Sum and difference of three normally distributed variables

Question

We are given three independent random variables $X, Y, Z$ with normal distribution $\mathcal{N}(1,2)$. Are $U=Z-Y+X$ and $V=X+Y$ independent?

I thought I would compute the joint density $f_{UV}$ and densities $f_U$ and $f_V$.

It would be easier if I also had another random variable $W = \varphi_3(X, Y, Z)$. Then I could write $(U, V, W) = (\varphi_1(X, Y, Z), \varphi_2(X, Y, Z), \varphi_3(X, Y, Z)) = (Z-Y+X, X+Y, ?)$

But I don't have $W$. If I introduce $W = Z$, then I get joint density of $U, V, W$ equal $$\frac{1}{2} \exp \left({\frac{(v+u-w)^2-(v-u-w)^2-w^2}{8}}\right) = \frac{1}{2} e^{(-w^2-4wv+4uv)/8}$$

I've also computed density of the sum $X+Y$. It is $f_{X+Y} = \frac{1}{2 \sqrt{\pi}} e^{y^2/4} $.

Could you tell me if this could lead me anywhere, because I am lost.

How can I check if $U$ and $V$ are independent?

Answer 1

Jointly normal random variables are independent iff they are uncorrelated. For convenience let $W = X-Y = U-Z$.
$$\text{Cov}(Z,V) = \text{Cov}(Z,X) + \text{Cov}(Z,Y) = 0 $$ $$\text{Cov}(W,V) = \text{Cov}(X,X) - \text{Cov}(Y,Y) = 0 $$ so $$\text{Cov}(U,V) = \text{Cov}(Z,V) + \text{Cov}(W,V) = 0 $$

Answer 2

We need to show Cov$[Z-Y+X,X+Y]=0$.

$$\text{Cov}[Z-Y+X,X+Y]$$ $$=E[((Z-Y+X) - (\mu_Z-\mu_Y+\mu_X))((X+Y) - (\mu_X+\mu_Y))]$$ $$=E[ZX-YX+X^2+ZY-Y^2+XY-\mu_ZX+\mu_YX-\mu_XX-\mu_ZY\mu_YY-\mu_XY$$ $$-\mu_XZ+\mu_XY-\mu_XX-\mu_YZ+\mu_YY-\mu_YX$$ $$+\mu_X\mu_Z-\mu_Y\mu_X+\mu_X^2+\mu_Z\mu_Y-\mu_Y^2+\mu_Y\mu_X]$$ Now $E[A+B]=E[A]+E[B]$ and $E[c\cdot X]=c\cdot E[X]$ for $c$ a constant. Thus, after a lot of cancelling we get $$=E[X^2-Y^2-\mu_ZX+\mu_YY-\mu_XX+\mu_X\mu_Z]$$ $$=(E[X^2]-\mu_X^2)-(E[Y^2]-\mu_Y^2)$$ $$=0$$