Vector of independent normal RV's - mgfs

**RoyalFlush** · August 29th 2009, 05:54 AM

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.

**matheagle** · August 29th 2009, 08:00 AM

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 $\chi_n^2$

**RoyalFlush** · August 29th 2009, 09:48 PM

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?

**matheagle** · August 29th 2009, 10:01 PM

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.

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