The number of accidents per day is a Poisson random variable with parameter $\lambda$, on 10 randomly chosen days the number of accidents were observed as 1,0,1,1,2,0,2,0,0,1, what will be an unbiased estimator of $e^{\lambda}$?
I tried to attempt in this way: We know that $E(\bar{x})=\lambda=0.8$, but $E(e^{\bar{x}})\neq\ e^{\lambda}$. Then what will be the required unbiased estimator?
If $X\sim \text{Pois}(\lambda)$, then $P(X = k) = \lambda^ke^{-\lambda}/k!$, for $k\geq 0$. It is hard to compute
$$E[X^n] = \sum_{k\geq 0} k^n P(X = k)\text{,}$$ but is is much easier to compute $E[X^{\underline{n}}]$, where $X^{\underline{n}} = X(X - 1)\cdots (X - n + 1)$: $$E[X^\underline{n}]=\lambda^n\text{.}$$ You can prove this by yourself - it is an easy exercise. Also, I will let you prove by yourself the following: If $X_1,\cdots, X_N$ are i.i.d as $\text{Pois}(\lambda)$, then $U = \sum_i X_i\sim \text{Pois}(N\lambda)$, hence $$E[U^{\underline{n}}] = (N\lambda)^n = N^n \lambda^n\quad\text{and}\quad E[U^\underline{n}/N^n] = \lambda^n\text{.}$$ Let $Z_n = U^{\underline{n}}/N^n$. It follows that
Since $e^\lambda = \sum_{n\geq 0}\lambda^n /n!$, we can deduce that
$$E\left[\sum_{n\geq 0}\frac{Z_n}{n!}\right] =\sum_{n\geq 0} \frac{\lambda^n}{n!} = e^\lambda\text{,}$$ hence, your unbiased estimator is $W = \sum_{n\geq 0} Z_n/n!$, i.e, $E[W] = e^\lambda$. However, to compute $W$, one must evaluate a sum that seems to be infinite, but note that $U\in \mathbb{N}_0$, hence $U^\underline{n} = 0$ for $n>U$. It follows that $Z_n = 0$ for $n>U$, hence the sum is finite.
We can see that by using this method, you can find the unbiased estimator for any function of $\lambda$ that can be expressed as $f(\lambda) = \sum_{n\geq 0}a_n\lambda^n$.
It follows that $Y=\sum_{i=1}^{10} X_i \sim \text{Pois}(10\lambda)$. We want to estimate $\theta=e^\lambda$. As you say, a possible estimator would be \begin{equation} \hat\theta = e^{\bar X} = e^{Y/10}. \end{equation} Using the moment generating function of $Y$, \begin{equation} M_Y(t)=e^{10\lambda(e^t - 1)}, \end{equation} we find that \begin{equation} E(\hat\theta) = E(e^{\frac1{10}Y}) = M_Y(\frac1{10}) = e^{10\lambda(e^{1/10} - 1)} = \theta^{10(e^{1/10}-1)}, \end{equation} so $\hat\theta$ is biased. Some guesswork suggest that \begin{equation} \theta^* = e^{aY}, \end{equation} may be unbiased for suitable choice of the correction factor $a$. Again, using the mgf of $Y$ we find that \begin{equation} E(\theta^*) = e^{10\lambda(e^a - 1)} = \theta^{10(e^a-1)}, \end{equation} so this is unbiased if $10(e^a - 1) = 1$ which leads to $a=\ln\frac{11}{10}$ and $\theta^* = (\frac{11}{10})^Y$ as an unbiased estimator of $\theta=e^\lambda$.
By the Lehmann-Scheffé theorem, since $Y$ is a sufficient statistic for $\lambda$, the estimator $\theta^*$ (a function of $Y$) is UMVUE for $e^\lambda$.
