# Math Help - Transforming the Normal distribution

1. ## Transforming the Normal distribution

$X$ is a standard normal random variable. Show that the PDF of $Z=(X+1)^2$ is

$(8\pi z)^{-\frac{1}{2}}e^{-\frac{1}{2}(z+1)}\left(e^{\sqrt{z}}+e^{-\sqrt{z}}\right)$

I've tried this loads of times but I can't get anywhere near the answer

2. I've already shown you the technique.
Why don't you post what you've done and we can find what's wrong.
There's a basic formula in most books for 2-1 transforms, so all you have to do is plug into that.
I'm sure you can find it online, but it's easy to do it from scratch.

3. I know you have, and I tried applying that (along with using a similar example we had in our notes) but didn't get anywhere near the required answer (that I'm aware of). I'll just type exactly what I've written.

$F_{Z}(z)=P(Z\leq z)$
$=P(X^2+1\leq z)$
$=P(X+1\leq\sqrt{z})$
$=P(X\leq\sqrt{z}-1)$
$=\Phi(\sqrt{z}-1)$

That's practically identical to what we wrote for the example in our notes. Then we wrote

$\frac{d}{dz}(\Phi(\sqrt{z}-1)$
$=\frac{1}{2}z^{-1/2}\Phi'(\sqrt{z}-1)$

Then I've got that $\Phi'(z)=\phi(z)=\frac{1}{\sqrt{2\pi}}e^{-z^2}$ so

$f_{Z}(z)=\frac{1}{2}z^{-1/2}\frac{1}{\sqrt{2\pi}}e^{-(\sqrt{z}-1)^2}$

And then this is where I start going off miles away from the answer.

4. no, look at your start

$x^2+1\ne (x+1)^2$

But that seems to be a typo

However the next line is wrong too

$x^2\le a$ means $-\sqrt a\le x\le \sqrt a$

AS I already said, its a 2 to 1 transformation.

$F_{Z}(z)=P(Z\leq z)$
$=P((X+1)^2\leq z)$
$=P(-\sqrt{z}\leq X+1\leq\sqrt{z})$

now continue

$=P(-\sqrt{z}-1\leq X\leq\sqrt{z}-1)$

$=F_X(\sqrt{z}-1)-F_X(-\sqrt{z}-1)$

NOW plug into the normal and differentiate

6. I know, $X^2+1$ isn't the same as $(X+1)^2$ it was indeed a typo was trying to be quick 'cause the homework was due in on the day I replied. It was an unassessed question so it didn't matter really, was just intrigued.