If have a random variables $X\in [0,1]$ for which i know: $$ F(R) = E(d(X) | X \leq R)$$ for any $R \in [0,1]$.
I would like to express $$ P = E(d(g(X))$$ in terms of $F(R)$. the function $g$ is well-behaved, i.e., monotonic, continuous and differentiable with g(0)=0, g(1)=1.
I expect the result to be of the form:
$$ P = \int_0^1 F(R)h(R) dR $$
I have the following intuitive direction: $$\frac{d}{dR}F(R) dR=f(R)dR \approx E(d(X) | X = R) dR$$
\begin{align} P &= E(d(g(X)) \\ &= \int_{R} E(d(g(X) | X = R)dR \\ &= \int_{R} E(d(g(X) | g(X) = g(R))dR \\ &= \int_{R} f(g(R))dR \end{align}
I am a bit stuck here. I expect some kind of change of variable here and even using integration by parts to get back to use $F$ instead of $f$.