^2 \\ \rho \sigma^2 & \sigma^2 \end{bmatrix} \right)$$

Evaluate when $$\rho = -1, 0, 1$$

The question actually gives you the form for the differential entropy for a multivariate normal, but I found a derivation which I thought was pretty nice and so I'm going to include it here (this is for a completely general MV normal of any dimension).

The pdf of a MV normal is:

$$p(\mathbf{x}) = \frac{1}{\sqrt{det(2 \pi \Sigma)}} e^{-\frac{(\mathbf{x}-\mathbf{\mu})^T \Sigma^{-1} (\mathbf{x}-\mathbf{\mu})}{2}}$$

Now the differential entropy is:

$$H(\mathbf{x}) = \int \int d\mathbf{x} \left[ \log(\sqrt{det(2 \pi \Sigma)}) + \frac{1}{2}(\mathbf{x}-\mathbf{\mu})^T \Sigma^{-1} (\mathbf{x}-\mathbf{\mu}) \right] \mathcal{N}(\mathbf{x}; \mathbf{\mu}, \Sigma)$$

Now the nice trick here is that, by definition, the covariance matrix $$\Sigma = \mathbb{E} \left[ (\mathbf{x}-\mathbf{\mu}) (\mathbf{x}-\mathbf{\mu}) ^T \right]$$, and the second term in the differential entropy is 1/2 times $$\mathbb{E} \left[ (\mathbf{x}-\mathbf{\mu})^T \Sigma^{-1} (\mathbf{x}-\mathbf{\mu}) \right]$$. Crucially, since this is a scalar we can say it's equal to it's trace, i.e.

$$\mathbb{E} \left[ (\mathbf{x}-\mathbf{\mu})^T \Sigma^{-1} (\mathbf{x}-\mathbf{\mu}) \right] = \mathbb{E} \left[ Tr\left( (\mathbf{x}-\mathbf{\mu})^T \Sigma^{-1} (\mathbf{x}-\mathbf{\mu}) \right) \right]$$

Then we can use the cyclic properties of traces, i.e. $$Tr(ABC) = Tr(BCA) = Tr(CAB)$$, such that this is equal to:

$$\mathbb{E} \left[ Tr\left( (\mathbf{x} -\mathbf{u})(\mathbf{x}-\mathbf{u})^T \Sigma^{-1} \right) \right] = Tr \left[ \Sigma \Sigma^{-1} \right] = d$$

where d is the number of random variables (i.e. the dimension). Now $$d = \log(e^d)$$ and $$det(2\pi \Sigma) = (2 \pi)^d det(\Sigma)$$, so we are left with:

$$H(\mathbf{x}) = \frac{1}{2} \log((2 \pi)^d det(\Sigma))$$

Now that we have this the rest of the question is relatively straightforward as we can rewrite the mutual information as $$I(X;Y) = H(X) + H(Y) - H(X,Y)$$, and so in this case:

$$I(X_1; X_2) = \log(2 \pi e \sigma^2) - \frac{1}{2} \log((2 \pi e)^2 det(\Sigma))$$

Now, $$det(\Sigma) = \sigma^4 - \rho^2 \sigma^4$$, and so making a few algebraic manipulations we arrive at:

$$I(X_1; X_2) = \frac{1}{2} \log(\frac{1}{1-\rho^2})$$

As $$\rho \to 0$$, $$I \to 0$$, and as $$\rho \to \pm 1$$, $$I \to \infty$$. Intuitively this makes sense - if $$\rho = 0$$, there is no correlation and so the variables give us no information about each other. If they are perfectly correlated, we learn everything about $$X_1$$ from $$X_2$$, and an infinite amount of information can be stored in a real number.

