Calculating variance, given g(x) and p(x)

Question

Given

• $g(x)$ defined for all real numbers $x$.
• $p(x)$ defined for $a\le x\le b$, which is the probability distribution of $x$

I have trouble calculating variance of $Var(g(x))$ because the two approaches below yield different results:

1. Calculate the expectation of $g(x)$ using $\mu =\int_a^b g(x)p(x) dx$. Then, calculate variance $Var=\int_a^b(g(x)-\mu)^2p(x)dx$. Note that when calculating variance, I plugged in $\mu$

2. Calculate variance directly using $$Var =\int_a^b\Big[g(x)-Exp(g(x))\Big]^2p(x)dx = \int_a^b\left[g(x)-\left(\int g(x)p(x)dx\right)\right]^2p(x)dx$$

Which of my approaches above (if any) is correct, if any, and why?

Answer 1

Both approaches are doing the same workload.   The second approach calculates the expectation inline, rather than calculating first then substituting.

$$\int_a^b \left(g(x)-\int_a^b g(x)~p(x)\mathsf d x\right)^2~p(x)~\mathsf d x ~=~\int_a^b\left(g(x)-\mu\right)^2~p(x)~\mathsf d x$$