I am working on a probability problem to solidify my knowledge and running into some difficulty. I have the following setup:

  • Let $W_{i}, W_{j}$ represent 2 i.i.d random variables, drawn from $N(0,1)$. $W_{i}$ and $W_{j}$ represent the individual skill of 2 chess players, $i$ and $j$ respectively.
  • Let $v$ ~ $Laplace(0, \frac{1}{2})$ represent some variability of game $g$ between player $i$ and $j$ (time of day, weather, current state of both players etc.).
  • Define $y = sign(W_{i} - W_{j} + v)$ to be the outcome of game $g$. $y = -1$ if player $i$ wins and $y = +1$ if player $j$ wins.

This is where I am stuck: how do I write out the likelihood $P(y | W)$? I know the difference of 2 independent normal variables $X_1$ ~ $N(\mu_1, \sigma_1)$ and $X_2$ ~ $N(\mu_2, \sigma_2)$ is $X_1 - X_2$ ~ $N(\mu_1 - \mu_2, \sqrt(\sigma_1^{2} + \sigma_2^{2}))$, but that's as far as I've gotten.

Additionally, I am trying to write the joint probability $P(y, W)$, but unsure how to.

Any pointers or ressources would be appreciated. Thanks!

  • 1
    $\begingroup$ You need to compute the density of $(X_1-X_2)+V$ which is a convolution of a Normal and a Laplace. The resulting distribution of $Y$ is immediate. $\endgroup$
    – Xi'an
    Dec 17, 2022 at 18:05

1 Answer 1


Let us discuss how to find $P(y|W)$, which is your first part of the question.

I imagine that when you write $P(y|W)$ you mean that $W = W_i - W_j$, which we should really call $W_1 - W_2$ since there are only two players. Let us say that $W$ (random variable) has a known value equal to $w$. We want to find the likelihood function of $y$ given that $W=w$. Since $y=\pm 1$, once we figure out where $y=+1$, then the complement is when it will be $-1$, so let us focus on the condition when $\text{sgn}(w+v) > 0$. This condition is satisfied when $v>-w$. Recall that $v\sim \text{Lap}(0,\tfrac{1}{2})$, so the probability that $v>-w$ is equal to $\int_{-w}^{\infty} e^{-2|x|} ~ dx $.

The value of that integral will depend on whether $w>0$ or $w<0$ (because of the presence of absolute values in the integral). If $w<0$ then that integral is equal to $\tfrac{1}{2}e^{2w}$. If $w>0$ then that integral is equal to $\int_{-w}^{\infty} = \int_{-w}^0 + \int_0^{\infty}$, which computes to be $(\tfrac{1}{2} - \tfrac{1}{2}e^{-2w}) + \tfrac{1}{2} = 1 - \tfrac{1}{2}e^{-2w}$.

We can summarize our answer, by saying that, $$ P(y=+1|W=w) = \tfrac{1}{2}\left(1+\text{sgn}(w) - \text{sgn}(w)e^{-2|w|}\right)$$

  • $\begingroup$ @nicolas-bourkaki Many thanks for your answer, this helps quite a bit. Adding a few details for clarification: There are more than one game and more than 2 players, and I initially simplified. My goal is to find $W_{MLE} = argmax_{W} P(y | W)$ (by eventually using MH). Using your methodology, I derived $P(y = -1 | W = w) = \frac{1}{2} (1 - sgn(w) + sgn(w)e^{-2|w|})$ $\endgroup$
    – LucDupont
    Dec 18, 2022 at 15:16
  • $\begingroup$ Would you know how I may go about the joint probability $P(y, W)$? Thank you. $\endgroup$
    – LucDupont
    Dec 18, 2022 at 15:22

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