Let $W$, $S$ be two independent, continuos r.v., both in $L^2$. Let $P = g(W,S)$ a certain function of both $W$ and $S$; moreover, $P$ is continuous and in $L^2$.
Let $l,k: \mathbb R\mapsto \mathbb R$ two continuous functions such that $l(W), k(S)$ are also in $L^2$.
I am interested in computing
$$E[l(W)k(S) \mid P = p]$$
My idea was to condition on $W = w$ and integrate wrt to the density of $W$, i.e. writing
$$E[l(W)k(S) \mid P = p] = \int_{\mathbb R} l(w) \ E[k(S) \mid g(w, S) = p] \ f_W(w) dw$$
This would help because I can compute $E[k(S) \mid g(w, S) = p]$.
But is the formula correct? Intuitively I think it should be, as $W$ and $S$ are independent, but I am a little confused about the definition of conditional expectation wrt to a probability zero event. The definition I encountered weren't so precise (usually just saying "use the density instead of the probability"), so I am not sure how to determine if the formula is correct or not (and how to fix it!)
Hints/references are also appreciated. Thanks!
By definition of conditional expectation, $$E(\ell(W)k(S)\mid P)=h(P)$$ where the measurable function $h$ is such that, for every bounded measurable function $m$, $$E(h(P)m(P))=E(\ell(W)k(S)m(P))$$ that is, $$E(h(P)m(P))=E(\ell(W)k(S)m(g(W,S)))$$ or, equivalently, $$\int h(x)m(x)f_P(x)dx=\iint \ell(w)k(s)m(g(w,s))f_W(w)f_S(s)dwds$$ In particular, for every $p$, $$\int_{-\infty}^p h(x)f_P(x)dx=\iint \ell(w)k(s)\mathbf 1_{g(w,s)\leqslant p}f_W(w)f_S(s)dwds$$ By identification, this holds true if $$h(p)=\frac1{f_P(p)}\frac{d}{dp}\iint \ell(w)k(s)\mathbf 1_{g(w,s)\leqslant p}f_W(w)f_S(s)dwds$$ Thus, for $P$-almost every $p$, $$E(\ell(W)k(S)\mid P=p)=\frac1{f_P(p)}\frac{d}{dp}\iint_{g(w,s)\leqslant p} \ell(w)k(s)f_W(w)f_S(s)dwds$$ or, equivalently, $$E(\ell(W)k(S)\mid P=p)=\frac1{f_P(p)}\frac{d}{dp}E(\ell(W)k(S);g(W,S)\leqslant p)$$
Edit: Above, we exposed the correct approach to every similar problem of conditional expectations but this does not disqualify, a priori, the approach in the question. We turn to this problem now and, for this, we first rewrite carefully what the claim in the question entails.
The claim is that $$E(\ell(W)k(S)\mid P=p)=E(\ell(W)h(W,p))$$ where $$ h(w,p)=E(k(S)\mid g(w,S)=p)$$ hence the function $h$ is characterized by the property that, for $W$-almost every $w$ and for every measurable function $r$, $$E(k(S)r(g(w,S)))=E(h(w,g(w,S))r(g(w,S)))$$ Thus, the claim holds true iff, for every measurable function $m$, $$E(\ell(W)k(S)m(P))=E(q(P)m(P))$$ where $$q(p)=E(\ell(W)h(W,p))$$ No idea why this should be true... So, let us turn to an example: assume that $P=W+S$, then $g(w,s)=w+s$ hence $$h(w,p)=E(k(S)\mid w+S=p)=k(p-w)$$ One sees that the claim becomes that, for every measurable function $m$, $E(\ell(W)k(S)m(P))$ is simultaneously $$\iint\ell(w)k(s)m(w+s)f_W(w)f_S(s)dwds$$ and $$\iint\ell(w)k(p-w)m(p)f_W(w)f_P(p)dwdp$$ thus, one requires that $$\iint\ell(w)k(s)m(w+s)f_W(w)f_S(s)dwds=\iint\ell(w)k(s)m(w+s)f_W(w)f_P(w+s)dwds$$ which can only hold in full generality if the densities coincide, that is, after some obvious simplifications, if, for $W$-almost every $w$, $$f_S(s)=f_P(w+s)$$ Well, at last we have a proof that the claim does not hold, right?