2.14 Normalised mutual information. Let X and Y be discrete and identically distributed RVs (so H(X) = H(Y)). Let:

$$r=1 - \frac{H(Y|X)}{H(X)}$$

(a) Show $$r = \frac{I(X;Y)}{H(X)}$$:

This is quite straightforward: $$r = \frac{H(X) - H(Y|X)}{H(X)} = \frac{I(X;Y)}{H(X)}$$

(b) Show $$0 \le r \le 1$$

Entropy is always positive, so clearly $$r \le 1$$ from it's initial definition. Then we know the mutual information is $$\ge 0$$ too, and so it's greater than 0. I guess we should prove that $$I(X;Y) \ge 0$$. We do this starting from the initial definition:

$$I(X;Y) = -\sum_{x,y} p(x,y) \log(\frac{p(x)p(y)}{p(x,y)})$$

Now, since the negative logarithm is convex we can apply Jensen's inequality $$\sum_i \lambda_i f(x_i) \ge f(\sum_i \lambda_i x_i)$$, where $$\sum_i \lambda_i = 1$$):

$$I(X;Y) \ge -\log \left( \sum_{x,y} p(x,y) \frac{p(x) p(y)}{p(x,y)} \right) = 0$$

(c) When is r=0?

When $$I(X;Y) = 0$$, i.e. when the variables give us no information about each other.

(d) When is r=1?

When $$H(Y|X) = 0$$, i.e. when the variables tell us everything about each other (i.e. once you know x, you get no information by learning y). I guess this is why r can be thought of as a normalised mutual information!

2.15 MLE minimises the KL divergence with empirical distribution

$$P_{emp}(x_i) = N_i / N$$ (sticking with the discrete case for now, where $$N_i$$ is the number of occurrences of $$x_i$$ and N is the total amount of data.

$$KL(p_{emp} || q(x;\theta)) = \sum_i p_{emp}(x_i) \log(\frac{p_{emp}(x_i)}{q(x_i;\theta)}) = \sum p_{emp} \log(p_{emp}) - \sum p_{emp} \log(q(x_i;\theta))$$

The first term here is fixed by the data and so it is clear that:

$$argmin_{\theta} KL (p_{emp} || q) = argmax_{\theta} \frac{1}{N} \sum_i N_i \log(q(x_i;\theta))$$

which is the same as maximising the log-likelihood, and hence the likelihood.

2.16 Derive the mean, mode and variance of the Beta(a,b) distribution.

$$Beta(x; a,b) = \frac{\Gamma(a+b)}{\Gamma(a) \Gamma(b)} x^{a-1} (1-x)^{b-1} \equiv \frac{x^{a-1}(1-x)^{b-1}}{B(a,b)}$$

To get the mode you just differentiate wrt x and set equal to zero - I really couldn't be bothered.

For the mean:

$$\text{mean} = \mathbb{E}[X] = \frac{1}{B(a,b)} \int_0^1 x^a (1-x)^{b-1} dx$$

Integrating by parts we find:

$$\mathbb{E}[X] = \frac{a}{B(a,b) b} \int_0^1 x^{a-1} (1-x)^b dx = \frac{a}{B(a,b) b} \left[ \int_0^1 x^{a-1} (1-x)^{b-1}dx - \int_0^1 x^a (1-x)^{b-1} dx \right]$$

However, one of these just integrates to 1 because it is a Beta distribution, and the second is exactly what we started with, i.e. $$\mathbb{E}[X]$$, and hence:

$$\mathbb{E}[X] = \frac{a}{b}(1-\mathbb{E}[X])$$

rearranging we find: $$\mathbb{E}[X] = \frac{a}{a+b}$$. You should be able to get the variance using the same kind of trick but it looked like a lot more algebra and I was too tired to attempt it!

2.17 Expected value of the minimum. Let X, Y be iid U(0,1) RVs. What is the expected value of min(X,Y).

If we let the CDF of X (and Y) be $$F(x) = x$$ (for x between 0 and 1), then:

$$P(\text{min}(X,Y) \ge x) = 1-P(\text{min}(X,Y) \le x) = P( X \ge x, Y \ge x) = (1-F(x))^2 = (1-x)^2$$

Therefore $$P(\text{min}(X,Y) \le x) = 1 - (1-x)^2$$, and so the can write the probability density of the minimum as:

$$p_{min}(x) = 2(1-x)$$

as such, the expected value of the minimum, which is what we want, is:

$$\mathbb{E}[min] = 2 \int_0^1 x(1-x) dx = 1/3$$

There are a couple of questions I skipped, maybe at some point I'll get back to them but I don't think they were especially interesting/informative. If there are any questions, or if you spot any errors, please leave a comment! I have done most of the CH3 exercises now, and so will be writing them up soon as well!

