# Vector of independent normal RV's - mgfs

Printable View

• Aug 29th 2009, 06:54 AM
RoyalFlush
Vector of independent normal RV's - mgfs
Let $X = (X_1,...,X_n)^T$ be a column vector of independent N(0,1) distributed random variables. Let A be an $n \times n$ orthogonal matrix.

Prove that $||X||^2 = X_1^2 + ... + X_n^2 \sim \mathcal{X}_n^2$ using moment generating functions.
• Aug 29th 2009, 09:00 AM
matheagle
And where does this A come in?
Do you want AX?

$X_1^2 + ... + X_n^2 \sim \chi_n^2$ via moment generating functions.

Each $X_i^2 \sim \chi_1^2$ and the sum is $\chi_n^2$
• Aug 29th 2009, 10:48 PM
RoyalFlush
Quote:

Originally Posted by matheagle
It's been proved many times on this site that

$X_1^2 + ... + X_n^2 \sim \chi_n^2$ via moment generating functions.

Each $X_i^2 \sim \chi_1^2$ and the sum is $\chi_n^2$

I cannot seem to find such a post. Could you please recommend one?
• Aug 29th 2009, 11:01 PM
matheagle
Let $X\sim\Gamma(\alpha_1,\beta)$ and $Y\sim\Gamma(\alpha_2,\beta)$

If they are independent, then the MGF of the sum is the product of the individual MGFs.

So $M_{X+Y}(t)=M_X(t)M_Y(t)= (1-\beta t)^{-\alpha_1}(1-\beta t)^{-\alpha_2}=(1-\beta t)^{-(\alpha_1+\alpha_2)}$

Thus $X+Y\sim\Gamma(\alpha_1+\alpha_2,\beta)$

NOW, let $\beta=2$ which shows that
the sum of any two independent $\chi^2$s
is a $\chi^2$, you just add the degrees of freedom.
This can easily be extended to n random variables.