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Let's say I know $X$ is a Gaussian Variable.
Moreover, I know $Y$ is a Gaussian Variable and $Y=X+Z$.

Let's $X$ and $Z$ are Independent. How can I prove $Y$ is a Gaussian Random Variable if and only if $Z$ is a Gaussian R.V.?

It's easy to show the other way around ($X$, $Z$ Orthogonal and Normal hence create a Gaussian Vector hence any Linear Combination of the two is a Gaussian Variable).


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The equation is essentially symmetric. I don't see what the issue is, you seem to already have it reasoned out. – Noldorin Jul 20 '10 at 19:57
Use whatever method you used to prove sum of two Gaussian is a Gaussian to prove Z = Y-X. – Chao Xu Jul 20 '10 at 21:21
I made a mistake and said they are all Independent. I only know X and Z are independent. Hence it's not a Symmetric case. Sorry and thank you. – Drazick Jul 21 '10 at 5:27
@Drazick: You need to clarify further. You said you know 'Y' is a Gaussian variable, then you asked us to prove a fact about 'Y' being a Gaussian variable – Casebash Jul 22 '10 at 13:37
@This question isn't easy. If X and Z are independent, then it X+Z and X will only be independent in very unusual and specific circumstances and so we won't be able to use the rule about linear combinations of Gaussians – Casebash Jul 22 '10 at 21:41
up vote 3 down vote accepted

Your question: given that X and Z are independent, X is Gaussian (I'll use "normal"), and Y = X+Z, prove that Y is normal iff Z is normal. Right? As you observed, one direction is easy: if Z is normal, then so is Y=X+Z. So for the other direction, assume that Y is normal. We need to prove that is Z normal too.

Perhaps there's an even easier way, but it's straightforward to use characteristic functions, which completely characterise distributions. Because X and Z are independent,

$ \varphi_Y(t) = E[e^{itY}] = E[e^{it(X+Z)}] = E[e^{itX}]E[e^{itZ}]$, and so,
$ \varphi_Z(t) = E[e^{itZ}] = E[e^{itY}]/E[e^{itX}] $

This means that Z has exactly the right characteristic function for a normal variable, and hence it's normal.

More interestingly and much more generally, there is a theorem of Cramer (e.g. see here) which says that if X and Z are independent and X+Z is normally distributed, then both X and Z are!

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Cramer is just what I was looking for! Thank You! Is there a place with different proofs of it? – Drazick Sep 20 '10 at 9:39
Note that you don't need Cramer's theorem for the question you asked: what I gave in my answer is a proof. For the more general case answered by Cramer's theorem, I don't know of any good reference online, but it's usually proved using characteristic functions, as I used above. (I haven't seen an actual proof myself; it may be hard.) – ShreevatsaR Sep 20 '10 at 10:04

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