# Big 2 Reinforcement Learning AI is Working!!

I've finally had a decent amount of time to invest in my Big 2 reinforcement learning AI and I've actually managed to get it working now really well (much, much better than I was ever expecting in fact!). At some point I will do a full detailed write up but for now I'll just make a few notes about the process I used and summarize the results so far but the main result is that in the initial testing the AI actually beat me (who has played the game a lot) pretty damn convincingly and showed clear signs of being able to formulate good plans to get rid of its cards! I totally wasn't expecting this to happen so I'm really pleased with this, particularly as I was recently reading this article recently about how deep reinforcement learning doesn't really work very well yet (or rather that it's very difficult to get it to work properly compared to supervised deep learning).

So in the end I decided to use the "Proximal Policy Optimization" algorithm which seems to be very popular atm (particularly at OpenAI) and from what I've read is one of the best in terms of sample efficiency and robustness to varying hyper-parameters. It's also relatively simple to implement and OpenAI have released an excellent implementation in their "Baselines" project which was incredibly useful to use as a basis for my own code (which can be found here). I won't go into the details of the theory behind the PPO algorithm (as I actually still need to read more about this myself to really understand why it works so well) but it's a policy where you policy gradient method of sorts but where you use a surrogate loss function which is clipped so as to stop updates which change the policy too much. The surrogate function requires being able to estimate the advantage of a given action in a given state given the current policy you have. There are a number of ways you can do this but I followed the original paper which suggests using "Generalized Advantage Estimation" which tries to balance as best as possible reducing the bias and reducing the variance of the advantage estimates. This requires being able to estimate the value function of each state under the policy being employed making this an actor-critic algorithm, i.e. you are trying to learn a policy $$\pi$$ and a value function $$v$$ at the same time.

I also read quite a bit about how a lot of deep reinforcement learning algorithms (e.g. deep Q-learning) often don't perform very well in the multi-agent setting where you have a number of agents who are competing against each other in some way because these environments are complex and non-stationary whereas typical policy gradient methods suffer from very high variance which increases rapidly with the number of agents. This worried me that a four-player game of imperfect information would be very tricky to get working (especially with a complicated action space) but I was somewhat encouraged by this paper which demonstrated that using PPO with huge training batch sizes seemed to work really well in complicated competitive environments.

### What I actually did:

The first thing I needed to do was to parallelize my Big2 game class to allow for multiple games to be ran at once on different cores of my processor. The way PPO works is by running a number of environments $$n_e$$ in parallel (for ATARI and MuJoCo they use $$n_e=8$$ and run them forward for some number of steps $$n_s$$ into the future with the current policy to generate a batch to train on. The reason for running many games at once is so as to have a batch of samples which aren't all correlated (which they would be if they all game from the same game) and so it provides a big speed up if you can run these games on different cores as much as possible. Then when the batch has been generated you train the neural network using the PC's GPU. There are a number of challenges and things I had to alter from the OpenAI Baselines implementation which are mainly to do with the fact that this is designed for training a single agent whereas I needed it to be set up for four players. The main difficulty here is that to use generalized advantage estimation you need to work backwards and use all of the states in the batch you simulate after the one you're estimating the advantage for to get this estimate but now the "next state" is actually the state four time steps after the current time step as the three other players have to make a decision first. Another issue is the fact that you don't know if a state is terminal (and what reward to assign) until the other three players have had their next turn because the game finishes at one point when a player plays their final cards. To account for this it's necessary to run an extra four steps after each batch, save these, and then put them in as the first four states on the next batch. It also makes vectorizing the whole batch processing a bit trickier to deal with but I was eventually able to get this working.

In terms of the neural network architecture I went for the following:

Here the input layer has size 412 and contains information about the player's current hand and the cards/hands that other players have already played and which I described in a previous blog post. This is fed into a fully connected layer of 512 neurons which use a "RelU" activation function. This layer is shared and fed into two further hidden fully connected layers of 256 neurons each (also with RelU activation) which are each in turn connected to an output layer (each of which is linear (i.e. has no activation function). The first of these outputs represents the probability weighting that the network gives to each possible action in this current state whilst the second outputs the value estimate of the current state. To get the actual probabilities we consider only the outputs of actions $$\{o_i \}$$ that are actually allowed in the current state and sample actions with probabilities: $$p_i = \frac{ e^{ o_i } }{\sum_j e^{o_j} }$$. This means there are nearly a million trainable parameters in this model! I chose the number of hidden neurons in a fairly arbitrary way really and have made no attempt to play around with this so far but Big 2 is a reasonably complicated game so I figured that having fairly large layers would be sensible. Also sharing the first layer between the probability output and the value output seemed sensible as there are likely to be a lot of features of the game state which are useful for calculating both things!

