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there is the exercise 5.30 from the book A First Course in Probability and Statistics from Rao.

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What I have read is that a joint normal PDF is more an exception than a rule here:

Is it possible to have a pair of Gaussian random variables for which the joint distribution is not Gaussian?

But I can't rule out to see how we can prove this. Does someone have potential solutions for this one?

kjetil b halvorsen
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Gabriel Amiot
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    Hint: you don't need to calculate anything. $f$ is a normal density to which a perturbation has been added; the perturbation is proportional to $xy\exp(-(x^2+y^2)/2) \exp(x^2+y^2-2)$ and that is proportional to $xy\exp((x^2+y^2)/2).$ This is manifestly an odd function of $x$ and $y$ (separately). Thus, all you have to show is that $f$ is a density (it's never negative and has finite integral--or so we hope!) because when you integrate out one variable to find the marginal of the other, that integral will necessarily be zero. – whuber Apr 22 '23 at 17:21

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The density itself is clearly not a Gaussian one. On the other hand, marginal density $f(x) = \int f(x,y) dy = \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^{2}} + \frac{x}{2\pi}\int y e^{-\frac{1}{2}(x^{2} + y^{2})} e^{x^{2} + y^{2}-2}dy = \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^{2}}$ is Gaussian (similarly, $f(y)$ is also Gaussian density), because the second integral is zero as integral of odd function.

Edit: function $f(x,y)$ in the previous post is not a probability density, since it can be negative (e.g. for $f(1,-5) < 0$). However, if we correct it to $f(x,y) := \frac{1}{2\pi} e^{-\frac{1}{2}(x^2 + y^2)}(1+xye^{-\frac{1}{2}(x^2 + y^2)})$, it is non-normal probability density, with with marginal normal distributions (by the above argument).

povidanius
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