I believe that the MLE of parameter $p$ in the geometric distribution, $\hat p = 1/(\bar x +1)$, is an unbiased estimator for $p$ and would like to prove it. So far, I have:

$E[\bar x + 1] = E[\bar x] + 1 = \frac{1}{p} - 1 + 1 = \frac{1}{p}$

I can see that the relationship is likely there, but I don't know how to work with the $(\bar x + 1)$ being in the denominator.

  • $\begingroup$ This seems to be a duplicate of 504914, 473190 and another question within the last few days with a less descriptive title that I cannot seem to track. Just put the number into the search box on this site, or search for 'unbiased estimator of geometric distribution' here or elsewhere. $\endgroup$ – BruceET Apr 28 '15 at 1:24
  • $\begingroup$ Thanks Bruce, but those are somewhat different from what I'm asking one discusses finding an unbiased estimator of variance, which is pretty easy to do), and one of them isn't entirely clear in its conclusion. $\endgroup$ – kathystehl Apr 28 '15 at 18:44

$1/(\bar{X}+1)$ is not an unbiased estimator of geometric$(p)$ distribution (number of failures before the first success), but $1/\left(\frac{n}{n-1}\bar{X}+1\right)$ is. Note that $Y_n:=n\bar{X}=\sum_{i=1}^nX_i$ has negative-binomial$(p,n)$ distribution (number of failures before $n$-th success). Then $E\left[1/\left(\frac{n}{n-1}\bar{X}+1\right)\right] = E\left[\frac{n-1}{Y+n-1}\right] =$ $\sum_{y=0}^{\infty} \frac{n-1}{y+n-1} \binom{y+n-1}{n-1} p^n(1-p)^y = p\sum_{y=0}^{\infty} \binom{y+n-2}{n-2}p^{n-1}(1-p)^y=p$ (as desired) as the sum denotes the mass function of negative-binomial$(p,n-1)$ distribution.

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