let $Y$ and $Z$ be two normal random independent variables. prove that: $Z$ and $\rho \frac{\sigma_X }{\sigma_Y } \cdot Y $ are also independent.

Question

let $ (X,Y) $ be a random Gaussian vector where: \begin{align*} X \sim \mathcal{N}(0,\sigma_X ^2 ) \end{align*} \begin{align*} Y \sim \mathcal{N}(0,\sigma_Y ^2 ) \end{align*}

and lets denote: $ Z = X - \rho \frac{\sigma_X }{\sigma_Y } Y $ where $ \rho $, the co-variance coefficient, is defined as: \begin{align*} \rho = \frac{E(XY)}{\sigma_X\sigma_Y} \end{align*}

I had been asked to prove that $Y$ and $Z$ are independent random variables. I proved it by finding matrix $A$ that satisfied the linear transofrmation:

$ \begin{pmatrix}Y\\ Z \\ \end{pmatrix} = \underbrace{\begin{pmatrix}0&1 \\ 1&- \rho \frac{\sigma_X }{\sigma_Y } \\ \end{pmatrix} }_{A} \begin{pmatrix}X\\ Y \\ \end{pmatrix} $

so that means $(Y,Z)$ is also a Gaussian vector.

by showing that $Cov(Y,Z)=0$ using: $ Z = X - \rho \frac{\sigma_X }{\sigma_Y } Y $ and $\rho = \frac{E(XY)}{\sigma_X\sigma_Y}$, I was able to conclude that $Y$ and $Z$ are statistically independent variables.

Now, for the rest of the exercise that I'm doing, I must have also concluded from what I've done here that $Z$ is also independent of $ \hat X \overset{\Delta}{=}\rho \frac{\sigma_X }{\sigma_Y } \cdot Y $

this might be an easy question but I can't quiet grasp why this is really true, although intuitively it sounds true, but I couldn't prove this formally. all I could think of is theorems and rules that assume independence at the first place. If I want to do prove this, it must be by definition somehow, but is this the only way? and if yes, how can I start doing this?

UPDATE:

actually by thinking about this, I think I can do the same as I did before, saying: $ \begin{pmatrix}Z\\ \hat X \\ \end{pmatrix} = \underbrace{\begin{pmatrix}1&0 \\ 0& \rho \frac{\sigma_X }{\sigma_Y } \\ \end{pmatrix} }_{A} \begin{pmatrix}Z\\ Y \\ \end{pmatrix} $
and showing that the $Cov(Z, \hat X) = 0$.

is this an accepted proof?

Answer 1

If you are aware of the following fact, it is easy:

If $X,Z$ are independent and $f:\mathbb{R}\to\mathbb{R}$ is a measurable function, then also $f(X),Z$ are independent.

(This follows from $\sigma(f(X),Z)\subset\sigma(X,Z)$ and the definition of independence for random variables via the generated sigma algebras.)

In your case $f:\mathbb{R}\to\mathbb{R}, x\mapsto \rho\frac{\sigma_X}{\sigma_Y}x$ is a linear function and hence measurable.

EDIT: Since you commented that you are not familiar with measure-theoretic probability theory, here is another way not using any measure theory. Let X,Z be continuous independent random variables. And let $a\neq0$ be some constant real number. We want to show the $aX,Z$ is independent. The joint cdf is for all $(x,z)\in\mathbb{R}^2$ $$F_{aX,Z}(x,z)=F_{X,Z}(x/a,z)= F_{X}(x/a)F_{Z}(z)=F_{aX}(x)F_{Z}(z),$$ where the second inequality follows from the independence of $X,Z$. Hence, $X,aZ$ is independent.

In your example $a=\rho\frac{\sigma_X}{\sigma_Y}$.

A third way is to again argue that $(\hat{X},Z)$ is multivariate normal (same idea as what you have done for $(Y,Z)$) and show that the covariance is zero.