Maximum likelihood estimator of $p$ for the binomial (truncated) distribution This question is from the book "Introduction To The Theory Of Statistics, Mood Alexander", Chapter 7, Problem 15. In genetic investigations one frequently samples from a binomial distribution, except that observations of $x=0$ are impossible; so, in fact, the sampling is from the conditional (truncated) distribution
\begin{equation*}
\dbinom{m}{x} \dfrac {p^x(1-p)^{m-x}}{1-(1-p)^m}\mathbf{I}_{\{1,2,\cdots,m\}}(x)
\end{equation*}
Find the maximum-likelihood estimator of $p$ in the case $m=2$ for samples of size $n$.
So, i did the calculations by find the zero of the first derivative from the likelihood function, then i got:
\begin{align*}
L(p|x_{1}, x_{2}, \cdots, x_{n}) & = \displaystyle\prod_{i=1}^{n} \dbinom{2}
{x_{i}}\dfrac{p^{x_{i}}(1-p)^{2-x_{i}}}{1-(1-p)^{2}} \mathbf{I}_{\{1,2\}}(x_{i}) \\
\mbox{log}\hspace{1ex}L(p|x_{1}, x_{2}, \cdots, x_{n}) & = \mbox{log} \left[\displaystyle\prod_{i=1}^{n} \dbinom{2}{x_{i}}\right] - n\mbox{log}[p(2-p)] + \displaystyle\sum_{i=1}^{n}x_{i}\mbox{log} (p)\\
&\qquad\qquad + \left( 2n - \displaystyle\sum_{i=1}^{n}x_{i} \right)\mbox{log}(1-p)\\
\dfrac{d}{dp}\mbox{log}\hspace{1ex}L(p|x_{1}, x_{2}, \cdots, x_{n}) & = - n \cdot\dfrac{(2-2p)}{[p(2-p)]} + \displaystyle\sum_{i=1}^{n}x_{i}\cdot\dfrac{1}{p}\\
&\qquad\qquad - \left( 2n - \displaystyle\sum_{i=1}^{n}x_{i} \right)\cdot\dfrac{1}{1-p}\\
 & = \dfrac{- 2n + 2\displaystyle\sum_{i=1}^{n}x_{i} - p \displaystyle\sum_{i=1}^{n}x_{i}}{p(1-p)(2-p)} \\
\end{align*}
Now,
\begin{align*}
\dfrac{d}{dp}\mbox{log}\hspace{1ex}L(p|x_{1}, x_{2}, \cdots, x_{n}) & = 0 \\
\dfrac{- 2n + 2\displaystyle\sum_{i=1}^{n}x_{i} - p \displaystyle\sum_{i=1}^{n}x_{i}}{p(1-p)(2-p)} & = 0 \\
\hat{p} & = 2\left( 1 - \dfrac{n}{\displaystyle\sum_{i=1}^{n}x_{i}} \right) = 2\left( 1 - \dfrac{1}{\bar{x}} \right) \\
\end{align*}
So I'd really appreciate it if anyone can see if the estimator I've found doesn't look weird, or give me any ideas on a way to verify this.

As a sanity check, you can think of this as an iid sample from a shifted Bernoulli distribution with parameter $q=p^2/(1-(1-p)^2)$.  This gives you the MLE of $q$.  You can then in turn use functional equivalence of MLEs to obtain the MLE of $p$.
