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Are there two distributions $X$ and $Y$ over $\mathbb{R}$ such that the distribution of the product $XY$ follows a Gaussian distribution?

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    $\begingroup$ Have $X$ having a half-normal distribution (or indeed a normal distribution) with mode $0$ and independently $Y=\pm1$ $\endgroup$
    – Henry
    Jul 12 '19 at 10:05
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If $X$ and $Y$ are both standard normal then $X^2 + Y^2 \sim \chi^2_2 = \mathcal{E}(1/2)$ and the angle of $(X,Y)$ in the plane whose sinus is given by $\frac{Y}{\sqrt{X^2 + Y^2}}$ is $\mathcal{U}_{[-\pi,\pi]}$.

Thus let $\theta \sim \mathcal{U}_{[-\pi,\pi]}$ and $Z \sim \mathcal{E}(1/2)$, then

$$ Y = \sqrt{Z} \text{sin}(\theta) \sim \mathcal{N}(0,1) $$

where $\sqrt{Z}$ follows a Rayleigh distribution with scale 1 while $\text{sin}(\theta)$ follows the Arcsine distribution.

This method is known as the Box-Muller transform which enables one to generate standard normal variable from independent uniform variables.

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  • $\begingroup$ Thank you for the answer. I will check that Box-Muller transform. If you know other ways to generate Gaussian distributions as products of other (non-trivial) distribution, please, let some links here. I am accepting and upvoting. Thank you again. $\endgroup$ Jul 17 '19 at 6:20

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