In terms of the parameters I used for the PPO algorithm I chose to run 48 games in parallel and run them all for 20 steps to generate a training batch. This leads to batch sizes of 960 samples (which is tiny in comparison to what is used in the OpenAI paper for multi-agent environments where they generated batches of around 400,000 samples apparently) but a lot more than was used for Atari where they had only 8 games running in parallel. These batches were then divided into four mini-batches of equal size to be trained on with 5 epochs of SGD per batch. For the generalized advantage estimation I chose $$\gamma = 0.995$$ and $$\lambda = 0.95$$ and for the main PPO algorithm I chose a learning rate of $$\alpha = 0.00025$$ and a clip range of 0.2 (but both of these are linearly annealed to zero as the training progresses) and set the value/entropy coefficients in the loss function to 0.5 and 0.01 respectively. I then trained the agent by making it play against itself (always using the current version of the neural network - I'd quite like to experiment with some kind of opponent sampling and see if that makes any difference) for 136500 updates (~130 million time steps). I made absolutely no attempt to tune these hyperparameters in anyway (although I know they are somewhat sensible from reading the results of the paper) so it's really pretty cool that it ends up working so well!

### Results

The main result is that the network which is trained learns to play Big 2 really well! I've only played 15 games against it so far because the GUI I've made is kind of clunky and needs improving a bit but I got well and truly embarrassed - and I've played the game a lot (although I am by no means a proper expert). From the 15 games I played against three of the final trained AIs my scores were: $$\{-3, -3, -1, -1, -10, -11, -1, -1, -4, -8, +16, -2, -5, +14, -8 \}$$ - I only won two games (and in one of those I had what was essentially an unbeatable hand). There were also situations where I could tell that the AI was playing really well and had clearly "planned" in some sense how to get rid of its cards by learning both simple things like the value of passing to save 2s to gain control at a later stage in the game but also in terms of when to play certain number of cards when it gets control. I have to say this was really surprising and cool to see! Obviously I need to test this a bit more rigorously and so I am planning to make a web app where people can play against it and record the results so that I can get a better idea of how good it really is. One thing I have looked at is just its performance against earlier versions of itself as well just against random opponents and I get the following results (each point is averaged over 10,000 simulated games):

Note that the first data point is not at zero updates but after 1000 updates - hence a lot is getting learned very quickly! More importantly we see that there seems to be steady improvement throughout training which suggests that training for even longer could yield further improvement still!

To Do:

• A proper write up fully explaining the game, the PPO algorithm etc.
• A web app that allows for people to play and save their results in a leaderboard against the currently trained neural net.
• Experiment with different batch sizes (e.g. try >100 games running in parallel) and other parameters to see if performance can be improved further. Also would like to experiment with some kind of opponent sampling, i.e. only having one player definitely being the most recent neural network with other players being sampled from versions of the NN earlier on in the training. The point of this would be to try and ensure good play vs. all opponents, not just the new one. I guess what I'm thinking about here is the way a poker pro plays vs. a new player vs. another pro is completely different so this could be important to try and get the network to try and understand this sort of thing.

EDIT

Code is available on Github here. The webapp is finished, so you can play against the AI here (it will take a while to load most likely, as I am using a free Heroku account).

# Big 2 Reinforcement Learner Progress

This is just a post mainly to document (to myself more than anything) the progress that I've made over the past couple of weeks with the "Big 2 reinforcement learning project" that I'm working on as well as to lay out my plan for what to do next. For anyone who has stumbled across this post it might make sense to check out the previous post and/or to try out a single player version of the game I wrote in Javascript here. The main goal of the project is to train an AI to learn how to play the four player card game "Big 2". This is a game of imperfect information with a reasonably complicated action space and so should be an interesting challenge particularly as I have only recently become interested in the field of reinforcement learning.

