evaluating moment generating functions

Question

Let $Z_1,Z_2,\ldots,Z_{14} $ be 14 independent N(0,1) variables, and let $Y=Z_1^2+Z_2^2+\cdots+Z_{14}^2$. Provide answers to the following to two decimal places.

Evaluate the moment generating function $M_{Z^2_1}(t)$ of $Z_1^2$ at the point $t=0.14$

First I tried to find the new MGF for $Z^2_n$

$$M_{Z^2_1}(t)=E(e^{X^2t})$$

$$=\left(\frac1{\sqrt{2\pi}}\right)\int_{-\infty}^{\infty}e^{\left(\frac12\right)x^2(1-2t)} \, dx$$

then I have no idea how to integrate it....I know the solved integral and answer but don't know how to solve it

Answer 1

You have $$ \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{x^2 t} e^{-x^2/2} \, dx. \tag 1 $$ The exponent is $$ x^2 t - \frac{x^2}{2} = \frac{-x^2}{2} \left(1 - 2t\right) = -\frac{w^2}{2} $$ with $$ \begin{align} w & = x\sqrt{1-2t} \\[6pt] dw & = \sqrt{1-2t}\ dx \\[6pt] \frac{dw}{\sqrt{1-2t}} & = dx \end{align} $$ As $x$ goes from $-\infty$ to $+\infty$, then so does $w$, since $\sqrt{1-2t} > 0$. So $(1)$ becomes $$ \frac{1}{\sqrt{2\pi}}\cdot\frac{1}{\sqrt{1-2t}} \int_{-\infty}^\infty e^{-w^2/2} \, dw = \frac{1}{\sqrt{1-2t}}. $$

Answer 2

The MGF is given by $$\begin{align*} M_{Z_1^2}(t) &= {\rm E}[e^{Z^2 t}] = \int_{z=-\infty}^\infty e^{z^2 t} \frac{1}{\sqrt{2\pi}} e^{-z^2/2} \, dz \\ &= \frac{1}{\sqrt{2\pi}} \int_{z=-\infty}^\infty e^{-z^2(1/2 - t)} \, dz \\ &= \sigma \frac{1}{\sqrt{2\pi} \sigma} \int_{z=-\infty}^\infty e^{-z^2/(2\sigma^2)} \, dz \\ &= \sigma, \end{align*}$$ where $\sigma$ satisfies the equation $$2\sigma^2 = \frac{1}{\frac{1}{2}-t},$$ or $\sigma = (1-2t)^{-1/2}$. Notice that we used the fact that the integral of a normal density with mean $\mu = 0$ and standard deviation $\sigma > 0$ over its support is simply $1$, so the MGF is valid only when $t < 1/2$. Since $0.14 < 0.5$, we can evaluate the MGF at this point accordingly.

To find the MGF of the sum of the squares of all 14 standard normal variables, we just look at the product of the individual MGFs for a single variable.