Let $X_1,\ldots,X_n$ to be sample distributed geometric with parameter $p$. Find MLE. Is it unbiased?
The distribution for each is $p(1-p)^{x_i-1}$ so the function is $$L(p)=\displaystyle\prod_{i=1}^np(1-p)^{X_i-1}.$$ After taking lns on both sides I got $$l(p)=\ln(L(p))=n\log(p)+\sum_{i=1}^n(X_i-1)\cdot \log(1-p).$$ I derivatied and found maximum in $p_m=\dfrac{n}{n+\sum_{i=1}^n(X_i-1)}$. Now I need to calculate $E[p_m]$: $$E[p_m]=nE\left[\frac{1}{\sum X_i}\right]$$ How can proceed?
Best Answer
Here is one way to answer this. Consider the case $n = 1$. The estimator in this case is $\hat{p} = 1/X_{1}$. Let us try to see what it is expectation is, $$ E[\hat{p}] = E\Big[ \frac1{X_1}\Big] = \sum_{k=1}^\infty \frac{1}{k} P(X_1 = k) = \sum_{k=1}^\infty \frac1k p(1-p)^{k-1} $$
Hint: Note that for $ \alpha \in (-1,1)$, we have $\sum_{k=1}^\infty \frac{\alpha^k}{k} = - \log(1-\alpha)$.
EDIT2: You can obtain the exact expression, or use the following simple bound $$ E(\hat{p}) = p + \sum_{k=2}^{\infty} \frac{1}{k} p (1-p)^{k-1} > p $$ for $p \in (0,1)$ since the sum above is strictly positive.