Derivative of an Expectation over an Indicator function

Question

I have a small question on how to compute the derivative of an expectation when an indicator function gets in the way. Let $x$ be a random variable. We are interested in computing the derivative wrt A of

$E_x [(A-x)1_{(x \leq A)}]$

where $1_n$ is the indicator function 1 when $n$ is true. A paper I am reading mentions that this is simply

$F(A)$

but I am unsure how to get that result. Your intuitive explanations are most appreciated.

It's pretty clear that $E[1_{(x\leq A)}]$ is $F(A)$, so I don't think it's also true for $E[(A-x)1_{(x\leq A)}]$. — Gregory Grant, May 14 '15 at 16:29
I posted an answer below, let me know if you need more details. — Gregory Grant, May 14 '15 at 16:39

score 3 · Accepted Answer · answered May 14 '15 at 16:36

3

$$E_x [(A-x)1_{(x \leq A)}]$$ $$=\int_{-\infty}^\infty(A-x)1_{(x\leq A)}p_x(x)dx$$ $$=\int_{-\infty}^A(A-x)p_x(x)dx$$ $$=\int_{-\infty}^AAp_x(x)dx-\int_{-\infty}^Axp_x(x)dx$$ $$=A\int_{-\infty}^Ap_x(x)dx-\int_{-\infty}^Axp_x(x)dx$$ $$=AF(A)-\int_{-\infty}^Axp_x(x)dx$$

So taking the derivative with respect to $A$ we get $$F(A) + AF'(A)-\frac{d}{dA}\int_{-\infty}^Axp_x(x)dx$$ $$=F(A)+Ap_x(A)-Ap_x(A)$$ $$=F(A)$$

answered May 14 '15 at 16:36

Gregory Grant

14,874

You're welcome glad to help – Gregory Grant May 15 '15 at 14:39

Derivative of an Expectation over an Indicator function

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