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 ?


For the first question only:


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

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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.