Good estimator of $p$ knowing $\operatorname{Bin}(2,p)$

Question

We consider the statistical sample ${1,0,2,1,1,0,1,0,0}$ and we assume that the values of this sample come from a binomial distribution $\operatorname{Bin}(2, p)$.

How can we estimate the value of $p$?

For me, an intuitive estimator is $\hat{p} = \frac{1}{n}\sum_{i=1}^n X_i$. So $\hat{p}$ should be just $\frac{2}{3}$. Where am I wrong?

Answer 1

If $X_1,X_2,\dots,X_n$ are binominal $(2,p)$, and $Y:=\sum_{i=1}^{n}X_i$ then $Y$ is binomial $(2n,p)$.

Since $E[Y]=2np$ and you observed $Y=6$ in your sample, the maximum likelihood estimator will arise from solving $$2n\hat p=6$$

The solution is $\hat p=\frac{6}{18}$, or $\frac{1}{3}$.

As for the proposed answer of $2/3$, just look at your sample. Most of the numbers are closer to $0$ than $2$ right?

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Another way to approach is that if you take the estimators for each of your individual sample points, you will get $\frac{1}{2}, 0, 1, \frac{1}{2}, \frac{1}{2}, 0, \frac{1}{2}, 0, 0$ (just dividing by two). The average of these numbers is $\frac{1}{3}$.

Answer 2

Your "intuitive estimator" is an unbiased estimator of the mean of a random variable. But the mean of $\text{Bin}(n,p)$ is $np$, and you want $p$.