Estimating the parameter of a geometric distribution from a single sample I was surprised not to find anything about this with Google.
Consider a geometric distribution with $\text{Pr}[X=k]=(1-p)^{k-1}p$, so the mean is $\sum_{k=1}^\infty k\,\text{Pr}[X=k]=\frac{1}{p}$.
Now assume we observe a single outcome (number of trials until success, including the success) $n$. What is our estimate of $p$? The "natural" estimate (whatever that means) seems to be $\frac{1}{n}$. However, this is a biased estimate of $p$. Indeed, we have $\sum_{k=1}^\infty \frac{1}{k}\text{Pr}[X=k]=\frac{p}{1-p}\log\frac{1}{p}$.
Can we find an unbiased estimate of $p$ from $n$? What is the MSE estimate of $p$?
Many thanks.
 A: By definition, an estimator is a function $t$ mapping the possible outcomes $\mathbb{N}^{+} = \{1,2,3,\ldots\}$ to the reals. If this is to be unbiased, then--writing $q=1-p$--the expectation must equal $1-q$ for all $q$ in the interval $[0,1]$.  Applying the definition of expectation to the formula for the probabilities of a geometric distribution gives
$$1-q = \mathbb{E}(t(X)) = \sum_{k=1}^\infty t(k) \Pr(X) = (1-q)\sum_{k=1}^\infty t(k) q^{k-1}.$$
For $q\ne 1$ we may divide both sides by $1-q$, revealing that $t(k)$ are the coefficients of a convergent power series representation of the function $1$ in the interval $[0,1)$.  Two such power series can be equal in that interval if and only if they agree term by term, whence
$$t(k) = \begin{cases}
 1 & k=1 \\
 0 & k \gt 1
\end{cases}$$
is the unique unbiased estimator of $p$.
A: UPDATE. Re-writing my previous sloppy answer.
There is no unbiased estimator for $p$, here is the proof. 
The estimator $p=\frac{1}{k}$ is biased, but it's best you can get in the sense of MLE or method of moments.  Here's  the derivation in Math SE.
