Mean and variance of two normal distributions.

Question

Assume X and Y are both normally distributed random variables.

• Assume X is distributed with means of Mx and variance of $\Sigma_x$
• Assume Y is distributed with means of My and variance of $\Sigma_y$

Let there by Z a random variable such as: Z = X - Y Does that make Z a random variable with the means of Mx - My and the variance of $\Sigma_x - \Sigma_Y$?

If no what are Z's parameters?

Thanks

Answer 1

You really should look at $-Y$ as Normal with mean $-M_y$, variance $\Sigma_y$ and use the formula for $Var (X+Y)$.

Alternatively, for any random variables

$E(X-Y) = EX - EY$ and $Var(X-Y) = Var (X) + Var (Y) - 2 Cov (X,Y)$.

Your expression for Variance is almost always wrong. The only obvious casee I can see it is correct is when $Y$ is a constant or $Y=-X$.

What is interesting here is whether $Z$ is normally distributed. The answer is

Yes if (X,Y) is jointly normal

Not necessarily if (X,Y) are only known to be marginally normal

Consider this construction:

take $X$ to be $N(0,1)$ distribution. Take $Y=X$ if $|X|\leq c$, $Y=-X$, if $|X|>c$, where $c>0$.

Convince yourself $Y$ is also distributed as $N(0,1)$!

here $X-Y=0$ if $|X|\leq c$, else it equals to $2X$, which is never not normal.

Answer 2

var(X+Y) = var(X) +var(Y) + 2*cov(X,Y)
so we have var(X-Y) = var(X) + var(Y) - 2*cov(X,Y)
If the variables are independent, then cov(X,Y) = 0, see Andre's comment.