I run into a problem that is surprising to me.
Suppose three r.v. $Y, U, V$ satisfy $Y = U + V$, $U$ is independent of $V$.
We compute the following conditional density of $V$ given $Y$:
$ f_{V | Y}(v|y) = f_{V|U}(v|y-v) = f_V(v)$.
It's surprising to me because the conditional density $V|Y$ equals the marginal density of $V$.
Is my understanding correct? Thanks!
On
Take $\begin{cases}P(U=1)=&\frac 1 3\\P(U=2)=&\frac 1 3\\P(U=3)=&\frac 1 3\end{cases}$ and $\begin{cases}P(V=-1)=&\frac 1 3\\P(V=0)=&\frac 1 3\\P(V=1)=&\frac 1 3\end{cases}$ and let $Y=U+V$ with U, V being independent.
Then $f_{V|Y=0}(-1)=1$ (and is 0 otherwise). But clearly $f_V\ne f_{V|Y=0}$. Thus $f_{V|Y}\ne f_V$
$$f_{Y,V}(y,v)=f_U(y-v)f_V(v)$$which indicates that the numerators of your first equality are equal but the denominators are not equal since $f_Y(y)\ne f_U(y-v)$ as $f_Y(y)$ does not have $v$. Take for example $U,V\sim B(n,p),Y\sim B(2n,p)$, then $f_Y(y)=\binom{2n}y p^y(1-p)^{n-y}\ne f_U(y-v)$.
Your second equality is true and follows from the independence of $U,V$.