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I am trying to derive a marginal probability distribution for $y$, and failed, having tried all methods to solve the following integral:

$$ \operatorname{p}\left(y\right) = \int_0^{1/\sqrt{\,2\pi\,}}\!\!\!\! \frac{\sqrt{2/\pi}\,\,\mathrm{e}^{-y/\left(2z\right)}}{\sqrt{y\, z}\,\, \sqrt{-\log \left(2\pi\,z^2\right)}} \,\mathrm{d}z\quad \mbox{with}\quad y>0. $$

It is easy to verify that

$$\int_0^\infty \int_0^{\frac{1}{\sqrt{2 \pi }}} \frac{\sqrt{\frac{2}{\pi }} e^{-\frac{y}{2 \,z}}}{\sqrt{y\, z} \sqrt{-\log \left(2 \pi \, z^2\right)}} \, \mathrm{d}z \,\mathrm{d}y=1$$

After some work, figured out that, remarkably, we can get the fist moment, $\mathbb{E}(y)=\frac{1}{2 \sqrt{\pi }}$ and $\mathbb{E}(y^2)=\frac{\sqrt{3}}{2 \pi }$ without the density.

Felix Marin
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Nero
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  • Sorry it was a typo in the posting that I fixed (an extra z term). I still can't find solution. – Nero Apr 03 '14 at 17:24
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    Looks as if conditioned on the value of $Z$, $Y$ is a Gamma random variable with parameters $\left(\frac{1}{2}, \frac{1}{2\sqrt{Z}}\right)$ and $Z$ had marginal pdf of the form $\frac{1}{\sqrt{-\log(2\pi z^2)}}$ on $(0,1/\sqrt{2\pi})$. – Dilip Sarwate Apr 03 '14 at 18:59
  • Indeed I am working on the product of 1) Chi-square distribution and 2) distribution of the density of a standardized Gaussian. – Nero Apr 03 '14 at 19:02
  • Bounty: Intuition for Conditional Expectation (1. by linking bounty question here, this question gets attention because the bounty question is linked to this question. 2. by linking bounty question here, nero gets to see bounty question and so might answer bounty question) – BCLC Sep 02 '18 at 17:46
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    I was able to simplify the integral to $$4(2\pi)^{-1/4}\frac{1}{\sqrt{z}} \int_0^\infty \exp\left(-ze^{2t^2}\right) e^{-t^2},dt,$$ where $z=\sqrt{\frac{\pi}{2}}y$. – Diffusion Nov 30 '20 at 17:14
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    @Zachary this kind of integrals is notoriously resistant to be evaluated in closed form. I'm pretty sure there isn't much to be done here :() – asgeige Nov 30 '20 at 17:36
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    Yes, that's true. I'll keep trying a few methods to evaluate it, but if they don't work, this is still a rather nice form I would argue. – Diffusion Nov 30 '20 at 23:05
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    I find $$\mathbb{E}(y^n) = \frac{2^{n/2}\Gamma\Big(n+\frac12\Big)}{\pi^{(n+1)/2}\sqrt{n+1}}.$$ – user26872 Dec 04 '20 at 23:15
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    Wolfram is also unable to solve it. – Dabed Dec 07 '20 at 19:42

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