# Writing the likelihood function for a linear combination of Normal and Laplacian random variables [duplicate]

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!

• 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. Dec 17, 2022 at 18:05

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, $$P(y=+1|W=w) = \tfrac{1}{2}\left(1+\text{sgn}(w) - \text{sgn}(w)e^{-2|w|}\right)$$
• @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 $P(y = -1 | W = w) = \frac{1}{2} (1 - sgn(w) + sgn(w)e^{-2|w|})$ Dec 18, 2022 at 15:16
• Would you know how I may go about the joint probability $P(y, W)$? Thank you. Dec 18, 2022 at 15:22