Finding the conditional distribution of 2 dependent normal random variables Here's the situation
$X \sim N(\mu, \sigma^2)$ and given $X=x$, $Y \sim N(x, \tau^2)$
I need to find the distribution of $X$ given $Y=y$
From what's given, I know the pdf's of $X$ as well as $Y|X=x$. By multiplying those pdf's together, I can find the joint distribution of $X$ and $Y$ ($f_{X,Y}=f_Xf_{Y|X=x}$). 
This seems useful, but I have no clue how to go about finding $f_Y$ to finish it off (since $f_{X|Y=y} = f_{X,Y}/f_Y$). The integral from $-\infty$ to $y$ of $f_{X,Y}$ with respect to $X$ doesn't have a closed form solution to my knowledge.
Is this possible? Am I on the right track? Any help would be tremendously appreciated! 
 A: Adapted from this previous answer of mine.
Suppose that $X \sim \mathcal N(\mu,\sigma^2)$ is the value of
a signal that we wish to observe, but what we can observe is
$X+N$ where $N \sim \mathcal N(0,\tau^2)$ is noise that is
independent of $X$. Let $Y$ denote $X+N$. Since $X$ and $N$
are independent, it is straightforward
to determine that $Y \sim \mathcal N(\mu, \sigma^2+\tau^2)$.
Now, the conditional
density of $Y$ given that $X$ has taken on value $x$ is
the density of $x+N$ which is clearly $\mathcal N(x,\tau^2)$.
What we are asked for, however, is the conditional density of 
$X$ given that $Y = y$, that is, the a posteriori
distribution of $X$ given that we have observed that $Y = y$.
We choke down the gorge that is rising in our throats
at this blatant Bayesian heresy
and note that $X$ and $Y = X+N$ are jointly normal random
variables, $X \sim \mathcal N(\mu,\sigma^2)$ and
$Y \sim \mathcal N(\mu, \sigma^2+\tau^2)$, and that their
covariance is
$$\operatorname{cov}(X,Y) = \operatorname{cov}(X,X+N)
= \operatorname{cov}(X,X)+ \operatorname{cov}(X,N) = \sigma^2.$$
Now, it is a standard result that when $X$ and $Y$ are
jointly normal random variables, the conditional density
of $X$ given that $Y = y$ is a normal density
with mean
\begin{align}E[X\mid Y = y] &= 
\mu_X + \left.\left. 
\frac{\operatorname{cov}(X,Y)}{\operatorname{var}(Y)}\right(y-\mu_Y\right)\\
&= \mu+\left.\left.\frac{\sigma^2}{\sigma^2+\tau^2}\right(y-\mu
\right)\\
&= \frac{\sigma^2}{\sigma^2+\tau^2}y +\frac{\tau^2}{\sigma^2+\tau^2}\mu,\end{align}
that is, a weighted blend of the observation $y$ and the
a priori mean of $X$.  Note that if $\sigma^2 \gg \tau^2$,
we give large weight to the observation $y$ and very little to
the known mean $\mu$, while if $\sigma^2 \ll \tau^2$, we tend
to ignore the observation and give heavy weight to the mean $\mu$.
Also, the variance of this conditional density is
$$\operatorname{var}(X)\left(1-\frac{(\operatorname{cov}(X,Y))^2}{\operatorname{var}(X)\operatorname{var}(Y)}\right)
= \sigma^2\left(1-\frac{\sigma^4}{\sigma^2(\sigma^2+\tau^2)}\right)= \frac{\sigma^2\tau^2}{\sigma^2+\tau^2}.$$
This allows us to write down the conditional density
$f_{X\mid Y}(x\mid y)$ directly without messing up the ranges
of integrations etc. as in the OP's approach. 
A: Once you know the joint distribution, just use the property of the multivariate normal that the conditional distributions are also normal with parameters as defined here: https://en.wikipedia.org/wiki/Multivariate_normal_distribution#Conditional_distributions. Just fix Y to some value rather than fixing X.
