Let's say $x \in \mathbb{R}$ is normally distributed so that $x \sim \mathcal{N}(\mu,\sigma^2)=p(x)$. I want to simplify the following integral, which includes arbitrary, but fixed integration limits $a\in \mathbb{R}$ and $b\in \mathbb{R}$.
$I := \int_{a}^{b} x^2 p(x)dx$.
In the case of $a=-\infty, b=\infty$ the integral $I$ equals $\sigma^2$, right? Does anyone have an idea how to simplify $I$ with different bounds?
Best Answer
Hint: Since $\frac{\mathrm d}{\mathrm dx}\phi(x) = -x\phi(x)$ where $\phi(x)$ is the pdf of the standard normal variable, it is possible to find the anti-derivative of $x^2\phi(x)$ by writing it as $x\cdot (x\phi(x))$ and then using the integration by parts formula: $$\int u\, \mathrm dv = uv - \int v \,\mathrm du$$ with $u = x$ and $v = \phi(x)$. Your mission, if you choose to accept it, is to figure out how this idea can be adapted for use with $p(x)$, the pdf of an arbitrary normal random variable. Good luck! This answer will self-destruct in thirty days.