Derivative conditional moments binary random variable

Question

Let $e$ be a continuously distributed RV with pdf $f$ and let $q( x )$ be a binary RV that depends on the former through the relation $q ( x ) = 1[h( x ) \geq e ]$, where $h$ is a well-behaved function.

Because the binary nature of $q(x)$, it holds that its first two uncentered moments are identical (this holds in fact for all higher moments): $E[q(x) \mid x] = E[q(x)^2 \mid x]$.

I am confused, however, when taking partial derivatives of these moments w.r.t. the conditioning variable $x$. For the first moment, supposing that dominated convergence holds, we have that \begin{equation} \frac{\partial}{\partial x} E[q(x) \mid x] = E[\delta(h(x) - e)\mid x] = f(h(x)), \end{equation} where $\delta$ is the Dirac delta function. For the second moment, using the chain rule, we have that \begin{equation} \frac{\partial}{\partial x} E[q(x)^2 \mid x] = E\left[2\delta(h(x) - e) 1[h(x) \geq e]\mid x\right]= 2f(h(x)). \end{equation} The partial derivatives being different seems to be at variance with the fact that both conditional moments are actually the same function. It is not clear to me, however, where I did make a mistake in the derivation of the partial derivates.

Answer 1

Try avoiding the delta function altogether:

$$\mathbb{E}q(x) = \int_{-\infty}^{\infty}1(e \leq h(x))f(e)de = \int_{-\infty}^{h(x)}f(e)de = F(h(x))$$

$$\mathbb{E}q^2(x) = \int_{-\infty}^{\infty}1^2(e \leq h(x))f(e)de = \int_{-\infty}^{h(x)}f(e)de = F(h(x))$$

with derivatives:

$${\partial\mathbb{E}q(x) \over \partial x} = h'(x)f(h(x))$$ $${\partial\mathbb{E}q^2(x) \over \partial x} = h'(x)f(h(x))$$

The derivative of $1^2(x\leq y)$ with respect to $y$ is not $2\cdot 1(x\leq y)\delta(y)$, but just $\delta(y)$. You can, perhaps, see this by analogy with the standard definition of a derivative:

$$\lim_{h \to 0}{f(x+h)-f(x) \over h}$$

In this case, $f(x+h)=0$ and $f(x)=1$. Now for the derivative of $f^2$:

$$\lim_{h \to 0}{f^2(x+h)-f^2(x) \over h}$$

and $f^2(x+h) = 0$ with $f^2(x)=1$, just as above, so the derivatives are the same.