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?

  • $\begingroup$ Check this answer: stats.stackexchange.com/questions/19948/… $\endgroup$
    – Tim
    Nov 28, 2014 at 8:42
  • $\begingroup$ @Tim That question and its answers are built upon "$X$ and $Y$ are jointly Gaussian", aren't they? $\endgroup$ Nov 28, 2014 at 8:57
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    $\begingroup$ 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. $\endgroup$
    – Glen_b
    Nov 28, 2014 at 10:58
  • $\begingroup$ 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. $\endgroup$
    – Glen_b
    Nov 28, 2014 at 11:08
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    $\begingroup$ Your 2. is a subset of 1. because if $X$ and $Y$ are marginally Gaussian and independent, then they are also jointly Gaussian. $\endgroup$ Nov 28, 2014 at 15:31

2 Answers 2


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.

Here is a nice explanation.

  • $\begingroup$ So if $X$ and $Y$ are orthogonal, this doesn't change the conclusion? Can you give an example illustrating this? $\endgroup$ Nov 28, 2014 at 8:44
  • $\begingroup$ @FarticlePilter based on what I remember orthogonal simply means that they are uncorrelated (which actually does not mean that they are independent). I am looking for some illustration to my example. $\endgroup$ Nov 28, 2014 at 8:50
  • $\begingroup$ No, orthogonality is not uncorrelatedness. They are orthogonal, if $E\{XY\}=0$. Only when $E\{X\}=0$ or $E\{Y\}=0$, orthogonality is uncorrelatedness. $\endgroup$ Nov 28, 2014 at 8:52
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    $\begingroup$ Isn't "independent or jointly normal" redundant here? If X and Y are independent and normally distributed then they're jointly normal with a diagonal covariance matrix, right? $\endgroup$ Nov 28, 2014 at 19:40
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    $\begingroup$ The site is no longer there. Does anyone have a copy of the intended PDF? $\endgroup$
    – adunaic
    Feb 4, 2020 at 16:26

You are saying in your second point “uncorrelated (hence independent)”. It’s not quite the case. You can have two Gaussian distributions both uncorrelated and dependent. Uncorrelated implies independence only if they are jointly Gaussian.


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