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I am trying to find the UMVUE for the parameter $p$ for an n i.i.d geometric distribution:

$(1-p)^{x-1}p$ for $x=1,2,…$ and $0<p<1$

and found that:

$P(X_1=1)$ is an unbiased estimator , so let $w=I[X_1=1]$ be my unbiased estimator and since $\sum_i X_i=t$ is complete and sufficient statistic for geometric distribution, I can improve my unbiased estimator as follows:

$E[w\mid\sum_i X_i=t] = P(X_1=1\mid\sum_i X_i=t) = \frac{P(X_1=1,\sum_i X_i=t-1)}{P(\sum_i X_i=t)}$

So I have two questions now: what is the pdf for $\sum_i X_i=t-1$ ? .. I know it is negative binomial but can't write it correctly and my second question is what is the variance of this modified unbiased estimator and does it achieve the Cramer-Rao lower bound ?

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For the first question only:

$P(X_1=1)=p$

$P(\sum_{i=2}^{n}X_i=t-1)={t-2\choose n-2}p^{n-1}(1-p)^{t-n}$, $t=n,n+1...$

$P(\sum_{i=1}^{n}X_i=t)={t-1\choose n-1}p^{n}(1-p)^{t-n}$, $t=n,n+1...$

So the UMVUE is $\hat p=\frac{n-1}{\sum_{i=1}^{n} X_i-1}$

For CRLB you may look here.

But for the variance of the UMVUE:

$Var(\hat p)=\sum_{t=n}^\infty \left(\frac{n-1}{t-1}-p\right)^2 {t-1\choose n-1}p^n(1-p)^{t-n}$

I'm afraid I was not able to get a closed form. Neither it worked for $E(\hat p^2)$

Maybe somebody else can step in.

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  • $\begingroup$ Can we find a closed form for the variance or E(p^2) when n=2 ? that is $ \hat p=\frac{n-1}{\sum_{i=1}^{2} X_i-1}$ and how did you develop the variance summation? thanks $\endgroup$ – Bassem Dec 1 '16 at 12:24
  • $\begingroup$ $Y=\sum_{i=1}^n X_i$ is Negative Binomial, $T=\frac{n-1}{Y-1}$ is unbiased ($E[T]=p$), so $Var(T)=E[(T-E[T])^2]=E\left[\left(\frac{n-1}{Y-1}-p\right)^2\right]=\sum_{t=n}^\infty\left(\frac{n-1}{t-1}-p\right)^2 P(Y=t)$ $\endgroup$ – Momo Dec 1 '16 at 14:33
  • $\begingroup$ Also, for $n=2$ you have $E[\hat{p}^2]=\frac{p^2\log(1/p)}{1-p}$ so $Var(\hat p)=E[\hat{p}^2]-p^2$ $\endgroup$ – Momo Dec 1 '16 at 14:39
  • $\begingroup$ Great .. thanks for the help .. but could you show what form did you use to find $E[\hat{p}^2]$ ? $\endgroup$ – Bassem Dec 1 '16 at 23:37
  • $\begingroup$ $E(\hat p^2)=\sum_{t=n}^\infty \left(\frac{n-1}{t-1}\right)^2 {t-1\choose n-1}p^n(1-p)^{t-n}$ So for $n=2$ $E(\hat p^2)=\sum_{t=2}^\infty \frac{1}{t-1} p^2(1-p)^{t-2}=\frac{p^2}{1-p}\sum_{t=2}^\infty \frac{1}{t-1} (1-p)^{t-1}=\frac{p^2}{1-p}\sum_{i=1}^\infty \frac{1}{i} (1-p)^i$ The last series needs $\sum_{i=1}^\infty \frac{x^i}{i}$, which is obtained by integrating $\sum_{i=1}^\infty u^{i-1}=\frac{1}{1-u}$ term by term from $u=0$ to $x$ You might consider upvoting and accepting the answer, if it was useful for you. $\endgroup$ – Momo Dec 1 '16 at 23:50

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