Finding conditional probability distribution (X|Y) from (Y|X) $(Y,X)$ has a joint distribution where the marginal of $X$ is a standard normal and $Y|X \sim U \left[|X|-\frac{1}{2},|X|+\frac{1}{2}\right] $ where $U[a,b] $ means uniform in the interval [a,b]. How can I find the distribution of $X$ given $Y$ and conditional expectation function $E(X|Y)$? I tried to multiply p.d.f. of $Y|X$ and $X$ to find joint distribution first, but support of $Y|X $ confused me. Thanks in advance :) 
 A: There are a number of issues here:
From A.S.'s comments $$f_{X\mid Y}(x\mid y)=\frac {f_X(x)f_{Y\mid X}(y\mid x)}{\int f_X(x)f_{Y\mid X}(y\mid x)\,dx}$$
where $f_X(x) =\phi(x)$ and $f_{Y\mid X}(y\mid x) = \mathbb 1_{\left[|x|-\frac12 \le y \le |x|+\frac12\right]}$
That last condition $|x|-\frac12 \le y \le |x|+\frac12$ can be rewritten as $y-\frac12 \le |x| \le y+\frac12$, and so when $x \ge 0$ as $\max\left(0,y-\frac12\right) \le x \le y+\frac12$ or when $x \lt 0$ as $-y-\frac12 \le x \le \min\left(0,-y+\frac12\right)$.
Here $y$ can take any value down to $-\frac12$, but when $-\frac12 \le y \lt \frac12$ the two possible intervals for $x$ meet giving $-y-\frac12 \le x \le y+\frac12$. Otherwise they are distinct and give two intervals for $x$ given $y$. So

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*When $-\frac12 \le y \lt \frac12$ and $-y-\frac12 \le x \le y+\frac12$ (one interval): $$f_{X\mid Y}(x\mid y)=\frac {\phi(x)}{\Phi\left(y+\frac 1 2\right)-\Phi\left(-y-\frac 1 2\right)}$$


*When $y>\frac12$ and $y-\frac12 \le |x| \le y+\frac12$ (two intervals): $$f_{X\mid Y}(x\mid y)=\frac {\phi(x)}{2\left(\Phi\left(y+\frac 1 2\right)-\Phi\left(y-\frac 1 2\right) \right)}$$
and otherwise $f_{X\mid Y}(x\mid y)=0$. Note the different signs and the factor of $2$ in one of the denominators. Here is what the conditional densities look like for $y=0.6$ (red), $y=0.4$ (blue), and $y=-0.3$ (green):

