Variance of min of r.v. and constant

Question

I am not a student of statistics, but need to compute an expression for my work. This is what I have so far:

I have a r.v. $D$ (pdf: $f$, support: $[0,\infty]$), and a positive constant $q$. I have a function $g=\min(q,D)$. I calculated $E[\min(q,D)]$ by using:

$$\int_0^\infty g(x)f(x)dx=\int_0^q xf(x)dx+q[1-F_D(q)]$$

Now I need to calculate $\text{Var}(g)$, and I am quite confused how to do it. If I try to expand $g$ using CDF of $D$, I end up with $E(g)$, which cant be correct. I need variance of $g$. It is not equal to variance of expectation of $g$, right?

I found this formula for var of a function of r.v.

$$Var[g(D)]=[g'E(D)]^2 Var(D)$$

But how do I compute g'? Any help is appreciated. Thanks.

Answer 1

Define $Z:=\min\{q,D\}$.

Since $\text{Var}(Z)=E[Z^2]-(E[Z])^2, $ you only have yet to compute $E[Z^2]:$

$$E[Z^2]=\int_0^\infty \min\{x^2,q^2\}f(x)dx=\int_0^q x^2f(x)dx+ q^2(1-F(q)).$$

Answer 2

You could try the variance formula $\text{Var}(g) = E(g^2) - [E(g)]^2$.

$D, q \geq 0$ so $g^2 = [\min(q, D)]^2 = \min(q^2, D^2)$. Thus you apply the same intuition as the expected value you did already, except with $x^2$ and $q^2$.