Unable to prove(may be under some condition) the identity: for random variables $X,Y,W$, we have $p(x|y) = \int p(x|w)p(w|y)dw$.

Question

Let $X,Y,W$ are random variables. Let $p(x)$ denotes the pdf $X$. We try proving the identity: $p(x|y) = \int p(x|w)p(w|y)dw$. We start with $$p(x|y) = \int p(x|y,w)p(w)\,dw,$$ $$p(x|y) = \int \int p(x|y,w)p(w|y')p(y')\,dy'\,dw.$$ Don't understand how to proceed further. May be the identity is true when we have some relationship between $W$ and $Y$. Any suggestions or hints are welcome!

Answer 1

What you need is "$X$ and $Y$ are independent given $W$." This means $$ P(X \in A, Y \in B \mid W) = P(X \in A \mid W) P(Y \in B \mid W). $$ Translated into densities, this becomes $$ p(x,y \mid w) = p(x \mid w) p(y \mid w). $$

If the last displayed equation is true, then $p(x,y,w) = p(x \mid w) p(y, w),$ so we can divide by $p(y)$ to reach $p(x, w \mid y) = p(x \mid w) p(w \mid y),$ and if we integrate over $w,$ you get $$ p(x \mid y) = \int dw\ p(x \mid w) p(w \mid y), $$ as you wanted.