Suppose we have discrete random variable given by \begin{align*} P(X=x_i)=\frac{1}{N}, i=1...N \end{align*} and Gaussian r.v. $Z \sim \mathcal{N}(0,1)$. Assume $Z$ and $X$ are independent. Suppose $X$ and $Z$ form an new r.v. $Y$ give by \begin{align*} Y=X+Z \end{align*}
I am interested in computing $E[X|Y]$?
Here are some of the distributions that I have computed: \begin{align*} f_Y(y)&=\frac{1}{N} \sum_{i=1}^N \frac{1}{\sqrt{2\pi}} e^{-(y-x_i)^2/2}\\ f_{Y|X}(y|x_i)&=\frac{1}{\sqrt{2\pi}} e^{-(y-x_i)^2/2}=f_Z(z)\\ f_{X|Y}(x_i|y)&=??? \end{align*}
But I am not sure how to proceed next. For example does density $f_{X|Y}(x_i|y)$ even exists? Thank you for any help or suggestions.
Firstly,$$f_{X,Y}(x,y) = \frac{1}{N\sqrt{2\pi}}\;e^{-(y-x)^2/2}.$$
Therefore, \begin{eqnarray*} f_{X|Y}(x|y) &=& \dfrac{f_{X,Y}(x,y)}{f_{Y}(y)} \\ && \\ &=& \dfrac{\frac{1}{N\sqrt{2\pi}}\;e^{-(y-x)^2/2}}{\frac{1}{N} \sum_{i=1}^N \frac{1}{\sqrt{2\pi}} e^{-(y-x_i)^2/2}} \\ && \\ &=& \dfrac{e^{-(y-x)^2/2}}{\sum_{i=1}^N e^{-(y-x_i)^2/2}}\qquad\qquad\qquad\text{for $x\in\{x_1,\ldots,x_N\},\; y\in\mathbb{R}$.} \\ && \\ \therefore \quad E(X\mid Y) &=& \dfrac{\sum_{i=1}^N x_i e^{-(y-x_i)^2/2}}{\sum_{i=1}^N e^{-(y-x_i)^2/2}}. \\ \end{eqnarray*}