Conditional mean and variance of $X$ given that $Y=6$ for $X$ normal $N(0,1)$ and $Y$ conditionally on $X=x$ normal $N(x,1)$

Question

Assume that $X\sim N(0,1)$ and that, given $X=x$, the conditional distribution of $Y\mid X=x$ is $N(x,1)$. Find the conditional mean and variance of $X\mid Y=6$.

I proceeded as follows: $$f_{X|Y}(X|Y=6)= \frac{f_X(x)f_{Y|X}(Y|X=x)}{\int^{\infty}_{-\infty}f_X(x)f_{Y|X}(Y|X=x)dx}$$ I know that $$f_X(x)f_{Y|X}(Y|X=x)=\frac{1}{2\pi}e^{\frac{-x^2}{2}}e^{\frac{-(y-x)^2}{2}}$$ But $${\int^{\infty}_{-\infty}f_X(x)f_{Y|X}(Y|X=x)dx}=\int^{\infty}_{-\infty}\frac{1}{2\pi}e^{\frac{-x^2}{2}}e^{\frac{-(y-x)^2}{2}}dx$$

And that is where I am stuck.

Answer 1

One is given that, for every $x$, the conditional distribution of $Y$ conditionally on $X=x$ is normal with mean $x$ and variance $1$, hence $Y=X+Z$ where $(X,Z)$ is i.i.d. standard normal. Thus, $X=\frac12Y+T$ where $T=\frac12(X-Z)$.

Since $(Y,T)$ is jointly normal, the fact that $\mathrm{Cov}(Y,T)=0$ alone, implies that $T$ is independent of $Y$.

Now, $T$ is centered normal with variance $\frac12$ hence, for every $y$, the conditional distribution of $X$ conditionally on $Y=y$ is normal with mean $\frac12y$ and variance $\frac12$.

This result can be recovered, using conditional PDFs and neglecting the inherent structure of families of normal random variables, only, more clumsily and more lengthily.