$Y= X+N$ what is ${\rm E}(X|Y=y)$ $Y = X+N$ where $X,N$ are normally distributed with zero mean. 
I know that $Y$ is also normally distributed with zero mean and sum of variances. 
I don't know how to get $f(X|y)$ in order to calculate ${\rm E}(X|y)$.
 A: You can find ${\rm E}(X|Y=y)$ immediately as follows.
If $X \sim {\rm N}(0,\sigma_1^2)$ and $N \sim {\rm N}(0,\sigma_2^2)$
(and $X$ and $N$ are independent), then
$$
{\rm E}(X|X + N) = \frac{{\sigma _1^2 }}{{\sigma _1^2  + \sigma _2^2 }}(X + N),
$$
so, in particular,
$$
{\rm E}(X|X + N = y) = \frac{{\sigma _1^2 }}{{\sigma _1^2  + \sigma _2^2 }}y.
$$
I will (probably) justify this (i.e., the first formula) later on today.
The point is that, in some respect, the normal distribution does not play a special role in this result.
For example, if $X$ and $N$ are independent Poisson random variables with means $\lambda_1$ and $\lambda_2$, respectively, then
$$
{\rm E}(X|X + N) = \frac{{\lambda _1 }}{{\lambda_1  + \lambda_2 }}(X+N).
$$
What is common to the normal distribution and the Poisson distribution in this context is that both are infinitely divisible. More details later on.
A: For some later purposes, I give here the complete solution based on svenkatr's approach (thus also verifying my previous answer). In what follows, I use  a different (somewhat more natural) notation, so don't get confused.
Suppose that $X$ and $Y$ are independent ${\rm N}(0,\sigma_1^2)$ and ${\rm N}(0,\sigma_2^2)$ variables, respectively.
We can find ${\rm E}(X|X+Y=z)$ as follows.
$$
{\rm E}(X|X + Y = z) = \int_{ - \infty }^\infty  {xf_{X|X + Y} (x|z)\,{\rm d}x} = \int_{ - \infty }^\infty  {x\frac{{f_X (x)f_{X + Y|X} (z|x)}}{{f_{X + Y} (z)}}\, {\rm d}x}
$$
(the notation should be clear from the context). Noting that $f_{X+Y|X}(\cdot|x)$ is the ${\rm N}(x,\sigma_2^2)$ density function, some algebra shows that the right-hand side integral is equal to
$$
\int_{ - \infty }^\infty  {x\sqrt {\frac{{\sigma _1^2  + \sigma _2^2 }}{{\sigma _1^2 \sigma _2^2}}} \frac{1}{{\sqrt {2\pi } }}\exp \bigg\{  - \frac{{[x - z \sigma _1^2 /(\sigma _1^2  + \sigma _2^2 )]^2 }}{{2[\sigma _1^2 \sigma _2^2 /(\sigma _1^2  + \sigma _2^2 )]}}} \bigg \} \,{\rm d}x.
$$
Now, from $\int_{ - \infty }^\infty  {x\frac{1}{{\sqrt {2\pi \sigma ^2 } }}{\rm e}^{ - (x - \mu )^2 /(2\sigma ^2 )} \,{\rm d}x}  = \mu $ (expectation of a ${\rm N}(\mu,\sigma^2)$ random variable), we can find that
$$
{\rm E}(X|X + Y = z) = \frac{{\sigma _1^2 }}{{\sigma _1^2  + \sigma _2^2 }}z.
$$
A: \begin{equation}
p(x|y) = \frac{p(y|x)p(x)}{p(y)} 
\end{equation}
You know $p(y)$ and $p(x)$.
\begin{equation}
p(y|x) = \frac{1}{\sqrt{2 \pi \sigma_N^2}} e^{\frac{-(y-x)^2}{2\sigma_N^2}}
\end{equation}
Using these expressions, you can easily get both $p(x|y)$ and $E(X|Y=y)$.
A: What follows is too long to be a comment. It is also very useful.
Suppose first that $X_i$, $i=1,2$, are independent ${\rm N}(0,\sigma_i^2)$ variables. Their characteristic functions are ${\rm E}[{\rm e}^{{\rm i}zX_i} ] = {\rm e}^{\sigma _i^2 ( - z^2 /2)}$. Suppose now that $X_i$, $i=1,2$, are independent Poisson variables, with means $\lambda_i$. Their characteristic functions are ${\rm E}[{\rm e}^{{\rm i}zX_i} ] = {\rm e}^{\lambda _i ({\rm e}^{{\rm i}z}  - 1)}$. For the normal distribution case
${\rm E}(X_1 | X_1 + X_2) = \frac{{\sigma _1^2 }}{{\sigma _1^2  + \sigma _2^2 }}(X_1  + X_2 )$, and for the Poisson case
${\rm E}(X_1 | X_1 + X_2) = \frac{{\lambda_1 }}{{\lambda_1  + \lambda_2 }}(X_1  + X_2 )$ (the later result is easy to prove). Finally, consider independent rv's $X_i$, $i=1,2$, having finite expectations and characteristic functions of the form ${\rm E}[{\rm e}^{{\rm i}zX_i} ] = {\rm e}^{c _i \psi(z)}$, $c_i > 0$. What would you expect ${\rm E}(X_1 | X_1 + X_2)$ to be, based on the previous observations? Though very informal, this is very useful, since the conclusion can be applied  to a vast number of rv's, e.g. gamma rv's (specifically, the conclusion applies to all integrable infinitely divisible rv's).
