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I am trying to determine $E\left[\frac{\overline X}{1-\overline X}\right]$, where distribution of $X_1,\ldots,X_n$ is

$$f(x;\theta)=\theta x^{\theta−1}\quad,\, 0 < x < 1\,,\, \theta > 0 $$

When I try it by definition of $E[X]$ how do I integrate $\frac{\overline X}{1-\overline X}$?

StubbornAtom
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qqq2
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    Please add the `[self-study]` tag & read its [wiki](http://stats.stackexchange.com/tags/self-study/info). – gung - Reinstate Monica Mar 18 '16 at 20:49
  • Are you only wondering how to integrate $\bar X / (1- \bar X)$? If so, this may be better suited for the [math.SE] SE site. – gung - Reinstate Monica Mar 18 '16 at 20:50
  • First try and figure out the distribution of $\bar{X}$. – Greenparker Mar 18 '16 at 20:51
  • Also, specify what is the support of the distribution, and the possible values of $\theta$? Is it all $\mathbb{R}$? – Greenparker Mar 18 '16 at 20:53
  • yes really the question wants me to find the bias of this estimator which is the MME for this specific distribution. But without this exected value i have no way to find the bias. Don't use the given distribution? find $\bar{X}$ pdf first? – qqq2 Mar 18 '16 at 20:54
  • x is bewteen 0 and 1 and theta is positive, sorry forgot to mention that, – qqq2 Mar 18 '16 at 20:55
  • In that case $X \sim Beta(\alpha, 1)$. So you can first figure out the distribution of the sum of i.i.d. Beta random variables. See [here](http://math.stackexchange.com/questions/965226/sum-of-two-beta-distributed-random-variables). – Greenparker Mar 18 '16 at 20:58
  • if X~Beta($\theta$,1) wouldnt the pdf be [($\theta$-1)/$\theta$]x^($\theta$-1)?? my book may have different pdf though i have seen many variations for beta pdf than my books, Statistical Inference 2nd ed. by Cassella and Berger – qqq2 Mar 18 '16 at 21:19
  • srry nvm i see how its Beta(a,1) ill stop commenting here i got a warning for too many commentys – qqq2 Mar 18 '16 at 21:39
  • @Greenparker finding the pdf of the summation of beta distributions seems almost impossible, by mgf method for sure, they seem to have issues in that post you posted over just 2 RVs, are you guys sure i should be trying to find the pdf of $\bar{X}$? – qqq2 Mar 19 '16 at 00:10
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    @Greenparker I don't think $\bar X$ has any standard distribution. – StubbornAtom Apr 13 '19 at 08:51
  • What we can [show](https://math.stackexchange.com/questions/3185959/moment-estimator-hat-theta-of-mathrmbeta-theta-1-and-bias-of-hat/) is that this estimator is biased for $\theta$. – StubbornAtom Apr 14 '19 at 16:34
  • An interesting case is when $n=1$. Then the mean is infinite. – Sextus Empiricus Nov 21 '20 at 09:38

2 Answers2

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  1. $\bar{X} = 1/n \sum X$, call $Y = \sum X$. $Y$ has a known distribution. What is it?
  2. $\bar{X} / (1-\bar{X}) = Y / (n - Y)$
  3. $E[Y/(n-Y)] = \int Y/(n-Y) dY$
AdamO
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  • on 3. should i have ∫Y/(n-Y) f(Y)dY? – qqq2 Mar 18 '16 at 21:25
  • Yes that should be there. We get the habit of relaxing that notation when we get into measure theoretic statistic, the dY is actually the measure induced by the RV Y. – AdamO Mar 18 '16 at 22:09
  • i noticed in my book that it used notations like E sub $\theta$ ($\theta$ hat) why do they have this E sub $\theta$? this is for the bias formula – qqq2 Mar 18 '16 at 22:26
  • @qqq2 a good read http://stats.stackexchange.com/questions/72613/subscript-notation-in-expectations for that notation I would read, "The expectation of theta-hat under the probability model induced by theta". – AdamO Mar 18 '16 at 22:55
  • let $\theta$=A then i shud read this as $\mathbb{bias}_A[A hat]$ = $\mathbb{E}_A[A hat]$ - A where $\mathbb{E}_A[A hat]$=$\int (Ahat) f(x) dx$ using the original f(x) since sub A implies with resp to my original $\theta$? – qqq2 Mar 18 '16 at 23:38
  • As long as $\hat{A}$ is some statistic then yes. – AdamO Mar 21 '16 at 15:08
  • Sorry, but what known distribution does $Y$ have? – StubbornAtom Aug 11 '18 at 15:27
  • In the case of $\theta=1$ then $\bar{X}$ follows the Bates distribution and in other cases something more complex. Finding $E[\frac{1}{1-\bar{X}}]$ by first finding the distribution of $\bar{X}$ and then integration is not gonna be simple. The OP is gonna need to be satisfied with some approximation instead. – Sextus Empiricus Nov 21 '20 at 01:43
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    Allow me ask. On 1, what is the well known distribution of the summations of $X_i\sim Beta(\alpha, 1)?$ – KU99 Nov 21 '20 at 01:48
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It is probably not easy to exactly compute the distribution. For instance, when $\theta = 1$ then $\bar X$ follows the Bates distribution. That is not easy to integrate over. However, we can approximate the mean by using a simulation or by using a function that approximates the distribution.

Using an approximation of the distribution

We can approximate the mean of $n$ variables $\bar{X}$ by another beta distribution that matches the mean and variance. For this approximate distribution we have the following parameters

$$\begin{array}{} \alpha'&=&\alpha\frac{\alpha n+2n-1}{\alpha+1}\\ \beta'&=&\frac{\alpha n+2n-1}{\alpha+1} \end{array} $$

The mean will be then

$$E\left[\frac{\bar X}{1- \bar X}\right] \approx \frac{\alpha'}{\beta'-1}$$

You should be able to derive this by using integration. I just looked it up from a table. We can do this because the variable $X/(1-X)$ is distributed as the beta prime distribution.

Simulation

With this method we simply simulate a large sample of $\bar{X}/(1-\bar{X})$ and compute the sample mean. If the variance of $\bar{X}/(1-\bar{X})$ is finite then this simulation will get as close as we want.

Coding and comparison

For $\theta = 3$ and $n=10$ we get the following distribution of $\bar{X}/(1-\bar{X})$.

plot

The histogram is the simulation and the curve is the approximate distribution with $\alpha'$ and $\beta'$.

The results give values

> sum(x/(1-x))/m
[1] 3.273982
> aprox/(bprox-1)
[1] 3.266667

which are very close.

set.seed(1)
theta = 3
m = 10^4  ### sample size for simulation
n = 10    ### number of variables to average

### average of 10 beta variables
x <- rowSums(matrix(rbeta(n*m,theta,1), ncol = n))/n

### approximated variable
aprox = theta * (theta*n+2*n-1)/(theta+1)
bprox = (theta*n+2*n-1)/(theta+1)
xs = seq(0,1,0.001)
y = dbeta(xs,aprox,bprox)

### plot histogram and compare with estimate
hist(x, breaks = seq(0,1,0.01), xlim = c(0.5,1), freq = 0)
lines(xs,y)

### compare approximates of average
sum(x/(1-x))/m
aprox/(bprox-1)
Sextus Empiricus
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