Is this estimator biased [Geometric Distribution]?

Question

Using the method of moments we get that $$\hat{\theta} = \frac{n}{\sum_{i=1}^{n}X_i},$$ where $X_i\sim \text{Geo}(p)$ for $i=1,2,3,\cdots,n$ and pmf $P(X_i=k) = (1-p)^{k-1}p$ for $k=1,2,3,\cdots.$

Then $\bar{X}=\sum_{i=1}^{n}X_i\sim \text{NB}(n,p)$ where NB stands for negative binomial distribution with pmf $$P(\bar{X}=k)=\binom{k+n-1}{k}(1-p)^{k-1}p^{n}$$ where $k=1,2,3,\cdots.$ Thus $$\mathbb{E}\left[\frac{n}{\bar{X}}\right] = n\sum_{k=1}^{\infty}\frac{1}{k}\binom{k+n-1}{k}(1-p)^{k-1}p^{n}.$$

I am not sure how to compute this sum analytically. Maybe I can say that $$E[n/\bar{X}]>np^n $$ and so $\hat{\theta}$ is not a biased estimator. Furthermore, I am also interested in the consistency of the estimator, but I am not sure how to do this. Any hints in this regard will also be much appreciated.

Answer 1

For consistency of the estimator, note that since the $X_i$ are i.i.d, we have by the SLLN, that $$ \hat {p}=\frac{1}{\sum_{i=1}^nX_i/n}\stackrel{\text{a.s}}{\to}\frac{1}{EX_1}=\frac{1}{1/p}=p $$ whence the estimator is (strongly) consistent.

Answer 2

Let $f$ be a strictly convex function and let $X$ be not constant a.s. Then by Jensen's inequality, $$\mathbb{E}f(X) > f(\mathbb{E}X).$$ Taking $f(x) = 1/x$ ($x>0$) and $X=\bar{X}$, you get: $$\mathbb{E}\left(\frac{1}{\bar{X}}\right)>\frac{1}{\mathbb{E}\bar{X}},$$ so indeed: $$\mathbb{E}\left(\frac{n}{\sum_i X_i}\right)>p.$$