(Warning, long text) I've been given the following instructions;
Let $\boldsymbol X = \mathit X_1, X_2, ..., X_n$ be a random sample from binomial distribution $b(N,p)$, with PMF $f(x;p) = {N\choose x} \cdot p^x \cdot (1-p)^{N-x}$, where $N$ is a known positive integer. Find the MVUE for the parameter $g(p)=p^k$
I've come up with a solution and a result, but I'm only starting getting into statistics, so I would really appreciate it if someone took the time to look at the following. Here it goes;
I'm going to use the Lehmann–Scheffé theorem, by finding an unbiased estimator of $g(p)$ that depends on the sample through a sufficient and copmplete statistic of the parameter $p$. Such a statistic is $T(\boldsymbol X)=\sum_{i=0}^N X_i$, which follows the binomial distribution $b(nN,p)$. So, I need to find a function $U(\boldsymbol X) = g(T(\boldsymbol X))$, such that $E\big[U(\boldsymbol X)\big] = E\big[g\big(T(\boldsymbol X)\big)\big] = p^k$. Solving this equation; $$E\big[g\big(T(\boldsymbol X)\big)\big] = \sum_{x=0}^{nN} {nN\choose x} \cdot p^x \cdot (1-p)^{nN-x} = \sum_{x=0}^{nN} {nN\choose x} \cdot \left(\frac{p}{1-p}\right)^x \cdot (1-p)^{nN} = p^k \implies$$ $$\sum_{x=0}^{nN} {nN\choose x} \cdot \left(\frac{p}{1-p}\right)^x \cdot p^{-k} \cdot \left(1-p\right)^k = \left(1-p\right)^{k-nN} \implies$$
$$\sum_{x=0}^{nN} {nN\choose x} \cdot \left(\frac{p}{1-p}\right)^{x-k} =\left(\frac{1}{1-p}\right)^{nN-k}$$ Now, if I denote $\frac{p}{1-p} = \rho$, I get to $1-p= \frac{1}{1+\rho}$, so $\frac{1}{1-p}=1+\rho$. Therefore, I have got to $$\sum_{x=0}^{nN} {nN\choose x} \cdot \rho^{x-k} = (1+\rho)^{nN-k} = \sum_{y=0}^{nN-k} {nN-k\choose y} \rho^y$$ using the binomial theorem, and substituting $y$ with $x-k$, I get to $$\sum_{x=0}^{nN} {nN\choose x} \cdot \rho^{x-k} = \sum_{x=k}^{nN-k} {nN-k\choose x-k} \cdot \rho^{x-k}$$ If I know equalize the coefficients of $\rho^{x-k}$ on each Sum, I have that;
$g(x)= \left\{ \begin{array}{ll} 0 & x=0, 1, ... k-1 \\ \frac{{nN-k\choose x-k}}{{nN\choose x}} & x=k, k+1, ..., nN)\\ \end{array} \right.$
This can be further simplified to $g(x)=\frac{(nN-k)! \cdot x!}{nN! \cdot (x-k)!}$, So I have fianally got to my result; The MVUE for the parameter $g(p)=p^k$ is the statistic $$U(\boldsymbol X) = \frac{(nN-k)! \cdot T(\boldsymbol X)!}{nN! \cdot \big(T(\boldsymbol X)-k\big)!}, x = k, k+1, ..., nN$$ (The statistic $T(\boldsymbol X) = \sum_{i=0}^N X_i$, as stated in the beginning).
First of all, I am not so confident about my result because, it's still the beginning of my journey through the course, and secondly, isn't this a very fussy function? Anyway, thanks in advance to anyone that may take the time to point out any mistake or confirm the solution, and again sorry for the long text!