# Computing expectation of a conditional Gaussian distribution

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.

• $P(X+Y=a)=0$. So this conditional expectation is not defined. – Kabo Murphy Sep 24 at 6:40
• @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. – KRL Sep 24 at 6:53
• @KaviRamaMurthy If we write $𝐸[𝑌|𝑋+𝑌,𝑋>𝑏]$ instead, would that make a difference? – KRL Sep 24 at 6:54

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