If $X\sim \mathcal{N}(\mu,\sigma^2)$ and $Y=X^2$, what is the moment generating function (MGF) of $Y$?
Answer: the MGF of Y is:
$$\mathbb{E}\left[e^{t Y}\right]$$
But, first I have to calculte the PDF of $Y$:
$$\mathbb{P}(Y\leq x) = 1-\mathbb{P}(X^2\geq x) = 1-2\mathbb{P}(X\geq \sqrt{x}) $$
I can continueue , but it is the long way. Is there a shorter way?
If $X$ was standard normal ($\mu=0$, $\sigma^2=1$), then $Y$ would be Chi Squared distribution whose MGF is (1-2t)^{-0.5} for $t<0.5$. How can I use this to obtian what I need?
Best Answer
Use the Law of the Unconscious Statistician. $X = \mu + \sigma Z$ where $Z$ is standard normal.
$$ M_Y(t) = \mathbb E\left[e^{tY}\right] = \mathbb E\left[e^{t (\mu + \sigma Z)^2} \right] = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{t(\mu + \sigma z)^2} e^{-z^2/2}\; dz$$ For convergence you want $\text{Re}(t\sigma^2) < 1/2$, and then $$ M_Y(t) = \frac{1}{\sqrt{1-2t\sigma^2}} \exp\left(\frac{\mu^2 t}{1-2t\sigma^2}\right)$$