Sum of Gaussian is Gaussian?

As a newbie in probability, I am recently cleaning my understandings about Gaussian distribution.

I know that

1. If $X$ and $Y$ are jointly Gaussian, then $aX+bY$ ($a$ and $b$ are both constant) is also Gaussian.
2. If $X$ and $Y$ are Gaussian and uncorrelated (hence independent), then $aX+bY$ ($a$ and $b$ are both constant) is also Gaussian.

My question

What about $X$ and $Y$ are only Gaussian themselves (NOT jointly Gaussian)? Without assuming uncorrelatedness (or independence), can we still say $aX+bY$ is Gaussian?

If not, does "$X$ and $Y$ are orthogonal" change our conclusion?

• Check this answer: stats.stackexchange.com/questions/19948/… – Tim Nov 28 '14 at 8:42
• @Tim That question and its answers are built upon "$X$ and $Y$ are jointly Gaussian", aren't they? – Sibbs Gambling Nov 28 '14 at 8:57
• What you know ain't so. Being uncorrelated isn't sufficient. For a counterexample see here Many other counterexamples can be found here on CV. – Glen_b Nov 28 '14 at 10:58
• Actually I think you should unfix it, since fixing it invalidates some of the responses you have. It's actually more useful to leave it that way. – Glen_b Nov 28 '14 at 11:08
• Your 2. is a subset of 1. because if $X$ and $Y$ are marginally Gaussian and independent, then they are also jointly Gaussian. – Dilip Sarwate Nov 28 '14 at 15:31

No and this is a common fallacy. People tend to forget that the sum of two Gaussian is a Gaussian only if $X$ and $Y$ are independent or jointly normal.
• So if $X$ and $Y$ are orthogonal, this doesn't change the conclusion? Can you give an example illustrating this? – Sibbs Gambling Nov 28 '14 at 8:44
• No, orthogonality is not uncorrelatedness. They are orthogonal, if $E\{XY\}=0$. Only when $E\{X\}=0$ or $E\{Y\}=0$, orthogonality is uncorrelatedness. – Sibbs Gambling Nov 28 '14 at 8:52