# pdf of f(Z) where Z is multivariate normal and f is its pdf

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.

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 $[0, \frac1{2\pi}]$.
• 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. – BiBi May 15 '17 at 22:08