I am trying to show that the sample variance is an unbiased estimator of $\lambda$ for a Poisson distribution.
Let $(X_1, \dots, X_n)$ be a random sample from a Poisson distribution with mean $\lambda > 0$.
The sample variance is given as
$$S^2 = \dfrac{\sum_{i=1}^{n}(X_i – \bar{X})^2}{n-1}$$
For $S^2$ to be an unbiased estimator, I need to show that $\mathbb{E}[S^2] = \lambda$
My attempt:
I know that $\mathbb{E}[\bar{X}] = \lambda$ and $\sum_{i=1}^{n}{X_i} \sim \text{Poisson}(n\lambda)$
Using linearity of expectations as in this similar question:
$$
\begin{align*}
\mathbb{E}[S^2] &= \mathbb{E}\left[ \dfrac{\sum_{i=1}^{n}(X_i – \bar{X})^2}{n-1} \right] \\
&= \dfrac{1}{n-1}\mathbb{E} \left[ \sum_{i=1}^{n} \left( {X_{i}}^{2} + {\bar{X}}^2 – 2X_i\bar{X} \right) \right] \\
&= \dfrac{1}{n-1} \left( n \lambda^2 + \lambda^2 – 2\lambda^2 \right) \\
&= \dfrac{1}{n-1}(n + 1 -2) \lambda^2 \\
&= \dfrac{1}{n-1} (n-1) \lambda^2 \\
&= \lambda^2
\end{align*}
$$
I end up with $\lambda^2$ but I need to end up with $\lambda$ to show that $S^2$ is an unbiased estimator of $\lambda$. Where have I gone wrong?
Best Answer
You asked where you went wrong. There are several mistakes you made. Firstly,
$$ \mathbb{E} \left[ \sum_{i=1}^{n} {X_{i}}^{2} \right] \ne n \lambda^2 $$
Handled correctly,
$$ \mathbb{E} \left[ \sum_{i=1}^{n} {X_{i}}^{2} \right] = n \mathbb{E}[X^2] = n\left(\mathbb{V}[X]+\mathbb{E}[X]^2\right)=n\lambda+n\lambda^2 $$
Also you incorrectly did this evaluation:
$$ \mathbb{E} \left[ \sum_{i=1}^{n} \left( {\bar{X}}^2 - 2X_i\bar{X} \right) \right] \ne \left(\lambda^2 - 2\lambda^2 \right) $$
The correct evaluation is:
$$ \mathbb{E} \left[ \sum_{i=1}^{n} \left( {\bar{X}}^2 - 2X_i\bar{X} \right) \right] =-n\mathbb{E} \left[ {\bar{X}}^2 \right] = -\lambda - n\lambda^2 $$ since $$ \mathbb{E}[\bar{X}^2] = \left(\mathbb{V}[\bar{X}]+\mathbb{E}[\bar{X}]^2\right)=\lambda/n+\lambda^2 $$