Cumulative distribution of Z=X^Y

Question

Given are $X$ a uniform distribution on $(0,1)$; Y a Pareto distribution with $\alpha=1$, with probability density function $f_Y (y)=\frac{1}{y^2}$ for $y \geq 1$ and equal to $0$ if $y \lt 1$.

$X$ and $Y$ are independent random variables.

I've tried to use the information provided in this answer: CDF of $X+Y$,$X−Y$,$XY$ for $(X,Y)$ Chosen Uniformly Inside Triangle but I didn't know how to use it in this problem.

How to get the cumulative distribution of $Z=X^Y$?

$F_Z (z)=P(Z\leq z)=P(X^Y\leq z)$

Answer 1

First observe that the derivative function of $t \mapsto \frac{z^\frac1t}{\ln z}$ is $t\mapsto -\frac1{t^2} z^\frac1t$ After that we have for $z \in (0,1)$ $$F_Z(z) = \mathbb P\left(Z \le z\right) = \mathbb P\left(X^Y \le z\right) = \mathbb E\left[1_{X \le \left(z^{\frac{1}{Y}}\right)}\right] = \mathbb E \left[\mathbb E \left[1_{X \le \left(z^{\frac{1}{Y}}\right)}\Big | Y\right]\right] = \mathbb E \left[z^{\frac1Y}\right] = \int_1^\infty \frac{z^{\frac1y}}{y^2} \mathrm d y = \left[-\frac{z^\frac1y}{\ln z}\right]_1^\infty = -\frac{1-z}{\ln z}$$

Answer 2

Given any random variable, the value of the PDF at a point is simply the integral of the probability densities over all events that result in the random variable obtaining that value.

In this case

P(Z=z) = P($X^Y$=z) = $\int_{x^y=z}P(X=x)P(Y=y)\,dxdy$

Since y has to greater than 1 for its probability to be greater than zero, we have to have X < Z.

You can rewrite $x^y = z$ as $y = \frac{log(z)}{log(x)}$

If x is in [0,1], then P(X=x) = 1, but if X > Z, then we have zero probability density, so if we take the limits of 0 to z, then P(X=x) = 1. This just leaves P(Y=y) = $\frac{1}{y^2}$. Substituting in $y = \frac{log(z)}{log(x)}$ gives $[\frac{log(x)}{log(z)}]^2$

So P(Z=z)= $\int_{0}^z [\frac{log(x)}{log(z)}]^2\,dx$

The CDF will be $\int_0^{z_0}\int_{0}^z [\frac{log(x)}{log(z)}]^2\,dxdz$