I have in my problem that $X_1,\ldots,X_n$ is a random sample from a distribution with probability density $f(x; \theta)=\theta x^{\theta-1}, 0

Question

I have in my problem that $X_1,\ldots,X_n$ is a random sample from a distribution with probability density $f(x; \theta)=\theta x^{\theta-1}, 0<x<1$. Furthermore, $-\log X_i \sim\text{EXP}(1/\theta), \,\, i=1,\ldots,n$.

How can I show, using the WLLN, that $$-\frac{1}{n}\sum_{i=1}^n \log X_i \xrightarrow{p}\frac{1}{\theta}$$

kirbykirby · Answer

The WLLN (Weak Law of Large Numbers) states: 
If $X_1,\ldots,X_n$ is a random sample from a distribution with $E(X_i)=\mu$ and $Var(X_i)=\sigma^2<\infty$, then $$\bar{X}_n=\frac{1}{n}\sum_{i=1}^n X_i \xrightarrow{p} \mu$$

anonymous · Answer

My instructor skipped over some of the details, so it's not as fresh in my mind as I'd like it to be... Here's an attempt at deciphering my notes.

It doesn't look like the density functions of the $X_i$ are important... (beta, by the way).

Denote $Y_i=-\log X_i$. Then you have to show that $\bar{Y}_n=\dfrac{1}{n}\displaystyle\sum_{i=1}^nY_i \xrightarrow{p}E(Y)=\dfrac{1}{\theta}$, i.e.
$$\lim_{n\to\infty}P\left(\left|\bar{Y}_n-\frac{1}{\theta}\right|>\epsilon\right)=0$$
You know that $E(Y_i)=E(\bar{Y}_n)=\dfrac{1}{\theta}$, $V(Y_i)=\dfrac{1}{\theta^2}$, and $V(\bar{Y}_n)=\dfrac{1}{n}V(Y_i)=\dfrac{1}{n\theta^2}$.

By Chebyshev's inequality, you have
$$P\left(\left|\bar{Y}_n-\frac{1}{\theta}\right|\ge\epsilon\right)\le\frac{V(\bar{Y}_n)}{\epsilon^2}=\frac{1}{n\theta^2\epsilon^2}$$
The right side then approaches 0 as $n\to\infty$.

anonymous · Answer

And that's all there is to it. I'm not sure where the law is explicitly applied, though.

kirbykirby · Answer

ohhhh I see thank you so much!!

kirbykirby · Answer

:D

anonymous · Answer

You're welcome!