1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

It might be helpful to make the transformation,

$$ Y = -\log\left(\prod\limits_{k=1}^nX_i\right) = \sum\limits_{k=1}^n(-\log(X_i))$$

You can show that $Y$ has a Gamma($n,\theta$) distribution.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.