Thread: Vector of independent normal RV's - mgfs

1. Vector of independent normal RV's - mgfs

Let $\displaystyle X = (X_1,...,X_n)^T$ be a column vector of independent N(0,1) distributed random variables. Let A be an $\displaystyle n \times n$ orthogonal matrix.

Prove that $\displaystyle ||X||^2 = X_1^2 + ... + X_n^2 \sim \mathcal{X}_n^2$ using moment generating functions.

2. And where does this A come in?
Do you want AX?

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

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

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

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

Each $\displaystyle X_i^2 \sim \chi_1^2$ and the sum is $\displaystyle \chi_n^2$
I cannot seem to find such a post. Could you please recommend one?

4. Let $\displaystyle X\sim\Gamma(\alpha_1,\beta)$ and $\displaystyle Y\sim\Gamma(\alpha_2,\beta)$

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

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

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

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