I am trying do find:
$$\frac{1}{E\left[\left(\frac{\sum_{i=1}^n X_i}{\alpha}-\frac{\sum_{i=1}^n (m-X_i)}{1-\alpha}\right)^2\right]}$$
But I can't seem to do it. Perhaps it isn't always possible.
$\def\a{\alpha}$ $X_1,X_2,\dots, X_n$ are randomly sampled from the binomial distribution with the pf:
$$P_X(x) = Pr(X=x) = \binom m x \a^{x} (1-\a)^{m-x},\quad x=0,1,2,\dots,m$$
$m$ is a given constant and $\alpha\in(0,1)$ is an unknown parameter.
The MLE is: $$\hat \alpha=\frac1n\sum_{i=1}^n \frac{X_i}m$$ and I have $E(\hat{\a}) = \a$, $\operatorname{Var}(\hat{\a})=\dfrac{\a(1-\a)}{mn}$
I think it best to try to work out $\left(\frac{\sum_{i=1}^n X_i}{\alpha}-\frac{\sum_{i=1}^n (m-X_i)}{1-\alpha}\right)^2$ first, but I am not sure how to expand such a thing, if it were $E(\sum_{i=1}^nX_i)$ and easy terms like that, I would see that it is equal to $n\a$, but here I have to square it first.
Start with rewriting the expression occurring in the exprectation: $$\begin{eqnarray} Y &=& \frac{1}{\alpha} \sum_{k=1}^n X_k - \frac{1}{1-\alpha} \sum_{k=1}^n \left(m-X_k\right) \\ &=& \frac{1}{\alpha} \sum_{k=1}^n X_k - \frac{1}{1-\alpha} \left(n \cdot m- \sum_{k=1}^n X_k\right) \\ &=& -\frac{n \cdot m}{1-\alpha} + \frac{1}{\alpha\left(1-\alpha\right)} \sum_{k=1}^n X_k \end{eqnarray} $$ Since $X_k$ are all iid, the random variable $Z=\sum_{k=1}^n X_k$ is known to also be Binomial distributed, specifically $Z \sim \mathrm{Binom}\left(n \cdot m, p\right)$. This is easy to understand, since each binomial random variable can be though of sum of $m$ independent Bernoulli trial outcomes.
Notice that $$ \mathbb{E}\left(Z\right) = n \cdot m \cdot \alpha $$ which implies that $\mathbb{E}\left(Y\right)=0$.
Now: $$\begin{eqnarray} \mathbb{E}\left(Y^2\right) &=& \mathbb{E}\left(\left(-\frac{n \cdot m}{1-\alpha} + \frac{1}{\alpha\left(1-\alpha\right)} Z \right)^2 \right) = \frac{\mathbb{E}\left(\left(Z- n\cdot m \cdot \alpha\right)^2 \right)}{\alpha^2 (1-\alpha)^2} = \frac{\mathbb{Var}\left(Z\right)}{\alpha^2 (1-\alpha)^2} \\& =& \frac{n \cdot m \cdot \alpha \cdot (1-\alpha)}{\alpha^2 (1-\alpha)^2} = \frac{n \cdot m}{\alpha \cdot (1-\alpha)} \end{eqnarray} $$