### An Island of Liars is an Ensemble of Experts

In my previous post I looked at how a group of experts may be combined into a single, more powerful, classifier which I call NaiveBoost after the related AdaBoost. I'll illustrate how it can be used with a few examples.

As before, we're faced with making a binary decision, which we can view as an unknown label $$L \in \{ +1, -1 \}$$. Furthermore, the prior distribution on $$L$$ is assumed to be uniform. Let our experts' independent probabilities be $$p_1 = 0.8, p_2 = 0.7, p_3 = 0.6$$ and $$p_4 = 0.5$$. Our combined NaiveBoost classifier is $C(S) = \sum_i \frac{L_i}{2}\log{\left( \frac{p_i}{1-p_i}\right)},$ where $$S = \{ L_i \}$$. A few things to note are that $$\log{\left( \frac{p_i}{1-p_i}\right)}$$ is $${\rm logit}( p_i )$$, and an expert with $$p = 0.5$$ contributes 0 to our classifier. This latter observation is what we'd expect, as $$p = 0.5$$ is random guessing. Also, experts with probabilities $$p_i$$ and $$p_j$$ such that $$p_i = 1 - p_j$$ are equivalent if we replace the stated label $$L_j$$ with $$-L_j$$.

Ignoring the last (uninformative) expert, we end up with the combined classifier $C(S) = \frac{L_1}{2} \log\left(4\right) + \frac{L_1}{2} \log\left(\frac{7}{3}\right) + \frac{L_3}{2} \log\left( \frac{3}{2} \right).$ If the overall value is positive, the classifier's label is $$L = +1$$; if it's negative, the classifier's label is $$L = -1$$. Note the base of the logarithm doesn't matter and we could also ignore the factor of $$\frac{1}{2}$$, as these don't change the sign of $$C(S)$$. However, the factor of $$\frac{1}{2}$$ must be left in if we want the ability to properly recover the actual combined probability via normalization.

Now say $$L_1 = -1, L_2 = +1, L_3 = +1$$. What's our decision? Doing the math, we get $$C(S) = \frac{1}{2} \log{ \left( \frac{7}{8} \right) }$$, and as $$7 < 8$$, $$C(S) < 0$$ and our combined classifier says $$L = -1$$. If we wanted to recover the probability, note $\exp\left( \frac{1}{2} \log \left( \frac{7}{8} \right) \right) = {\left( \frac{7}{8} \right)}^{1/2},$ hence our classifier states ${\rm Pr}(L = +1 | S ) = \frac{ {\left( \frac{7}{8} \right)}^{1/2} }{ {\left( \frac{7}{8} \right)}^{1/2} + {\left( \frac{7}{8} \right)}^{-1/2} } = \frac{ \frac{7}{8} }{ \frac{7}{8} + 1 } = \frac{7}{15},$ and of course $${\rm Pr}(L = -1 | S ) = \frac{8}{15}$$.

As a second example, consider the @CutTheKnotMath puzzle of two liars on an island. Here we have A and B, each of which lies with probability 2/3 and tells the truth with probability 1/3. A makes a statement and B confirms that it's true. What's the probability that A's statement is actually truthful? We can solve this in a complicated way by observing that this is equivalent to an ensemble of experts, where $$L \in \{ +1, -1 \}$$, the prior on $$L$$ is uniform and $$L_1 = L_2 = +1$$. The probability that $$L = +1$$ is precisely the probability that A is telling the truth.

Following the first example, $L_{+1} = \frac{L_1}{2}\log\left( \frac{1}{2} \right) + \frac{L_2}{2}\log\left( \frac{1}{2} \right) = \log\left( \frac{1}{2} \right).$ Continuing as before, we get ${\rm Pr}(L = +1 | S ) = \frac{ \frac{1}{2} }{ \frac{1}{2} + 2 } = \frac{1}{5}.$

1. In the event tree there are two cases where B says the response is true, one with probability of 1/9 (A says it's try and B also say's it is try, 1/3 * 1/3) and the other with probability of 4/9 (A lies and says it is false and B also lies and says it is true, 2/3 * 2/3). So the probability that A is really telling the truth given B said it is true is (1/9)/(1/9+4/9) or 1/5.

1. Yes, that's simpler. I used this approach to illustrate how a puzzle such as this is actually related to a more complicated idea such as ensembles in machine learning.

2. Understood. Just checking the math. I don't know anything about machine learning...

### A Bayes' Solution to Monty Hall

For any problem involving conditional probabilities one of your greatest allies is Bayes' Theorem. Bayes' Theorem says that for two events A and B, the probability of A given B is related to the probability of B given A in a specific way.

Standard notation:

probability of A given B is written $$\Pr(A \mid B)$$
probability of B is written $$\Pr(B)$$

Bayes' Theorem:

Using the notation above, Bayes' Theorem can be written: $\Pr(A \mid B) = \frac{\Pr(B \mid A)\times \Pr(A)}{\Pr(B)}$Let's apply Bayes' Theorem to the Monty Hall problem. If you recall, we're told that behind three doors there are two goats and one car, all randomly placed. We initially choose a door, and then Monty, who knows what's behind the doors, always shows us a goat behind one of the remaining doors. He can always do this as there are two goats; if we chose the car initially, Monty picks one of the two doors with a goat behind it at random.

Assume we pick Door 1 and then Monty sho…

### Mixed Models in R - Bigger, Faster, Stronger

When you start doing more advanced sports analytics you'll eventually starting working with what are known as hierarchical, nested or mixed effects models. These are models that contain both fixed and random effects. There are multiple ways of defining fixed vs random random effects, but one way I find particularly useful is that random effects are being "predicted" rather than "estimated", and this in turn involves some "shrinkage" towards the mean.

Here's some R code for NCAA ice hockey power rankings using a nested Poisson model (which can be found in my hockey GitHub repository):
model <- gs ~ year+field+d_div+o_div+game_length+(1|offense)+(1|defense)+(1|game_id) fit <- glmer(model, data=g, verbose=TRUE, family=poisson(link=log) ) The fixed effects are year, field (home/away/neutral), d_div (NCAA division of the defense), o_div (NCAA division of the offense) and game_length (number of overtime periods); off…

### Gambling to Optimize Expected Median Bankroll

Gambling to optimize your expected bankroll mean is extremely risky, as you wager your entire bankroll for any favorable gamble, making ruin almost inevitable. But what if, instead, we gambled not to maximize the expected bankroll mean, but the expected bankroll median?

Let the probability of winning a favorable bet be $$p$$, and the net odds be $$b$$. That is, if we wager $$1$$ unit and win, we get back $$b$$ units (in addition to our wager). Assume our betting strategy is to wager some fraction $$f$$ of our bankroll, hence $$0 \leq f \leq 1$$. By our assumption, our betting strategy is invariant with respect to the actual size of our bankroll, and so if we were to repeat this gamble $$n$$ times with the same $$p$$ and $$b$$, the strategy wouldn't change. It follows we may assume an initial bankroll of size $$1$$.

Let $$q = 1-p$$. Now, after $$n$$  such gambles our bankroll would have a binomial distribution with probability mass function \[ \Pr(k,n,p) = \binom{n}{k} p^k q^{n-k…