Suppose X ~ N(0,1), i.e. X is normally distributed with mean 0 and variance 1.

Let Y = X^2

Find the MGF of Y DIRECTLY...

Can't seem to get a good answer out. Any help would be appreciated.

Also...

If f(z) = e^(-z/2)/sqrt(2*pi*z) for z > 0
and f(z) = 0 for z <= 0

Find the MGF of Z.

THANKS ALOT guys!

2. Originally Posted by chakim
Suppose X ~ N(0,1), i.e. X is normally distributed with mean 0 and variance 1.

Let Y = X^2

Find the MGF of Y DIRECTLY...

Can't seem to get a good answer out. Any help would be appreciated.

Also...

If f(z) = e^(-z/2)/sqrt(2*pi*z) for z > 0
and f(z) = 0 for z <= 0

Find the MGF of Z.

THANKS ALOT guys!
$m(t) = \frac{1}{\sqrt{2 \pi}} \int e^{tx^2} e^{-x^2/2} \, dx = \frac{1}{\sqrt{2 \pi}} \int e^{-\left( \frac{1 - 2t}{2}\right) x^2} \, dx$.

Now make the substitution $u = x \sqrt{\frac{1 - 2t}{2}}$ and use a well known result.

It's simple to confirm the answer, see here: Chi-Squared Distribution -- from Wolfram MathWorld

I ended up getting to this stage

But wasn't sure how to transform that into a recognizable chi-squared distn.

Any help would be appreciated cheers.

4. Also, are the limits negative infinity to infinity, or from 0 to infinity considering that Y = X^2 accounts for only positive numbers?

Cheers

5. Originally Posted by chakim
Also, are the limits negative infinity to infinity, or from 0 to infinity considering that Y = X^2 accounts for only positive numbers?

Cheers
Since $m(t) = E\left(e^{X^2t}\right)$, what do you think ....?

Originally Posted by chakim

I ended up getting to this stage

But wasn't sure how to transform that into a recognizable chi-squared distn.

Any help would be appreciated cheers.
The integral limits are not correct.

The function of t can be taken out of the integration. You will then be left with a well known integral - I suggest you research it ....

6. Well firstly, considering

$m(t) = E\left(e^{X^2t}\right)$

Can't x take on any value between negative infinity and infinity? Since X ~ N(0,1)?

So the new limits from the substitution would be negative infinity to positive infinity?

And also, when I take the function of t out, i get...

I've researched a bit, and i've tried to integrate the function directly (cause its only an exponential) but the limits give me weird answers...

I'm really stuck mr fantastic...

7. Originally Posted by chakim
Well firstly, considering

$m(t) = E\left(e^{X^2t}\right)$

Can't x take on any value between negative infinity and infinity? Since X ~ N(0,1)?

So the new limits from the substitution would be negative infinity to positive infinity?

And also, when I take the function of t out, i get...

I've researched a bit, and i've tried to integrate the function directly (cause its only an exponential) but the limits give me weird answers...

I'm really stuck mr fantastic...
Your integral limits are correct. As far as the final integral, read this for example: Gaussian Integral -- from Wolfram MathWorld

(You really should know this at this level).

8. I'm in 1st year university, studying an actuarial science degree.

I haven't seen this "Gaussian Integral" before mr fantastic.

So I couldn't really understand where you were coming from...

I thought I could just integrate the integral directly considering it wasn't a tricky integral, but the limits don't give a suitable answer.

Is there any other method to integrate, say by recognition of a statistical distribution?

P.S you've been really helpful, and I appreciate it very much.

Thank you

9. Originally Posted by chakim
I'm in 1st year university, studying an actuarial science degree.

I haven't seen this "Gaussian Integral" before mr fantastic.

So I couldn't really understand where you were coming from...

I thought I could just integrate the integral directly considering it wasn't a tricky integral, but the limits don't give a suitable answer.

Is there any other method to integrate, say by recognition of a statistical distribution?

P.S you've been really helpful, and I appreciate it very much.

Thank you
Try comparing the integral with the integral of the pdf for a standard normal distribution and you will get the same well known result.

10. ahhh, cheers!