Question

Let Z be a standard Normal distribution, w = z^2, find the moment generating function of W

Answer 1

please help and explain

Answer 2

?

Answer 3

how do i apply this i am new to this?

Answer 4

We'll probably want the continuous integral rather than a discrete sum...

Answer 5

$$
\Large{
M_x(t) = E(e^{tx})= \int e^{tx} f(x)
}
$$

Answer 6

what is that \[M _{x} (t) = \int\limits_{-\infty}^{\infty} e ^{tx} f(x) dx\]

Answer 7

ok

Answer 8

now you can plug in
\( \Large f(x) = \frac{1}{\sigma\sqrt{2\pi}}\, e^{-\frac{(x - \mu)^2}{2 \sigma^2}} \)

Answer 9

lets use the variable z

Answer 10

okay I'm following so far

Answer 11

$$
\large
{ \text{Let Z be a random variable with mean = 0 , st. dev = 1 }
\\
\Large M_z(t) = E(e^{tZ^2})= \frac{1}{\sqrt{2\pi}}\int_{-\infty }^{ \infty } e^{tz^2}e^{-\frac12 z^2} ~dz
}

$$

Answer 12

okay so we plugged in for x with the z variable

Answer 13

right

Answer 14

and plug in mu = 0, sigma = 1 into f(x) above , to get the simplified expression

Answer 15

ok

Answer 16

can some one help me finish it after i plug in the mean and std. deviation is that it?

Answer 17

?

Answer 18

Recall the distribution function for the normally distributed random variable \(X\) with mean \(\mu\) and variance \(\sigma^2\):
\[f_X(x)=\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)\]
where \(\exp(\cdots)=e^{\cdots}\), in case you're unfamiliar with the notation. For a standard normal distribution, we set \(\mu=0\) and \(\sigma^2=1\), call it \(Z\), and so
\[f_Z(z)=\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{z^2}{2}\right)\]
By the definition of the moment generating function, you have
\[\begin{align*}
\mathrm{M}_W(t)&=\mathbb{E}\left(\exp(tZ^2)\right)\\\\
&=\int_{-\infty}^\infty \exp(tz^2)\left(\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{z^2}{2}\right)\right)\,dz\\\\
&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \exp\left(\left(t-\frac{1}{2}\right)z^2\right)\,dz\\\\
&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \exp\left(-\frac{1}{2}\left(1-2t\right)z^2\right)\,dz\\\\
&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \exp\left(-\frac{z^2}{2\left((1-2t)^{-1/2}\right)^2}\right)\,dz
\end{align*}\]
Nearly done. Let's substitution \(\xi=\dfrac{z}{(1-2t)^{-1/2}}\), then \(d\xi=\dfrac{dz}{(1-2t)^{-1/2}}\), and so the change of variables gives us the equivalent integral,
\[\frac{\color{red}{(1-2t)^{-1/2}}}{\sqrt{2\pi}}\int_{-\infty}^\infty \exp\left(-\frac{\xi^2}{2}\right)\,d\xi\]
What can you say about the integral above, not colored red?

Answer 19

They cancel out ? @sithsandgiggles

Answer 20

Let's be more careful with what you're trying to say... What exactly "cancels"? And what do you mean by "cancel"?

Answer 21

they have similar terms? i guess thats what i noticed

Answer 22

Here's a hint: we know that
\[f_Z(z)=\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{z^2}{2}\right)\]
is the standard normal distribution's density function defined over \(-\infty<z<\infty\). Replacing \(z\) with \(\xi\) doesn't change that fact.

The definite integral over \((-\infty,\infty)\) is the total area under the curve (the density function above). What's the total area (i.e. probability)?

Answer 23

is it 1?

Answer 24

Right! This means
\[\frac{\color{red}{(1-2t)^{-1/2}}}{\sqrt{2\pi}}\int_{-\infty}^\infty \exp\left(-\frac{\xi^2}{2}\right)\,d\xi=\color{red}{(1-2t)^{-1/2}}=\frac{1}{\sqrt{1-2t}}\]
and this is the mgf for \(W=Z^2\).

Answer 25

wow awesome ! great explanation

Answer 26

you should also list the values of t for which this result holds.