I have a question about conditional probability.
Let's say
$y = x + e$
where $x$ ~ $U[0,1]$ and $e$ ~ $N(0,1)$ and both are independent.
What is $p_x(x | y = 10)$ ?
In particular why isn't it the same as $p_e(10 - x)$, i.e. $N(10-x, 1)$? I'm trying to see where the constraint of x being in the range $[0,1]$ comes into play here. I'm likely missing something fundamental, would appreciate any pointers on this.
$p_x(x\mid y=10)$ is a probability density function, so its integral over the support of $[0,1]$ must equal $1$. This is not the case for a normal distributed random variable, so it is not that.
$$\int_0^1 p_x(x\mid y=10)\,\mathrm d x=1\\\int_0^1 p_e(10-x)\,\mathrm d x\neq 1$$
Use Bayes' Rule to find what actually is the conditional distribution.
$$\begin{align} p_x(x\mid y=10) &=\dfrac{p_x(x)\,p_e(10-x)}{p_y(10)}\\[1ex]&=\dfrac{p_x(x)\,p_e(10-x)}{\displaystyle\int_0^1p_x(s)\,p_e(10-s)\,\mathrm d s} \end{align}$$