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I have encountered the following problem: Let $X_1$, $\ldots$, $X_n$ be an independent random sample from the distribution with p.d.f. $$f(x;p)=p(1-p)^x, ~~x=0,1,2,\ldots,~~0<p<1.$$ (a) Find a complete and sufficient statistic of $p$;

(b) Find the UMVUE of $p$.

I notice that the geometric distribution is in the exponential family and successfully find that $T=\displaystyle\sum_{i=1}^nX_i$ is a complete and sufficient statistic.

However, for part (b), my original idea is to find a function $\phi(\cdot)$ such that $E(\phi(T))=p$ since we already know that T is complete and sufficient.

But the expectation for each $X_i$ is $\frac{1-p}{p}=\frac{1}{p}-1$, in order to get the expectation of phi T to be p, there are some issues.

I think $E\left(\dfrac{1}{\dfrac{1}{n}\displaystyle\sum_{i=1}^nX_i+1}\right)\ne\dfrac{1}{E\left(\dfrac{1}{n}\displaystyle\sum_{i=1}^nX_i+1\right)}$ in general, so this may not be easily applied.

So how can I solve this problem? The only methods we learned are this one and CRLB.

Lumos
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1 Answers1

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Let $Y=I(X_1 = 1)$ be the indicator that the first trial was a success. Then $EY =p$. In order to create a minimum variance unbiased estimator, we can consider $\phi(T) = E(Y\mid T)$ which is MVUE by the Lehmann-Scheffe theorem.

Sri-Amirthan Theivendran
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    Your $Y$ is not unbiased for $p$. – StubbornAtom Oct 14 '24 at 17:01
  • I saw the link here: https://stats.stackexchange.com/questions/401314/umvue-geometric-distribution-where-x-is-the-number-of-failures-preceding-the I think this is the proceeding part goes. – Lumos Oct 14 '24 at 17:24