0
$\begingroup$

I came accross this exercise when studying statistics, but I can't get to what's the solution. The exercise simply asks to show whether the reciprocal of the sample mean is an unbiased estimation for the unknown parameter $p$ if the sample is taken from the distribution $Geo(p)$.

I've tried the following: Let $\underline{x}$ be our sample, with size $n$. Let our estimation be $T(\underline{x}) = \frac{n}{\sum_{i=1}^{n} x_i}$. We need to show that $E(T(\underline{x})) = p$.

The sum of $n$ geometric distributions follows a $NegBinom(n, p)$ distribution, so with this, if I'm not mistaken the mean of our estimation goes like this:

$$E(T(\underline{x})) = n\sum_{i=n}^{\infty}\frac{1}{i}\binom{i}{n}(1-p)^{i-n}p^n = \sum_{i=n}^{\infty}\binom{i-1}{n-1}p^n(1-p)^{i-n}$$

Since our estimation has to be equal $p$, I arranged the two sides as: $\sum_{i=n}^{\infty}\binom{i-1}{n-1}p^n(1-p)^{i-n} = p$. From this, $\sum_{i=n}^{\infty}\binom{i-1}{n-1}p^{n-1}(1-p)^{i-n} = 1$ should be shown (with $p \neq 0$). It obviously reminds of a binomial series, but I can't progress from this spot. Any help would be appricated.

$\endgroup$
0
$\begingroup$

By Jensen inequality, $$\mathbb E\left(\frac{n}{\sum_{i=1}^{n} x_i}\right)>\frac{n}{\mathbb E(\sum_{i=1}^{n} x_i)}=p$$ So this estimator cannot be unbiased.

The exact value is $$ \mathbb E(T(\underline{x})) = n\sum_{i=n}^{\infty}\frac{1}{i}\binom{i-1}{n-1}(1-p)^{i-n}p^n $$ and this value leads to a hypergeometric function and does not have a simple form.

Form the other side, if you consider $\frac{n}{\sum_{i=1}^n x_i -1}$ instead of $T(\underline x)$, then the expected value can be found: $$ \mathbb E\left(\frac{n}{\sum_{i=1}^n x_i -1}\right) = n\sum_{i=n}^{\infty}\frac{1}{i-1}\binom{i-1}{n-1}(1-p)^{i-n}p^n $$ $$= \frac{n}{n-1}p \sum_{i=n}^\infty \binom{i-2}{n-2}p^{n-1}(1-p)^{i-n} = \frac{n}{n-1}p, $$ since the last sum equals to $1$ as the sum of probabilities of all possible values of $NegBinom (n-1,p)$ distribution: $$ 1=\sum_{j=n-1}^\infty \binom{j-1}{n-2}p^{n-1}(1-p)^{j-n+1}=\bigl[i=j+1\bigr]=\sum_{i=n}^\infty \binom{i-2}{n-2}p^{n-1}(1-p)^{i-n}. $$ And from $\mathbb E\left(\frac{n}{{\sum_{i=1}^n x_i} -1}\right) = \frac{n}{n-1}p$ you can design unbiased estimate if needed.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.