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I'm thinking about coding a ranking engine, and am unclear about how I can combine probability distributions for two (or more) entities.

I think this is easier to ask with an example. Say I have two students, A and B, and their expected scores on an exam are distributed as follows:

$A \sim \mathcal{N}(75, 5)$

and

$B \sim \mathcal{N}(70, 5)$

How do I calculate the probability A beats B in the exam?

How would I generalise the approach to handle more students (ie for each student, determine the probability they top the exam)?

What considerations would I need to take into account for other probability distributions?

Hobo
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  • Are you willing to assume the scores to be independent? (Independence may be violated, for example, if part of the random variation is your uncertainty about how difficult the exam was, or about how effective their shared teacher was)? – Juho Kokkala Aug 09 '16 at 04:31
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    $X=A-B$ is normally distributed with $\mu=5$ and $\sigma^2=10$. The answer to your question is $P(X>0)$. Hopefully that's enough of a hint. – gammer Aug 09 '16 at 04:35
  • @JuhoKokkala I'm assuming the scores are independent, yes – Hobo Aug 09 '16 at 05:45
  • @gammer - thanks. That's excellent - I figured I was missing a simplification somewhere. So the 10 is just the sum of the variances, nothing fancier? – Hobo Aug 09 '16 at 05:48
  • Yes, under independence, you can just sum the variances. If there were dependence you'd have to subtract twice the covariance between the two. – gammer Aug 09 '16 at 06:39
  • For the Gaussian case, http://stats.stackexchange.com/questions/74091 and http://stats.stackexchange.com/questions/44139/ seem to be duplicates – Juho Kokkala Aug 09 '16 at 06:55

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$A$ beats $B$ when $Y = A - B > 0$ . From the given information, and assuming independence, $Y$ is normally distributed with mean $\mu = 5$ and variance $\sigma^2 = 10$. So the probably that $A$ wins is

$$ P(Y > 0) = P \left( Z > \frac{0 - 5}{\sqrt{10}} \right) = 1 - \Phi(-1.58) = \Phi(1.58) = .943 $$

gammer
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