Is there a nice way (closed form) to represent the distribution of $Z$ that results from taking two independent normal draws $X_1, X_2 \sim N(0,1)$ and determining the outcome using $Y \sim U(0,1)$ as follows: $$Z = \begin{cases} X_1\ \ \mbox{if } Y\le\frac{1}{1 + e^{X_2 - X_1}} \\ X_2 \ \ \mbox{otherwise} \end{cases}$$ That is, we select two players from a normal distribution of skill and take the winner of a match played between them by using an Elo model. It would be nice if we could iterate the solution to determine the players left after conducting $n$ rounds of elimination matches. It would be fine to use sigmoid-type functions other than the logistic to determine win probability and skill distributions other than the normal if this would make this problem more tractable.


2 Answers 2


Looking at the MGF of $Z$ \begin{align*} \mathbb E[e^{tZ}] &= \mathbb E[\mathbb E[e^{tZ}|X_1,X_2]]\\ &= \mathbb E\left[e^{tX_1} \frac{e^{X_1}}{e^{X_1}+e^{X_2}} + e^{tX_2} \frac{e^{X_2}}{e^{X_2}+e^{X_1}}\right]\\ &= \mathbb E\left[e^{tX_1} \frac{e^{X_1}}{e^{X_1}+e^{X_2}}\right] + \mathbb E\left[e^{tX_2} \frac{e^{X_2}}{e^{X_2}+e^{X_1}}\right]\\ &=2 \mathbb E\left[e^{tX_1} \frac{e^{X_1}}{e^{X_1}+e^{X_2}}\right] \end{align*} which is not a Normal MGF.


A proof will have to wait, but some simulations indicate that the distribution of the Elo score of the winner is normal:

histogram of simulated elo score of the winner

Here is the R code:

N <- 1e7
X1 <- rnorm(N); X2 <- rnorm(N)
Y <- runif(N)
Z <- ifelse(Y<= 1/(1+exp(X2-X1)), X1, X2) 

 (mu <-mean(Z))
[1] 0.363302
 (sigma <- sd(Z))
[1] 0.9316078

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