Let $X_1,X_2,...,X_n$ be a random sample from the distribution


where $x=1,2,...$ and $0<p<1$.

I know that the sufficient statistic is $Y=\sum X_i$. Now I have to find a function $f(Y)$ so that this is an unbiased estimator of $\theta=\frac{1}{p}$.

I have computed $\frac{d}{dp}\ln (L(p))=0$ and got $\frac{1}{\hat p}=\frac{Y}{n}$.

Is that the function? If yes, why is that an unbiased estimator?


1 Answer 1


Your statistic $ Y$ is the sum of $ n $ independent geometric distributions, each of which has an expected value of $1/p $. The expected value of $ Y/n $ is therefore also $1/p $, and your estimator is therefore unbiased.

  • $\begingroup$ I would understand it, if the probability of each sample value is $p$. Why is the expected value (of each $X_i$) $\frac{1}{p}$? Edit: $\frac{1}{p}$ is the mean of this distribution. Thank you! $\endgroup$
    – Mike90
    Mar 7, 2014 at 14:25

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