Let $Z\sim N(0,a)$ with variance $a>0$. Show that the mgf of $Z^{2}$ is $M_Z^{2}(t)=\frac{1}{\sqrt{1-2at}}$

Question

Let $Z\sim N(0,a)$ with variance $a>0$. Show that the mgf of $Z^{2}$ is $M_Z^{2}(t)=\frac{1}{\sqrt{1-2at}}$

My Working:

Now using the density function of normal random variable with mean $0$ standard deviation $\sqrt{a}$ and the definition of mgf I get:

$M_Z^{2}(t)=\int_{-\infty}^\infty \frac{1}{\sqrt{2a\pi}} e^{z^{2}t} e^{-\frac{1}{2a}z^{2}} \,dz$

The exponent is $z^{2}t-\frac{z^{2}}{2a}=-\frac{z^{2}}{2a}(1-2at)=-\frac{w^{2}}{2a}$;

Where; $w=z \sqrt{1-2at} \Rightarrow dz = \frac{dw}{\sqrt{1-2at}}$

Putting the above values I get:

$M_Z^{2}(t)=\frac{1}{\sqrt{1-2at}}(\int_{-\infty}^\infty \frac{1}{\sqrt{2a\pi}} e^{-\frac{w^{2}}{2a}} \,dw)$

Now I don't know how to make the expression inside parenthesis equal to $1$; because the integrand doesn't seem the density function to me; in case $a=1$; it would have been density of exponential random variable but it isn't. What do I do? Please guide me

Answer 1

If $W \sim \mathcal{N}(0, a)$, its density function is given by $$f_{W}(w) = \dfrac{1}{\sqrt{a} \cdot \sqrt{2\pi}} \cdot e^{-\frac{w^2}{2a}}$$ for $w \in \mathbb{R}$. Then, use the fact that $\sqrt{a}\cdot \sqrt{2\pi} = \sqrt{2a\pi}$.