# Distribution for an operation of variables with identical distributions

I have this doubt:

Consider $X$~$N(\mu_X,\sigma^2)$, $Y$~$N(\mu_Y,\sigma^2)$ and $Z=X-Y$

I know that $E(Z)=E(X)-E(Y)=\mu_X-\mu_Y$ because the expected value is a linear operator.

And I know that $V(Z)=V(X)-V(Y)=0$

But which distribution does $Z$ have? Can I say that $Z$ has a normal distribution because $X$ and $Y$ have a normal distribution?

• I believe the question you are really asking concerns how to compute variances of differences of (independent) random variables. That has good answers at stats.stackexchange.com/questions/26886. – whuber Jun 23 '15 at 20:16
• That's true too. And I just assumed that $V(X-Y)=V(X)-V(Y)$ which isn't right. Thank you for your advice. – Élio Pereira Jun 23 '15 at 20:21

The variance of Z is equal to V(X) + V(Y) if X and Y are independent (You have to add the correlation term if X and Y are not independent) because V(-X) = V(X). Thus, Z has a normal distribution with $\mu_Z = \mu_X - \mu_Y$ and $\sigma_Z^2 = 2\sigma^2$.