product distribution of two uniform distribution, what about 3 or more

Solution 1:

We can at least work out the distribution of two IID ${\rm Uniform}(0,1)$ variables $X_1, X_2$: Let $Z_2 = X_1 X_2$. Then the CDF is $$\begin{align*} F_{Z_2}(z) &= \Pr[Z_2 \le z] = \int_{x=0}^1 \Pr[X_2 \le z/x] f_{X_1}(x) \, dx \\ &= \int_{x=0}^z \, dx + \int_{x=z}^1 \frac{z}{x} \, dx \\ &= z - z \log z. \end{align*}$$ Thus the density of $Z_2$ is $$f_{Z_2}(z) = -\log z, \quad 0 < z \le 1.$$ For a third variable, we would write $$\begin{align*} F_{Z_3}(z) &= \Pr[Z_3 \le z] = \int_{x=0}^1 \Pr[X_3 \le z/x] f_{Z_2}(x) \, dx \\ &= -\int_{x=0}^z \log x dx - \int_{x=z}^1 \frac{z}{x} \log x \, dx. \end{align*}$$ Then taking the derivative gives $$f_{Z_3}(z) = \frac{1}{2} \left( \log z \right)^2, \quad 0 < z \le 1.$$ In general, we can conjecture that $$f_{Z_n}(z) = \begin{cases} \frac{(- \log z)^{n-1}}{(n-1)!}, & 0 < z \le 1 \\ 0, & {\rm otherwise},\end{cases}$$ which we can prove via induction on $n$. I leave this as an exercise.

Solution 2:

If $X_1$ is uniform, then $-\log X_1 \sim \textrm{Exp}(1)$. Therefore, $$- \log X_1 \dots X_n = -\log X_1 + \dots -\log X_n$$ is a sum of independent exponential random variables and has Gamma distribution with parameters $(n,1)$ and density $g(y) = \frac{1}{(n-1)!} y^{n-1}e^{-y}$ for $y\geq 0$. Let $f$ be the density of the product $X_1 \dots X_n$, then the Jacobi's transformation formula yields $$ f( h^{-1}(y) ) | \partial h^{-1}(y) | = g(y), $$ with $h(x) = -\log x$ and $h^{-1}(y) = \exp(-y)$. The substitution $y=h(x)$ in the above equation gives $$ f(x) = \frac{1}{(n-1)!}(-\log x)^{n-1} \, 1_{ (0,1]}(x).$$