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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 $\frac{\overline X}{1-\overline X}$?

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  • $\begingroup$ Please add the [self-study] tag & read its wiki. $\endgroup$ – gung - Reinstate Monica Mar 18 '16 at 20:49
  • $\begingroup$ Are you only wondering how to integrate $\bar X / (1- \bar X)$? If so, this may be better suited for the Mathematics SE site. $\endgroup$ – gung - Reinstate Monica Mar 18 '16 at 20:50
  • $\begingroup$ First try and figure out the distribution of $\bar{X}$. $\endgroup$ – Greenparker Mar 18 '16 at 20:51
  • $\begingroup$ Also, specify what is the support of the distribution, and the possible values of $\theta$? Is it all $\mathbb{R}$? $\endgroup$ – Greenparker Mar 18 '16 at 20:53
  • $\begingroup$ 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? $\endgroup$ – qqq2 Mar 18 '16 at 20:54
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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. $E[Y/(n-Y)] = \int Y/(n-Y) dY$
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  • $\begingroup$ on 3. should i have ∫Y/(n-Y) f(Y)dY? $\endgroup$ – qqq2 Mar 18 '16 at 21:25
  • $\begingroup$ 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. $\endgroup$ – AdamO Mar 18 '16 at 22:09
  • $\begingroup$ 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 $\endgroup$ – qqq2 Mar 18 '16 at 22:26
  • $\begingroup$ @qqq2 a good read stats.stackexchange.com/questions/72613/… for that notation I would read, "The expectation of theta-hat under the probability model induced by theta". $\endgroup$ – AdamO Mar 18 '16 at 22:55
  • $\begingroup$ 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 $\theta$? $\endgroup$ – qqq2 Mar 18 '16 at 23:38

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