Derivative of $H(c)=\mathbb E[(c-X)\chi_{X\leq c}]$

Question

Let $c$ be a real consant, $X$ a real random variable and $H(c)=\mathbb E[(c-X)\chi_{X\leq c}]$. Show that $$H'(c)=P[X\leq c].$$ I already proved this if $X$ has a density function. But what if $X$ has no density? I thought about the weak derivative and started with $$\int \mathbb E[(c-X)\chi_{X\leq c}] \phi'(c)dc$$ for all test functions $\phi$. How can I go on?

Answer 1

Fix $c$. Consider the case $h > 0$, and compute $$ \dfrac{H(c + h) - H(c)}{h} = \dfrac{1}{h}\mathbb{E}[(c + h - X)\chi_{\{X \le c + h\}} - (c - X)\chi_{\{X \le c\}}] $$ In case $h > 0$, we have $$ \dfrac{(c + h - X)\chi_{\{X \le c + h\}} - (c - X)\chi_{\{X \le c\}}}{h} = \begin{cases} 1, \text{ if } X \le c, \\\\ 1 + \dfrac{c - X}{h}, \text{ if } c < X \le c + h,\\\\ 0, \text{ if } X > c + h \end{cases} $$ Thus, $$ \dfrac{H(c + h) - H(c)}{h} = \mathbb{P}(X \le c) + \mathbb{E}\left[\left(1 + \dfrac{c - X}{h}\right)\chi_{\{c < X \le c + h\}}\right] $$ Notice that $$ 0 \le \left(1 + \dfrac{c - X}{h}\right)\chi_{\{c < X \le c + h\}} \le 1 $$ and $$ \left(1 + \dfrac{c - X}{h}\right)\chi_{\{c < X \le c + h\}} \xrightarrow{h \downarrow 0} 0 $$ An application of Lebesgue's dominated convergence theorem implies $$ \lim_{h \downarrow 0} \dfrac{H(c + h) - H(c)}{h} = \mathbb{P}(X \le c) $$ The case $h < 0$ is similar and I'll leave it to you.