1

I am following the Wikipedia Proof but something goes wrong: $$ f_X(x) = a x^{a-1} \text{ for } x \in [0,1]$$ If $X$ and $Y$ are iid, I want the distribution of $XY$. Actually I want it for the product of $n$ such iid variables, but I thought I would derive that by starting with two. Yet when I try to follow the procedure:

$$f_{XY}(z) = \int_{0}^{1}f_X(x)\cdot f_Y(z/x)\cdot \frac{1}{|x|} dx $$ it diverges.
$$\int_0^1 ax^{a-1} \int_0^{z/x}ay^{a-1} dy dx = \int_0^1 az^a\cdot \frac{1}{x} dx$$ I tried following it step by step but I don't see my mistake. Yet it seems obvious the distribution should be something reasonable. Where is my error?

Ultimately I am trying to find $$E\Big[\frac{-n}{\ln(\Pi X_i)}\Big]$$ but I got it down to needing the distribution of $\Pi X_i$.

RobertTheTutor
  • 7,515
  • 1
    I believe it should be something like $f_{XY}(z)=\int_z^1 f_X(x) f_Y(z/x) \frac{1}{x} dx$, because for $x < z$, $z/x > 1$ and $f_Y(z/x)=0$. – Ivan Kaznacheyeu Oct 26 '21 at 14:19
  • @IvanKaznacheyeu : I think the O.P. is interested in the calculation of a particular expectation, thus the density $f_{XY}(z)$ is not needed. See my answer – tommik Oct 26 '21 at 14:25
  • 1
    I believe the O.P. is also interested in recognizing error source in his work. – Ivan Kaznacheyeu Oct 26 '21 at 14:33
  • @IvanKaznacheyeu : ok sorry for my comment. Let's wait the O.P. feedback – tommik Oct 26 '21 at 14:37
  • @StubbornAtom it is the exact question I needed. Thank you. I've never mastered the art of actually finding a previous question that answers what I ask. I'll sit down and try to decipher it later; clearly it assumes knowledge of special distributions and transformations. – RobertTheTutor Oct 26 '21 at 23:17

1 Answers1

1

but I got it down to needing the distribution of $\prod_i X_i$.

If your goal is to find the expectation you showed, the distribution of you are trying to derive is not needed.

In fact, observe that the distribution of

$$Y=-\log X$$

is....after easy calculations

$$f_Y(y)=a e^{-ay}$$

a negative exponential (nice result :)).

now let's take you expectation

$$\mathbb{E}\left[ \frac{-n}{\log\prod_i X_i} \right]=\mathbb{E}\left[ \frac{n}{\sum_i(-\log X_i)} \right]=n\mathbb{E}\left[\frac{1}{Z}\right]=\frac{n}{n-1}\alpha$$

this because $(1/Z)\sim \text{Inverse Gamma}$

If you do not want to use the known result of the inverse gamma you can calculate $\mathbb{E}\left[\frac{1}{Z}\right]$ by integration using the fact that $Z\sim\text{Gamma}(n;a)$

Z in fact is the sum of $n$ iid negative exponentials

Expectation's calculation by integration

$$n\mathbb{E}\left[\frac{1}{Z}\right]=n\int_0^{\infty}\frac{1}{z}\frac{\alpha^n}{\Gamma(n)}z^{n-1}e^{-\alpha z}dz=$$

$$=\frac{n \alpha\Gamma(n-1)}{\Gamma(n)}\underbrace{\int_0^{\infty}\frac{\alpha^{n-1}}{\Gamma(n-1)}z^{(n-1)-1}e^{-\alpha z}dz}_{=1}=\frac{n}{n-1}\alpha$$

Further conclusion (not requested but probably yes...)

your estimator is biased for $\alpha$ but asymptotically unbiased.

tommik
  • 33,201
  • 4
  • 17
  • 35