$f(Y|X,Y<X + E_X - E_Y)~= ~?$, where $X,Y\sim N(\mu,\sigma^2)$ and $E_x,E_y\sim N(\epsilon,s^2)$ are mutually independent

Question

I have four random variables: $x,y\sim N(\mu,\sigma^2)$ and $\epsilon_x,\epsilon_y\sim N(\epsilon,s^2)$. They all are mutually independent. $~X,Y,E_X,E_Y$ are their realizations.

I observe $X$ and also the fact that $Y<X+E_X-E_Y$, but do not observe $Y,E_X,E_Y$. What is the distribution of $Y$, conditional on the information that I observe?

i.e. $f(Y|X,Y<X + E_X - E_Y)~= ~?$

Any help is highly appreciated.

Answer 1

First of all, let us strongly simplify the question:

1. Since you condition on $X$ (and by the assumption of independence), you do not have to treat $x$ as a random variable any longer.

2. You do not need two variables $\epsilon_x,\epsilon_y$, since only their difference appears in the question and $\epsilon_x-\epsilon_y \sim N(0,2s^2)$.

Thereby, you are asking for the probability density $f_{\tilde y}$ of $\tilde y\sim (y | y<\epsilon)$, where the new random variable $\epsilon := X + \epsilon_x-\epsilon_y \sim N(X,2s^2)$ is independent of $y$. This is much simpler already.

Now, denoting the density of $\epsilon$ by $\rho_\epsilon$ (a well-known Gaussian density as stated above), we obtain for the cumulative distribution function (CDF) $F_{\tilde y}$ of $\tilde y$ by the law of total probability $$ F_{\tilde y}(Y) = \mathbb P(\tilde y\le Y) = \mathbb P(y\le Y | y<\epsilon) = \int \mathbb P(y\le Y | y<E) \rho_\epsilon(E)\, \mathrm dE. $$ By the definition of conditional probability, we can simplify the integrand further: $$ \mathbb P(y\le Y | y<E) = \frac{\mathbb P(y\le Y \text{ and } y<E)}{\mathbb P(y<E)} = \frac{\mathbb P(y\le \min(Y,E))}{\mathbb P(y<E)}, $$ which is $1$ for $Y\ge E$ and can be expressed in terms of the error function for $Y<E$.

What is left is to take the integral in order to obtain the CDF $F_{\tilde y}$ and then take its derivative to get the probability density $f_{\tilde y} = F_{\tilde y}'$. The mentioned integral splits up into two integrals corresponding to the two cases $Y\ge E$ and $Y<E$. I am not sure, whether the second one can be computed exactly, I would try Mathematica and see how far I get.