# Unbiased estimator of poisson parameter

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

• $Z_n$'s are functions of your measurements $X_1$, $\dots$, $X_N$
• $E[Z_n] = \lambda^n$,

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 $$e^\lambda$$.