So far I've mainly just been working on implementing the game logic in python and writing the code that generates what will be the input to the neural network. I spent some time creating a GUI in python with tkinter which allows you to generate a big 2 game, see what all of the options available to each player are, and essentially play through a full game. It also has a separate window that shows you what the input to the neural network for each player will be given the current state of the game. If you want to play around with it yourself you can find it here on Github or if you don't have python installed I made a standalone executabe you can run (although it's quite a big file). I did this partly because I thought it would be interesting to learn how to make a GUI with python (and tkinter turns out to be really cool and easy to use!) but it's also been extremely useful in debugging the game logic. Below is a screenshot of it in action:

In the middle are the three most recently played hands. The green circle shows that it is player 1's turn and that they have control (it is red otherwise) and then in the top right are the card indices (from 0 to 12) that are playable hands in the current situation. When I have a neural network to play against it should be easy to add this in to the GUI so that I can play against it as well as display statistics on things like the value it associates with each option that it has available to it, so I think this has definitely been a good use of my time. The second window it shows is the input state that will be provided to the neural network when it has to make a decision:

This may not be absolutely final but it's going to be what I try out initially. The first part of the input is the player's actual hand where each card (up to a maximum of 13 - the size of the initial hand you're dealt) is given a value and a suit. I also decided to include whether or not each card is part of a hand that uses more than one card (e.g. pair, straight, flush etc). In principle this could be learned by the network but I suspect/hope that it will be easier for the network to learn somewhat advanced strategies that involve saving cards for later on if you provide this information manually. Next we have the information about what the other players have/what they have played previously. One of the things which makes Big 2 interesting is that it is a game of imperfect information which of course means that we cannot provide the neural network with complete information about what is in each of the other player's hands, but instead can only include what is "allowed" to be known. So far I have included the number of cards they have left, and whether at some previous point they have played any ace or any two (the highest cards), as well as whether they've played any pairs, three of a kinds, two pairs, straights, flushes or full houses. It's at this point that I'm injecting the most "prior information" about the game of Big 2 because I know from experience that noting who's played certain high cards/hands is important in figuring out what decision you should make. Ideally you'd just like to include a full history of every hand that each player has played so far and try and learn directly from that but that simply isn't practical here and so until I can think of a better approach I'll stick with this for the time being. We then have an input which describes the previous hand which needs to be beaten - the value/suit as well as the type and then finally a "cards played" input which tracks whether anyone (so not specifically who) has played any of the 16 most valuable cards. Again this is injecting some prior knowledge about the game because I am not tracking whether the least valuable cards have been played as this is relatively unimportant (as the game is really all about gaining/maintaining control at the right times) and the input vector is already big enough as it is - 412 elements at the moment.

So this means everything is set up quite nicely meaning that I now need to start thinking about exactly how I'm going to try and train a neural network to play. Hopefully this weekend I will have a test set up working where I can get two untrained neural networks to play against each other in an essentially randomly manner so that I have an idea of how long it will take to simulate a single game, and hence have some idea about the amount of data I can generate in a reasonable time frame. This will have some effect on what I decide to do because unfortunately I don't work for Google and have virtually unlimited computational resources at my disposal and certain algorithms are much more computationally expensive than others!

### PLAN

Deep-Q Network

Probably the first thing I'm going to try is just a normal Deep Q Network like the one used in the Deepmind paper that learned to play Atari from raw pixel inputs. This involves training a deep neural network to try and learn the value function $$Q(s,a)$$ for any given state s and choice of action a. The reason that I'll try this first is because it is relatively simple (although there are a few tricks needed such as using experience replay and target networks to ensure that the training is stable), and there exist a number of good online tutorials for implementing this algorithm that I can use as a template (e.g. here and here). The obvious difficulty that I can see with using this approach is that the number of possible actions is quite high (see my previous post for a discussion of the Big 2 action space). The usual approach is to learn a function which outputs a Q-value for each possible action, i.e. you only provide the state s as input to the neural network rather than the state and the action. This has the big advantage that you then only need to carry out one forward pass of the network to obtain the Q-values for each action so that you can easily choose the one which is largest. The disadvantage is that it means that the network does nothing in terms of generalizing over actions which might be similar, i.e. you learn a different function for each action essentially independently. In Big 2 the number of possible actions is 1694 (in the way that I have decided to represent them) so it seems likely to me that this approach simply will not be possible and that the action will also have to be provided as an input to the network, such that the output is just a single number $$Q(s,a$$. This will cause an issue because then to choose the most "valuable" action you will have to evaluate the neural network for however many actions are available to the player in the current state each time requiring a full forward pass of the neural network. The only reason this might be possible is that although the number of possible actions is 1694 the actual number of actions that are allowable in any given situation is usually significantly lower meaning that this approach could be feasible. However being completely realistic I am expecting this to be too slow to get really good performance so the target here will be to hopefully train a network that can reliably beat a group of players who make random moves.

