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

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 unbiased. But how can I find an estimator out of this which is unbiased? At first, is this computation right?

Thanks

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

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.

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 $$\theta^2$$