# Work on transformation of variables

I have $X,Y \sim Uniform(0,1).$ I want to find the PDF for $X^Y$. I imagine that I should start with the CDF $$F_{X^Y}(x) = P(X^Y \leq x) = P(X^y \leq x | Y = y)P(Y = y),$$ But I seem to be having issues with this since $Y \sim Uniform(0,1),$ and thus $P(Y = y) = 0$ given the fact that $Y$ is realized from a continuous distribution. Any recommendation on how to alleviate these kinds of problems?

• Are $X$ and $Y$ independent?
– user9464
Commented Jun 8, 2016 at 0:28

Assuming that $X$ is independent of $Y$ we get the following by conditioning:
$$\mathbb{P}(X^Y \leq x) = \int_0^1 \mathbb{P}(X^Y\leq x|Y=y)f_Y(y)dy = \int_0^1\mathbb{P}(X^y \leq x)f_Y(y)dy = \int_0^1\mathbb{P}(X\leq x^{1/y})f_Y(y)dy =\int_0^1x^{1/y}dy.$$ Where $f_Y(y)$ is the pdf of $Y$.
I should start with the CDF $$F_{X^Y}(x) = P(X^Y \leq x) = P(X^y \leq x | Y = y)P(Y = y),$$ But I seem to be having issues with this since $Y \sim Uniform(0,1),$ and thus $P(Y = y) = 0$ given the fact that $Y$ is realized from a continuous distribution.
More than that, you don't have any particular value of $y$.   You have to measure over all the supported values (re: Law of Total Probability), and since $Y$ is a continuous random variable, that means you need to integrate a product of the conditional CDF of $X$ given $Y$ and the probability density function of $Y$.
\begin{align}F_{X^Y}(z) =&~ \int_\Bbb R \mathsf P(X^y\leq z\mid Y=y)~f_Y(y)\operatorname d y\quad\Big[z\in(0;1]\Big] & \textsf{Iverson Brakets}\\[1ex] =&~ \int_\Bbb R \mathsf P(-\sqrt[y]z\leq X\leq +\sqrt[y] z)~f_Y(y)\operatorname d y\quad\Big[z\in(0;1]\Big] \\[1ex] =&~ \int_0^1 (\sqrt[y] z)\cdot 1\operatorname d y\quad\Big[z\in(0;1]\Big]\\[1ex] =&~ \int_0^1 \mathsf e^{(\ln z)/y} \operatorname d y\quad\Big[z\in(0;1]\Big]\\[1ex] =&~ \big(z+\ln(z)\;\Gamma(0;-\ln(z))\big)\;\Big[z\in(0;1]\Big]&\textsf{Incomplete Gamma Function}\end{align}