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Given a random variable $Z\sim\mathcal{N}(\textbf{0}_2, \textbf{I}_2)$ and its pdf $f$, what is the density of $f(Z)$?

When I draw samples and plot the transformation it looks like a uniform distribution.

Usually, I would go about it calculating the Jacobian of the transformation but the transformation is not invertible.

Any help would be much appreciated.

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The easiest way to figure out the distribution of $f(Z)$ for an arbitrary $f$ is to determine the cumulative density function of $f(Z)$.

To determine the cumulative density function of $f(Z)$, we compute $\Pr[f(Z) < t]$ for all $t$ in the range of $f$. In this case, actually, we'd rather compute $\Pr[f(Z) > t]$, but that's not a significant difference.

When $f$ is the density function of $Z$, $f(Z) > t$ occurs if and only if $\|Z\| < r_t$, where $r_t$ is the radius at which $f(Z)$ is equal to $t$. We can find $\Pr[\|Z\| < r_t]$ by a polar integral: $$\Pr[\|Z\| < r_t] = \int_0^{2\pi} \int_0^{r_t} f(r,\theta) r\,dr\,d\theta$$ where $f(r,\theta) = \frac1{2\pi} e^{-r^2/2}$ is the density function of $Z$ in polar coordinates. This simplifies to $\Pr[\|Z\| < r_t] = 1 - e^{-r_t^2/2} = 1 - 2\pi t$. So $\Pr[f(Z) > t] = 1 - 2\pi t$, and the cumulative density function of $f(Z)$ is $\Pr[f(Z) < t] = 2\pi t$ for $0 \le t \le \frac1{2\pi}$, which is precisely the cumulative density function of the uniform distribution on $[0, \frac1{2\pi}]$.

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  • $\begingroup$ I think instead of $\Pr[\|Z\| < r_t] = ... = 1 - t \sqrt{2\pi}$ we should have $\Pr[\|Z\| < r_t] = ... = 1 - 2 \pi t$ without the square root. $\endgroup$ – BiBi May 15 '17 at 22:08
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    $\begingroup$ Whoops! Fixed that. $\endgroup$ – Misha Lavrov May 15 '17 at 23:00

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