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Let $X_1, X_2, \dots , X_n$ be a random sample from the PDF $$f(x;\theta)=\theta x^{\theta -1},\;\; 0<x<1,\;\; \theta > 0$$ What is the probabilty that $\prod_{i=1}^n X_i > t$?

The joint pdf is $$f(x_1,\dots,x_n;\theta)=\theta^n \left(\prod_{i=1}^n x_i\right)^{\theta -1},\;\; 0<x_i<1,\;\; \theta > 0$$ So \begin{align}\mathbb{P}\left(\prod_{i=1}^n X_i > t\right) &= \int_{\prod_{i=1}^n x_i > t}f(x_1,\dots,x_n;\theta)dx_1\dots dx_n\\ &= \int_{\prod_{i=1}^n x_i > t\\0<x_1,\dots,x_n<1}\theta^n \left(\prod_{i=1}^n x_i\right)^{\theta -1}dx_1\dots dx_n\\ \end{align} Is evaluating this integral the best way of solving this problem, and if so, what's the fastest way to compute the integral? It seems like we should use the fact that the boundary of the integral is similar to the integrand.

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This looks like a self-study question, so I will only give some hints:

  1. To study the distribution of a product of independent random variables, you can turn it into a problem of sums, by taking logs. Then at the end, antilog to get the answer,

  2. Since in your problem, $0\le X \le 1$, logs are negative, so to get a nonnegative random variable with a density function which can be compared to standard distributions, we study the distribution of $-\log X = \log X^{-1}$.

  3. Doing that you will find that $-\log X$ has an exponential distribution.

  4. Sum of iid exponential random variables has a gamma distribution.

  5. So the answer will be that $\prod_i X_i$ has a log-gamma distribution.

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