Working on a project that involves conditional expectation and derivatives, I came up with an identity, but don't know how to interpret/explain the reason behind.
Let $X$ be an RV over the unit interval $[0,1]$, distributed according to some CDF $F$ and pdf $f$. Let $A=[t,1]$, $t\geq 0$.
The derivative of the product of $P[A]$ and $\mathbb E[x|x\in A]$ with respect to $P[A]$ is
\begin{align}\frac{d P[A]\mathbb E[x|x\in A] }{d P[A]}=&\frac{d \int^1_t zf(z)dzP[A]}{d t}\frac{1}{d P[A]/dt}\\=&\frac{d \int^1_t zf(z)dzP[A]}{d t}\frac{1}{d (1-F(t))/dt}\\=&\frac{-tf(t)}{-f(t)}\\=&t.\end{align}
I was initially expecting to have an answer that is a function of $F$ and $f$, but, to my surprise, the answer is actually independent of the distribution $F$. Anyone have an idea/intuition why this should be true?
One way to see it is the following. If $f>0$ and $zf(z),f(z)\in L^1(\lambda)$ you have by Lebesgue differentiation theorem that for any such $f$ it holds that ($\lambda$ is the Lebesgue measure): $$\frac{\lim_{\varepsilon \downarrow 0}\frac{\int_{t-\varepsilon}^{t+\varepsilon}zf(z)dz}{\lambda([t-\varepsilon,t+\varepsilon])}}{\lim_{\varepsilon \downarrow 0}\frac{\int_{t-\varepsilon}^{t+\varepsilon}f(z)dz}{\lambda([t-\varepsilon,t+\varepsilon])}}=\frac{tf(t)}{f(t)}=t$$ at least $\lambda(dt)$-a.e. To see that this is our case, note that since $P(X\leq t)=\int^t_{-\infty} f(z)dz$ we have $$\frac{dP(X\geq t)}{dt}=-\frac{dP(X\leq t)}{dt}=-\lim_{\varepsilon \downarrow 0}\frac{P(X\leq t+\varepsilon)-P(X\leq t-\varepsilon)}{2\varepsilon}=-\lim_{\varepsilon \downarrow 0}\frac{\int_{t-\varepsilon}^{t+\varepsilon}f(z)dz}{\lambda([t-\varepsilon,t+\varepsilon])}$$ and similarly $$\frac{dE[X\mathbf{1}_{\{X\geq t\}}]}{dt}=\frac{d(E[X]-E[X\mathbf{1}_{\{X\leq t\}}])}{dt}=-\frac{dE[X\mathbf{1}_{\{X\leq t\}}]}{dt}=-\lim_{\varepsilon \downarrow 0}\frac{\int_{t-\varepsilon}^{t+\varepsilon}zf(z)dz}{\lambda([t-\varepsilon,t+\varepsilon])}$$