Skip to main content

Probability and Cumulative Dice Sums

Combining Expert Opinions: NaiveBoost

In many situations we're faced with multiple expert opinions. How should we combine them together into one opinion, hopefully better than any single opinion? I'll demonstrate the derivation of a classifier I'll call NaiveBoost.

We'll start with a simple situation, and later gradually introduce more complexity. Let each expert state a yes or no opinion in response to a yes/no question (binary classifiers), each expert be independent of the other experts and assume expert \(i\) is correct with probability \(p_i\). We'll also assume that the prior distribution on whether the correct answer is yes or no to be uniform, i.e. each occurs with probability 0.5.

Label a "yes" as +1, and "no" as -1. We ask our question, which has some unknown +1/-1 answer \(L\), and get back a set of responses (labels) \(S = \{L_i \}\), where \(L_i\) is the response from expert \(i\). Observe we have \[ \Pr(S | L=+1) = \prod_{i} {p_i}^{\frac{L_i+1}{2}} \cdot {(1-p_i)}^\frac{-L_i+1}{2}\] and also \[ \Pr(S | L=-1) = \prod_{i} {(1-p_i)}^{\frac{L_i+1}{2}} \cdot {p_i}^\frac{-L_i+1}{2}. \] As \( \Pr(L=+1 | S) = \frac{\Pr(S | L=+1)\cdot \Pr(L=+1)}{\Pr(S)}\) and \( \Pr(L=-1 | S) = \frac{\Pr(S | L=-1)\cdot \Pr(L=-1)}{\Pr(S)}\), and given our assumption that \( \Pr(L=+1) = \Pr(L=-1) \), we need only compute \( \Pr(S | L=+1) \), \( \Pr(S | L=-1) \) and normalize.

We'll now take logs and derive a form similar to AdaBoost. Note for \( L_{+1} = \log\left( \Pr(S | L=+1) \right) \) this gives us \[ L_{+1} = \sum_i \frac{L_i+1}{2}\log{(p_i)} + \frac{-L_i+1}{2}\log{(1-p_i)}.\] Rearranging, we get \[ L_{+1} = \sum_i \frac{L_i}{2}\log{\left( \frac{p_i}{1-p_i}\right)} + \frac{1}{2}\log{\left( p_i(1-p_i)\right)}.\] Similarly, for \( L_{-1} = \log\left( \Pr(S | L=-1) \right) \) we get \[ L_{-1} = \sum_i -\frac{L_i}{2}\log{\left( \frac{p_i}{1-p_i}\right)} + \frac{1}{2}\log{\left( p_i(1-p_i)\right)}.\] Note that each of these includes the same terms \( \sum_i \frac{1}{2}\log{\left( p_i(1-p_i)\right)}\). Upon exponentiation these would multiply \( \Pr(S | L=+1) \) and \( \Pr(S | L=-1) \) by the same factor, so we can ignore them as to recover the probabilities we'll need to normalize anyway. Thus we end up with a linear classifier with the AdaBoost form \[ C(S) = \sum_i \frac{L_i}{2}\log{\left( \frac{p_i}{1-p_i}\right)}. \] If \( C(S) \) is positive, the classifier's label is +1; if \( C(S) \) is negative, the classifier's label is -1. Furthermore, we may recover the classifier's probabilities by exponentiating and normalizing.

Comments

Popular posts from this blog

Simplified Multinomial Kelly

Here's a simplified version for optimal Kelly bets when you have multiple outcomes (e.g. horse races). The Smoczynski & Tomkins algorithm, which is explained here (or in the original paper): https://en.wikipedia.org/wiki/Kelly_criterion#Multiple_horses Let's say there's a wager that, for every $1 you bet, will return a profit of $b if you win. Let the probability of winning be \(p\), and losing be \(q=1-p\). The original Kelly criterion says to wager only if \(b\cdot p-q > 0\) (the expected value is positive), and in this case to wager a fraction \( \frac{b\cdot p-q}{b} \) of your bankroll. But in a horse race, how do you decide which set of outcomes are favorable to bet on? It's tricky, because these wagers are mutually exclusive i.e. you can win at most one. It turns out there's a simple and intuitive method to find which bets are favorable: 1) Look at \( b\cdot p-q\) for every horse. 2) Pick any horse for which \( b\cdot p-q > 0\) and mar...

Probability and Cumulative Dice Sums

Let a die be labeled with increasing positive integers \(a_1,\ldots , a_n\), and let the probability of getting \(a_i\) be \(p_i>0\). We start at 0 and roll the die, adding whatever number we get to the current total. If \({\rm Pr}(N)\) is the probability that at some point we achieve the sum \(N\), then \(\lim_{N \to \infty} {\rm Pr}(N)\) exists and equals \(1/\rm{E}(X)\) iff \((a_1, \ldots, a_n) = 1\). The direction \(\implies\) is obvious. Now, if the limit exists it must equal \(1/{\rm E}(X)\) by Chebyshev's inequality, so we only need to show that the limit exists assuming that \((a_1, \ldots, a_n) = 1\). We have the recursive relationship \[{\rm Pr}(N) = p_1 {\rm Pr}(N-a_1) + \ldots + p_n {\rm Pr}(N-a_n);\] the characteristic polynomial is therefore \[f(x) = x^{a_n} - \left(p_1 x^{(a_n-a_1)} + \ldots + p_n\right).\] This clearly has the root \(x=1\). Next note \[ f'(1) = a_n - \sum_{i=1}^{n} p_i a_n + \sum_{i=1}^{n} p_i a_i = \rm{E}(X) > 0 ,\] hence this root is als...

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 an...