V is modeled as a normal random variable with mean 0 and variance σ 2. What is E[X^2]

Question

I am trying to find the expected value of the function $Z = \frac 12mV^2$, in terms of sigma and $m$, where $m$ is a constant greater than $0$.

Answer 1

Because $\mathbb E[V]=0$, the variance is given by $\text{Var}\ V=\mathbb E[V^2]-(\mathbb E[V])^2=\mathbb E[V^2]$. Therefore $$ \mathbb E\left[\frac{mV^2}{2}\right]=\frac{m}{2} \mathbb E[V^2] = \frac{m}{2}\ \text{Var}\ V=\frac{m\sigma^2}{2}. $$