# Joint pdf of standard normal and chi sq.

• Aug 8th 2009, 04:12 AM
Zenter
Joint pdf of standard normal and chi sq.
$X_{1},...,X_{N}$ is a random sample from a $N(\mu,\sigma^2)$ distribution, where $\mu$ and $\sigma$ are unknown.

I'm given:

$Z = \frac{\sqrt{n}(\overline{X} - \mu)}{\sigma}$

$U = \frac{(n - 1)S^2}{\sigma^2}$

where

$\overline{X} = \sum\frac{X_{i}}{n}$

$S^2 = \frac{\sum(X_{i} - \overline{X})^2}{n-1}$

Now, I think that Z is the standard normal dist, and U is dist. Chi Sq with n-1 degrees of freedom.

I'm now asked to find the joint pdf of Z and U; how would I do this?

I'm also given:

$T = \frac{\sqrt{n}(\overline{X} - \mu)}{S}$

and asked to deduce the joint pdf of T and U; again, I think T has the t dist. with n-1 degrees of freedom, but I'm not sure how to go about doing this question.
• Aug 8th 2009, 04:31 AM
Moo
Hello,

It may be easier to deal with standard normal distribution... : $Z_1,\dots,Z_n \sim \mathcal{N}(0,1)$

By knowing that if A follows a normal distribution (mu,sigmaČ), aA+b follows a normal distribution (a*mu+b , aČ sigmaČ), you have :

$Z=\sqrt{n}\cdot \bar Z$

and $U=\sum_{i=1}^n (Z_i-\bar Z)^2$

where $\bar Z=\frac 1n \sum_{i=1}^n Z_i$

And it can be proved that U and $\bar Z$ are independent. But the only proof I know uses Gaussian vectors...
• Aug 8th 2009, 06:18 AM
Zenter
I think I've found a proof for it, just trying to get my head round it :)

And... for the joint pdf of T and U, I've said:

$R = U$

$T = \sqrt{n-1}U^{-\frac{1}{2}}Z$

Therefore $Z = TR^{\frac{1}{2}}(n-1)^{-\frac{1}{2}}$

So I can find the joint pdf of T and R, using Jacobian transformation...

However, my Jacobian keeps coming up as 0.

I've got:

$\frac{dz}{dt} = r^{\frac{1}{2}}(n-1)^{-\frac{1}{2}}$

$\frac{dz}{dr} =\frac{1}{2}tr^{-\frac{1}{2}}(n-1)^{-\frac{1}{2}}$

$\frac{du}{dt} = -2z^2t^{-3}(n-1)$

$\frac{du}{dr} = 1$

And then I calculate J and I get 0 :S... so I'm assuming one or more of my differentials is wrong; are they? Or am I just going about this the wrong way?