Large Action Spaces Paper

If this turns out to be too computationally expensive (or even if it doesn't) I would then like to have a look at using the technique described in this paper by some of the folks at Deepmind that I found recently and which can be applied to situations where you have large discrete action spaces. I haven't read through it in complete detail but the basic idea seems to be to describe actions as vectors $$\mathbf{a} \in \mathbb{R}^n$$ such that there is some (predefined) way of measuring how close actions are to each other. This will be quite natural given the way that I am choosing to represent the possible actions available in Big 2. The idea then seems to be to use an actor-critic approach whereby the actor chooses a "proto-action" $$\mathbf{\hat{a}} = f(s)$$ in continuous space and from which you evaluate the k-nearest actually allowable actions. Essentially the aim of the paper is to introduce an architecture that can generalize over actions without having the heavy cost associated with evaluating the neural network for every action that occurs when you explicitly include the action as an input to the neural network. Obviously I need to think more about the details (and read the paper properly!!) but it seems like it should be possible to apply this to the Big 2 action space so this will be interesting to try!

Alpha Zero and Generalizing the Monte Carlo Tree Search

The recent papers on "Alpha Go Zero" and then more generally "Alpha Zero" (which was applied to Chess and Shogi) are really, really cool! (and I should also mention this paper which uses a very similar technique and was developed at the same time!) The reason is because they learn to play these games completely from self-play, i.e. without any domain specific knowledge provided to them they just play games against each other and work out what moves lead to the best results. Doing this they beat their previous Alpha Go program with significantly less training time required and also beat the computer "world chess champion" which is pretty damn awesome if you ask me!

A key component of this algorithm is the Monte Carlo Tree Search (MCTS) which is used both during in training the neural network and afterwards once the neural network has been trained. I think the MCTS is a really cool algorithm and had a go at implementing my own version in c++ a while ago, applying it to the extremely complicated game of Tic Tac Toe. If you've never heard of it before then I would recommend reading this introductory blog post on the subject. Unfortunately despite it being a very cool algorithm it can only really be applied directly to games of perfect information. The basic idea is that you gradually build up a game tree:

(Image taken from "Recent Advances in General Game Playing")

in a way that sensibly balances exploration and exploitation, i.e. you don't build up a full game tree but just focus on "promising" branches. The actual implementation used by Alpha Zero is slightly different from the standard approach of using rollouts (and is again something I need to read up on in more detail - luckily there is an interesting post with code that implements the Alpha Zero algorithm on the much simpler game of Othello here. I am going to read through this and try to replicate their results when I get a chance). The reason this can only be applied to games of perfect information is that to properly build a game tree you need to know what state you will end up in when you take any particular action. In Big 2 you do not know this because you do not know what your opponents hand is and the number of possible things they could possibly do in their next move with all the possible cards they could have is far too large. Nevertheless I feel like it should be possible to do something similar where in the simplest case you just sample a certain number $$k$$ of possible hands that your opponents could have and use them to generate future states. This would be a very crude way of doing it - a more interesting way could be to train a neural network to try and learn how likely it is that a player has each card given what has happened during the game so far and then use these probabilities to sample the hands rather than doing so at random. Again I don't have any details for how I could go about implementing this at the moment but it seems possible that something like this should be possible. Then if I have a neural network which has learned to play I would expect that using some generalization of MCTS on top of this would lead to much better play (as you explicitly explore a large number of possible futures for each option available to you and are just guided by the neural network - in fact even without a neural network you could expect that this might lead to decent play if you let it think for long enough. This would be interesting to check!). In an ideal world I would want to use the MCTS generalization to train the network too but this has got to be hugely expensive and I don't have 5000 TPUs at my disposable so I suspect that this will not be realistic!

Anyway this has been a bit of a brain dump - it will be interesting to look back on this in a few months time and see whether anything I've thought about here ends up working out!

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