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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?

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  • $\begingroup$ 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. $\endgroup$ – whuber Jun 23 '15 at 20:16
  • $\begingroup$ 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. $\endgroup$ – Élio Pereira Jun 23 '15 at 20:21
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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 $\sigma_Z^2 = 2\sigma^2$.

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You should unaccept bassir's answer because it only addresses the case in which X and Y are independent, and his statement that Z is Normal may not be true if X and Y are not independent.

If X and Y are independent, then EZ and Var(Z) are as stated by bassir.

If X and Y are jointly Normal (i.e., Bivariate Normal), but not necessarily independent, then Z is a Normal random variable, and EZ is the same as for independent case, and Var(Z) = 2*σ^2 -2*Cov(X,Y) = 2*σ^2 - 2*(correlation coefficient between X and Y) * σ^2, where Cov(X,Y) is the covariance between X and Y.

If X any Y are not independent, Z does not necessarily even have a Normal distribution. EZ and Var(Z) are per the immediately above case which assumes X and Y are jointly Normal, but in fact, Z may not be Normal. There are several such examples in http://www.amazon.com/Introduction-Probability-Theory-Applications-Vol/dp/0471257095 .

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  • $\begingroup$ +1 There are examples in answers here on CV also (up to a flip of sign on the second variate to convert a sum to a difference). e.g there's some here; and another here. Nice images here of bivariates with the property. $\endgroup$ – Glen_b Jun 24 '15 at 1:06

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