 Solution

The moment-generating function of a random variable $X$ is $M_X(t)=\mathbb{E}(e^{tX})$, defined for real $t$ where that expectation is finite. For a standard normal distribution, completing the square givesIt is finite for every real $t$. Derivatives at zero give the moments: the expansion $e^{t^2/2}=1+t^2/2+t^4/8+O(t^6)$ therefore givesFor $Y$ with the normal distribution of mean $\mu$ and variance ${\sigma }^2$, write $Y=\mu +\sigma X$ in distribution. Its moment-generating function is consequentlyThis formula also covers the constant random variable when $\sigma =0$.