It's about the demostration of the quadratic forms and chi-squared distribution.
Let's split the problem:
- We have a $n$ vector with n standardized normal distribution called $z={[z_1,z_2...z_n]}$. Obviously $z'z$ is a $\chi^2_n$
- If we have a symmetric, idempotent matrix (i.e. called $A$) then $z'Az \thicksim \chi^2_p $ where $p$ is the rank of $A$. I got this splitting the $A$ matrix with the spectral theorem into: $Q'\Lambda Q$ hence: $z'Az = z'Q'\Lambda Qz$ we call $w= Qz$ hence $z'Az =w'\Lambda w$. Because it's a idempotent matrix we have only 1s or 0s as eigenvalues and the number of 1's is exactly the rank of $A$. Then we have $w'w$ with with different length now as the number of the rank of $A$. We backward the process and we get: $z'Q'Qz$ but $Q'Q= I_{rank(A)}$ (they are orthogonal since we consider the eigenvectors) so we have the summation of $p$ $Z^2$.
- Now the question comes: if we have a symmetric, positive definite matrix $\Sigma_{nxn}$ my teacher told me that $z'\Sigma z$ is a $\chi^2_n $ distribution. I was trying to find a proof, but I can't find one.
My efforts are: use the spectral theorem again, hence we have: $z'Q' \Lambda Qz$ but now $\Lambda$ is a diagonal matrix with real positive numbers (cause it's positive definite).
$\Lambda$= \begin{bmatrix} \lambda_{1} & 0 & 0 & \dots & 0 \\ 0 & \lambda_{2} & 0 & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & \lambda_{n} \end{bmatrix}
We have called, as before, $w=Qz$ hence we have: $w'\Lambda w$. Let's suppose, that w is 3-dimensional vector with 3 elements called a,b and c. So $w=$ \begin{bmatrix} a \\ b \\ c \\ \end{bmatrix}
so $w'\Lambda w= a^2 \lambda_1+b^2 \lambda_2+c^2 \lambda_3$ And here I'm stuck. Maybe this last point is wrong. Therefore I'm asking for clarification and a clear proof of what my teacher said.
[EDIT]
I found this:
Let $X$ be a $ K \times 1$ standard multivariate normal random vector, i.e., $X \thicksim N(0,I)$. Let $A$ be an orthogonal $K \times K$ real matrix. Define $Y=AX$ Then also $Y$ has a standard multivariate normal distribution, i.e., $Y \thicksim N(0,I)$. I'm okay with this, the proof is here: https://www.statlect.com/probability-distributions/normal-distribution-quadratic-forms So my problem would come something like this: $w'\Lambda w$ where $w=QZ$ and by this theorem $w \thicksim N(0,I)$ since Q is an orthogonal matrix. So, if w is a 3 dimensional vector I'll have something like this:
$\lambda_1 \chi^2_1 + \lambda_2 \chi^2_1 + \lambda_3 \chi^2_1 $ A lineare combination of chi-squared distribution with weights the value of the eigenvalues. Am I right? If I am, does this sum has a particular distribution?