1
$\begingroup$

I know that the sum of two normally distributed random variables is also normal.

But what about the opposite? Can the sum (or substraction) of two non-normally distributed random variables be normal?

X-Y=Z Can Z be normal if X and Y are not? Any example?

I came to this question while thinking about the need of checking normallity of the differences of paired measures.

The question Is joint normality a necessary condition for the sum of normal random variables to be normal? is not the same than mine, it speaks about joint normallity.

And this one Sum of independent non-normal random variables says it's a duplicate but I can't find the original one.

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ Your question is unclear - in what way (if any) are the two random variables related? Can they be dependent? Or only independent? $\endgroup$
    – Glen_b
    Oct 4, 2016 at 9:20

1 Answer 1

Reset to default
5
$\begingroup$

As you've quoted, joint normality is a sufficient condition for $X+Y$ to be normal. The general case is not true. We can have $X+Y$ to be non-normal with $X$ and $Y$ normal. The simple example is to have $X=-Y$ where $X,Y$ will be normal and $X+Y$ isn't.

Similarly, to make a sum normal out of two non-normal, you just need $X = \frac12A+B$ and $Y = \frac12A-B$ where $A$ is normal and $B$ follows whatever strange distribution.

And the linked question answers the independent $X$ and $Y$ case. Not sure why you can't find the link, but it's here

Improve this answer
$\endgroup$
1
  • 4
    $\begingroup$ $X+Y$ indeed is Normal when $X=-Y$: it has a standard deviation of zero. It would be awful if we had to modify all the theorems that assert linear combinations of jointly Normal variates are Normal just to rule out this special case! $\endgroup$
    – whuber ♦
    Oct 4, 2016 at 18:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.