Given the pdf $f(y\mid\theta) = \theta y^{\theta - 1}1_{0<y<1}$, show $E\left(\frac{1}{2\theta\sum_{i = 1}^n W_i}\right) = \frac{1}{2(n-1)}$

Question

Let $Y_1, \ldots, Y_n$ be a random sample from the pdf $$f(y\mid\theta) = \theta y^{\theta - 1}, 0 < y < 1, \theta > 0 .$$ show that $$E\left(\frac{1}{2\theta\sum_{i = 1}^n W_i}\right) = \frac{1}{2(n-1)}$$

I have showed that $\sum_{i = 1}^n -\ln(Y_i)$ is sufficient for $\theta$, and if $W_i = -\ln(Y_i)$, I have showed that $W_i$ has an exponential distribution with mean $1/\theta$. I have also showed that $2\theta \sum_{i=1}^n W_i $ has a $\chi^2$ distribution with $2n$ degrees of freedom.

I am wondering how to show it?

Answer 1

If you know that the chi-square distribution with $k$ degrees of freedom is $$f(x)\,dx=\frac{1}{\Gamma\left(\frac k 2\right)}\left(\frac x 2\right)^{k/2-1} e^{-x/2}\,\frac{dx}{2}\text{ for }x>0,$$ then it's just evaluating an integral:

\begin{align} \int_0^\infty \frac 1 x f(x)\,dx & = \int_0^\infty \frac 1 x \frac{1}{\Gamma\left(\frac k 2\right)}\left(\frac x 2\right)^{k/2-1} e^{-x/2}\,\frac{dx}{2} \\[12pt] & = \frac{1}{\Gamma(k/2)}\cdot\frac 1 2 \int_0^\infty \left(\frac x 2\right)^{k/2-2} e^{-x/2} \,\frac{dx}{2} \\[12pt] & = \frac{1}{\Gamma(k/2)}\cdot\frac 1 2 \int_0^\infty u^{k/2-2} e^{-u}\,du \\[12pt] & = \frac{1}{\Gamma(k/2)}\cdot\frac 1 2 \cdot \Gamma\left( \frac k 2 - 1 \right) \\[12pt] & = \frac{1}{\frac k 2 - 1} \cdot\frac 1 2 = \frac{1}{k-2}. \end{align}