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