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?


  • 1
    $\begingroup$ If the random variables are independent, the variance of the difference is the sum of the variances. More generally if $X$ and $Y$ are any independent random variables with variances $\sigma^2$ and $\tau^2$, then $aX+bY$ has variance $a^2\sigma^2+b^2\tau^2$. (Your expression for the mean of the difference is right. And $X-Y$ has normal distribution if $X$ and $Y$ are independent normal.) $\endgroup$ Feb 3, 2014 at 19:03
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    $\begingroup$ One way to tell that the variance of the difference isn't equal to the difference of the variances is that the difference of two variances can be negative, while the variance of any random variable is always nonnegative. $\endgroup$ Feb 3, 2014 at 19:27

2 Answers 2


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.


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.


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