I am a software engineer by trade doing stats in my free time. I am playing around with an implementation of Microsoft's TrueSkill rating system for ranking players and openings in from a data set of ~700,000 Dominion games. I want to measure the log loss of predictions that the system gives for multiplayer games so that I can optimize hyperparameters of the system as well as explicitly model the inherent turn order advantage in the game.

In general, the TrueSkill builds a Bayesian Graphical Model, assuming a normal distribution on player skills as well as a normal distribution for player luck per game, and produces a mean/variance for the independent performance of players. For a two player game, the probability of one beating the other is encoded in the difference of the performance distributions.

However, predicting the probability of a player winning a 3 player game seems to requires estimating the probability that given player will exceed the maximum performance of the other two players. Is the maximum of two Gaussian also Gaussian?

I've read these notes which seem to solve a more general version of problem that I want on page 27 with equation (8-47).

Basically it says that for random variables X and Y, then if

$Z = \max(X, Y)$

With X and Y independent,

(1) $F_z(z) = F_x(x) F_y(y)$

and hence

(2) $f_z(z) = F_x(z)f_y(z) + f_x(z)F_y(z)$

Should I then plug in the formulas for the cdf/pdfs and pray that I get something useful?


The answer to this is basically "Yes". The formulae for the cdf/pdfs won't simplify very far (which seems to be what you're "praying" for). Further discussion is given in the following paper, in which they also consider the case when the gaussians are correlated:

Exact Distribution of the Max/Min of Two Gaussian Random Variables by Nadarajah, S. and Kotz, S. (2008).

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.