Unbiased estimator of binomial parameter Let 4,3,5,2,6 are 5 observations of the $\text{binomial}(10,p)$ random variable.  What will be an unbiased estimate of $(1+p)^{10}$?
I have tried to solve the problem in this way.
We know that $E[\frac{\bar{(X)}}{n}]=p=0.8$, also $\frac{(x)!}{(x-r)!}\frac{(n-r)!}{n!}=\hat{p}^{r}$
$(1+\hat{p})^{n}=1+\dbinom{n}{1}\hat{p}+\dbinom{n}{2}\hat{p^2}+...+\dbinom{n}{n}\hat{p^n}$
Is this the right way to proceed?But it will be difficult to calculate by putting all the values of $\hat{p}$
Please provide an easier way to calculate this. 
 A: Just notice that the probability generating function of $X\sim\mathsf{Bin}(m,p)$ is
$$E(a^X)=(1-p+pa)^m$$
Setting $1-p+ap=1+p$ gives $a=2$.
So for $X_i\sim \mathsf{Bin}(m,p)$ we have $$E(2^{X_i})=(1+p)^m$$
This also means $$E\left(\frac{1}{n}\sum_{i=1}^n 2^{X_i}\right)=(1+p)^m$$
Hence an unbiased estimator of $(1+p)^m$ based on a sample of size $n$ is $$T=\frac{1}{n}\sum\limits_{i=1}^n 2^{X_i}$$ 
Here an unbiased estimate of $(1+p)^{10}$ is therefore the observed value of $T$, which is $24.8$.
A: Let $n$ be the parameter of the binomial, $n=10$ in your case, and $m$ the sample size, $m=5$ in your case. I think there is an approximate answer to this that avoids long explicit summations in the case that $n$ is large and if additionally $np$ (or $n(1-p)$) is also large enough so that the normal approximation to the binomial applies. Unfortunately, $5$ and $10$ are likely too small for the following approximation to be useful, but perhaps it may lead to further ideas.
The sample is $X_1,\ldots,X_m\sim\text{Bin}(n,p)\approx\text{N}(np,np(1-p))$, with sample sum $m\bar{X} \sim \text{Bin}(mn,p)\approx \text{N}(mnp,mnp(1-p))$, so that approximately, the sample mean $\bar{X}\sim \text{N}(np,\frac{np(1-p)}{m})$ and the sample variance $S^2$ is unbiased with $\text{E}[S^2] = np(1-p)$. 
Let's use the conventional unbiased estimator for $p$, that is $\hat{p}=\frac{\bar{X}}{n}$, and see what that the bias is of the estimator
$$
\hat{\theta} = (1+\hat{p})^n
$$
for $\theta=(1+p)^n$. Now if $n$ is large, then approximately
$$
  \theta = (1+p)^n = (1+\frac{np}{n})^n \approx e^{np}\,,\ \ \text{and}\ \ \hat{\theta} = (1+\frac{\bar{X}}{n})^n \approx e^{\bar{X}}\,.
$$
Because $\bar{X}$ is normally distributed, $\hat{\theta}=e^{\bar{X}}$ is lognormally distributed. From the properties of the lognormal distribution we easily obtain, with $\mu=np$ and $\sigma^2=\frac{np(1-p)}{m}$ the mean and variance of $\bar{X}$, that
$$
  \text{E}[\hat{\theta}] = e^{\mu+\sigma^2/2} = \exp(np + \frac{np(1-p)}{2m})
$$
The bias of $\hat{\theta}$ is therefore
$$
  \text{E}[\hat{\theta}] - \theta = \exp(np + \frac{np(1-p)}{2m}) - \exp(np)
  = e^{np} [\exp(\frac{np(1-p)}{2m})-1]
$$
By replacing $p$ by its estimate $\hat{p}$, this can be used to eliminate the bias of $\hat{\theta}$. You can also use $S^2$ to estimate $np(1-p)$, for example I think 
$$
  e^{n\hat{p}} [e^{S^2/2m} -1]
$$
will be a pretty good estimation of the bias and thus
$$
  \hat{\theta}' = e^{n\hat{p}+S^2/2m}
$$
will be a reasonably unbiased estimate of $(1+p)^n$.
