PDF of $(AB)^C$ where $A, B, C \sim U[0,1]$

Question

Let $A$, $B$, $C$ be i.i.d uniform random variables on $(0,1)$.

What is the distribution of $(AB)^C$?

What is a good way to go about answering a question like this?

Trying to find the CDF of this distribution and then differentiating to find the PDF doesn't seem to be a particularly nice way to go about this problem.

Instead I have been trying to use this change of variables method (example here). In particular:

Let $X = A$, $Y = B$, $Z = (AB)^C$.

(So the inverse transformation is $A = X$, $B = Y$, $C = \frac{\log(Z)}{\log(XY)}$.)

Then, the joint probability distribution of $X$, $Y$, $Z$ is given by:

\begin{align} f_{X,Y,Z}(x,y,z) & = f_{A,B,C}(a(x),b(y),z(x,y,z))\cdot|\frac{\partial (a,b,c)}{\partial (x,y,z)}| \\ & = |\frac{\partial (a,b,c)}{\partial (x,y,z)}| \\ & = \frac{-1}{z \cdot \log(xy)} \\ \end{align}

In order to find the distribution of $Z$, I would now need to integrate this w.r.t x and y over $[0,\frac{z}{y}]\times[0,1]$:

$$f_Z(z) = \int_0^1 \int_0^{\frac{z}{y}} \frac{-1}{z \cdot \log(xy)} \,dx \,dy \,\,\,\,\,\,\,\, (\text{for} \, z \in (0,1))$$

This is not easy (i.e. I don't know how to do this - maybe I am missing a trick as it is a (improper) definite integral?).

The issue seems to be: either I am going about this question in the wrong way, or, I am using the wrong change of variables. Indeed, since finding the distribution of Z using the change of variables method only involves integrating the Jacobian, using a different, more clever set of new variables may lead to a much nicer integral - I just have not been able to find such a set of variables.

Any help would be much appreciated.

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Edit: In light of the comment below that the distribution of $(AB)^C$ is just $U[0,1$, it seems that a good way to go forward is to consider the MGF. (This did not originally seem like a particularly good option as I had guessed that the distribution would be somewhat esoteric, rather than a ‘known’ one.)

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Note: Wolfram Alpha says that the integral of $\frac{-1}{\log(xy)}$ over $[0,1]^2$, for example is 1 - am I missing something obvious here?

Answer 1

Here is a brutal computation: let $w \in (0, 1)$ and set $W = (AB)^C$. Then

\begin{align*} \mathbb{P}(W \leq w) &= \int\limits_{\substack{(xy)^z \leq w \\ x,y,z\in[0,1]}} dxdydz = \int\limits_{\substack{z(u+v) \geq -\log w \\ u\geq 0, v\geq 0, z\in[0,1]}} e^{-(u+v)} \, dudvdz \end{align*}

where we utilized the substitution $(x,y) = (e^{-u},e^{-v})$. Further substitution with $(t, \lambda) = (u+v, \,\frac{u}{u+v})$ gives

\begin{align*} \mathbb{P}(W \leq w) &= \int\limits_{\substack{zt \geq -\log w \\ t\geq 0, z\in[0,1]}} t e^{-t} \, dtdz = \int_{0}^{\infty} t e^{-t} \max\left\{0, 1+\frac{\log w}{t} \right\} \, dt \\ &= \int_{0}^{\infty} e^{-t} \max\left\{0, t+\log w \right\} \, dt \\ &= \int_{\log w}^{\infty} e^{-s+\log w} \max\left\{0, s \right\} \, ds \qquad (s = t+\log w) \\ &= \int_{0}^{\infty} w e^{-s} \, ds = w. \end{align*}

Answer 2

Knowing a priori that the distribution of $(AB)^C$ is uniform on $[0,1]$ suggests that this problem can be attacked by utilizing moment generating functions:

Let $X = \log(Y), \, \text{ where } Y \sim U[0,1]$.

The MGF of $X$ is given by \begin{align} M_X(t) & = \mathbb{E}(e^{tX}) \\ & = \mathbb{E}(e^{t\log(Y)}) \\ & = \mathbb{E}(Y^t) \\ & = \frac{1}{t + 1} \\ \end{align}

Now consider $Z = \log((AB)^C) = C\log(AB)$

The MGF of $Z$ is given by \begin{align} M_Z(t) & = \mathbb{E}(e^{tZ}) \\ & = \mathbb{E}(e^{t\,C\log(AB)}) \\ & = \mathbb{E}(\, \mathbb{E}((AB)^{\,tC} \, | \, C \,)) \\ & = \mathbb{E}(\,( \mathbb{E}(A^{\,tC} \, | \, C \,))^2) \,\text{ as } A, B \text{ are i.i.d.}\\ & = \mathbb{E}(\frac{1}{(tC + 1)^2}) \\ & = \int_{0}^{1}\frac{1}{(tc + 1)^2} \, dc \\ & = \frac{1}{t + 1} \end{align}

Hence, $X$ and $Z$ have the same distribution, so $e^X$ and $e^Z$ have the same distribution.

In other words, $(AB)^C \sim U[0,1]$.

Had we not known beforehand that the distribution was going to be uniform, we would have had to notice that $\frac{1}{t + 1} = \mathbb{E}(Y^t)$ (which is not too hard to see). However, had the final distribution not been as nice, then this method would have failed; it seems that the method mentioned in the question, or the method used in Sangchul Lee's answer is more reliable in the general case.

Perhaps a more intuitive idea of how we may have spotted/deduced more simply that this has a relatively simple distribution is given in the top answer here; I highly rate this answer, as it shows how some good experience can make the path to an answer more natural.