Suppose that $X$ and $Y$ are i.i.d. Gaussian random variables with mean $\mu$ and variance 1. I am trying to compute the following conditional expectation

$E[Y|X+Y = a, X > b]$.

I tried to compute the density by using the Bayes rule

$p(Y|X+Y = a, X>b) = \frac{p(Y)p(X>b|Y)p(X+Y=a|Y, X>b)}{p(X>b)p(X+Y|X>b)}$

Since $X$ and $Y$ are independent we have $p(X>b) = p(X>b|Y)$

$p(Y|X+Y = a, X>b) = \frac{p(Y)p(X+Y=a|Y, X>b)}{p(X+Y|X>b)}$

Let $f(X;\mu,\sigma)$ and $F(X;\mu,\sigma)$ be the Gaussian pdf and CDF, respectively.

$\frac{p(Y)p(X+Y=a|Y, X>b)}{p(X+Y|X>b)} = \frac{f(Y;\mu, 1)F(X;\mu+Y,1)1{X>b}}{1-F(b;\mu+Y,1)p(X+Y|X>b)}$.

I don't know how to proceed from here and how to compute $p(X+Y|X>b)$. This density is basically the density of sum of a normal rv and a truncated normal rv but according to the answer here, it is difficult to obtain a closed-form to.

  • $\begingroup$ $P(X+Y=a)=0$. So this conditional expectation is not defined. $\endgroup$ – Kavi Rama Murthy Sep 24 '19 at 6:40
  • $\begingroup$ @KaviRamaMurthy I'm confused. Suppose that we have a 2D Gaussian distribution over $(X,Y)$ and we want to compute $E[X|Y=y]$. Here $P[Y=y] = 0$ but the conditional expectation exists. For example, consider the trivial case that $X$ and $Y$ are independent. $\endgroup$ – KRL Sep 24 '19 at 6:53
  • $\begingroup$ @KaviRamaMurthy If we write $𝐸[π‘Œ|𝑋+π‘Œ,𝑋>𝑏] $ instead, would that make a difference? $\endgroup$ – KRL Sep 24 '19 at 6:54

$$ \int_{-\infty}^\infty y \, P(Y\in dy|X+Y\in da, X>b) = \int_{-\infty}^\infty y \, \frac{P(Y\in dy, X+Y\in da,X>b)}{P(X+Y\in da,X>b)} = \int_{-\infty}^\infty y \, \frac{f_{\mu,1}(y) \, f_{\mu,1}(a-y) 1_{\lbrace a-y>b\rbrace}}{\int_b^\infty f_{\mu,1}(x)\, f_{\mu,1}(a-x) \, dx} \, dy = \frac{\int_{-\infty}^{a-b} y \, f_{\mu,1}(y) \, f_{\mu,1}(a-y) \, dy}{\int_b^\infty f_{\mu,1}(x) \, f_{\mu,1}(a-x) \, dx} = \frac{\int_{-\infty}^{a-b} y \, f_{\mu,1}(y) \, f_{\mu,1}(a-y) \, dy}{\int_{-\infty}^{a-b} f_{\mu,1}(x) \, f_{\mu,1}(a-x) \, dx} $$

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