Jtdylktuy

Firstly let's explain some signs used later:

1. 
   $y_1, ldots, y_2$ are random variables,


2. 
   $mathbb{E}(y_i) = mu_i$,


3. 
   $text{Cov}(y_i, y_j) = sigma_{ij}$,


4. 
   $Y = (y_1 ldots y_n)^{T}$,


5. 
   $mathbb{E}(Y) = mu = (mu_1 ldots mu_n)^{T}$,


6. 
   $text{Cov}(Y) = mathbb{E}[(Y- mu)(Y - mu)^{T}]$.

Now we can define a random vector $Y$ with normal distribution - $N(mu, Sigma)$.

Let's consider a random variable:
$$Z = (Y-mu)^{T} Sigma^{-1}(Y-mu).$$
What is the distribution of $Z$? How can it be found?

I know that one method would be to find expected value and covariance but I don't know how. Are there any other possibilities?

You have that $X$ is distributed $N(mu,Sigma)$ so:

$(X-mu)sim N(0,Sigma)$
and

$Sigma^{-1/2}(X-mu)sim N(0,Sigma^{-1/2}cdotSigmacdot{Sigma^{-1/2}}^{'})$

$Sigma^{-1/2}(X-mu)sim N(0,Sigma^{-1/2}cdotSigma^{-1/2}({Sigma^{1/2}}^{'}cdotSigma^{1/2})cdot{Sigma^{-1/2}}^{'})$ Since $Sigma^{-1/2}$ is symmetric by construction (use an orthogonal eigendecomposition) this leads to:

$Sigma^{-1/2}(X-mu)sim N(0,I)$ i.e. a standard normal.

Calculating the square of this new random variable gives us:
$(Sigma^{-1/2}(X-mu))^{'}cdot Sigma^{-1/2}(X-mu)=(X-mu)^{'}Sigma^{-1}(X-mu)$ Which is precisely $Z$

Since it's the square of a Standard Normal, the distribution will be Chi-Squared with k (dimension of the vector $y$) df