Show that this estimator is not unbiased poisson($\theta$)

Question

Say, $X_1, ..., X_n$ is Poisson($\theta$) distributed. I have to find an unbiased estimator for $\theta^2$. First I have to show that the estimator $T = \bar{(X)}$$^2$ is not unbiased:

$\mathbb{E}\bar{(X)}$$^2$ $ = \mathbb{E}(\frac{1}{n} \sum_{i=1}^{n}X_i)^2 = \frac{1}{n^2}\sum_{i=1}^{n}\mathbb{E}(\bar{X}^2) = \frac{1}{n^2}\cdot n(\theta^2 + \theta) = \frac{\theta^2 + \theta}{n}$. So it is biased. But how can I find an estimator out of this which is unbiased? At first, is this computation right?

Thanks

Answer 1

Assuming the $X_i$ are independent,

we have

\begin{align} E\left[ \bar{X}^2 \right] &= \frac1{n^2}E\left[ \left(\sum_{i=1}^nX_i\right)^2 \right]\\ &= \frac1{n^2}E\left[ \left(\sum_{i=1}^nX_i^2\right)+ 2\sum_{i<j} X_i X_j\right]\\ &= \frac1{n^2}\left[ \left(\sum_{i=1}^nE[X_i^2]\right)+ 2\sum_{i<j} E[X_i]E[ X_j]\right]\\ &= \frac1{n^2} \left[ \left(\sum_{i=1}^n(Var[X_i]+E[X_i]^2)\right)+ n(n-1) \theta^2\right]\\ &= \frac1{n^2} \left[ \left(\sum_{i=1}^n(\theta+\theta^2)\right)+ n(n-1) \theta^2\right]\\ &= \frac1{n^2}(n \theta + n^2 \theta^2)\\ &= \theta^2 + \frac{\theta}{n} \end{align}

Hence it is biased.

To make it unbiased,

note that we have

$$E\left[ \bar{X}^2- \frac{\theta}{n}\right] = \theta^2$$

If $U$ is an unbiased estimator for $\theta$, then

$$E\left[ \bar{X}^2- \frac{U}{n}\right] = \theta^2$$

I will leave the task of finding an unbiased estimator for $\theta$ as an exercise.

Answer 2

Since you are trying to estimate $\theta^2$, using $T=(\bar X)^2$ should not be too far away. $\frac{\theta^2 + \theta}{n}$ is too far away since it tends to $0$ for large $n$, so you probably have an error

I would have thought $$E\left[\left(\sum\limits_{i=1}^n X_i\right)^2\right] = n E[X_i^2] + n(n-1) (E[X_i])^2 = n(\theta^2+\theta)+n(n-1)\theta^2 = n^2 \theta^2 + n \theta$$ so dividing this by $n^2$ would give $E[(\bar X)^2]=\theta^2+\frac{\theta}{n}$ which looks much more plausible, though still biased

You know or can find that $E[\bar X]=\theta$ so $E[\frac1n\bar X]=\frac{\theta}{n}$

and thus $(\bar X)^2 - \frac1n\bar X$ should be an unbiased estimator of $\theta^2$