Let $X_1,X_2,\ldots,X_n$ be a random sample from a $\operatorname{Poisson}(\theta)$ distribution with probability function $$ P(X = x) = \frac {\theta^xe^{-\theta}}{x!} $$

Show that $\hat \theta$ is the minimum variance unbiased estimator of $\theta$ i.e. that is unbiased and attains the Cramer–Rao bound.

My attempt:

Showing that $\hat \theta$ is unbiased is easy; $$ E[\hat \theta] = \theta $$ $$ \hat \theta = \bar X $$ $$\operatorname E[\bar X] = \operatorname E\left[ \frac{1}{n} \sum_{i=1}^n X_i \right] = \frac{1}{n} E\left[\sum_{i=1}^n X_i\right] = \frac{1}{n}n\theta = \theta $$

Now to show that the variance attains the Cramer-Rao Bound is what I'm having trouble with.

$$\operatorname{Var}(\hat \theta) = \frac{1}{E[(\frac{d\ell(\theta)}{d\theta})^2]} $$ $$\operatorname{Var}(\hat \theta) = \frac{1}{E[(\frac{\sum_{i=0}^n(x_i)}{\theta} - n)^2]} $$

And now I'm not sure how to expand this further. Is the square of a sum just itself?

  • $\begingroup$ The Lehmann–Scheffe theorem can be used, if you know what it says. $\endgroup$ Sep 7, 2018 at 21:28

1 Answer 1


Consider the denominator $E\left[\left(\frac{\sum_{i=1}^n X_i}{\theta} - n\right)^2\right]$. Since $$ E\left[\frac{\sum_{i=1}^n X_i}{\theta}\right] = \frac{\sum_{i=1}^n E[X_i]}{\theta}=\frac{n\theta}{\theta}=n $$ and since the variance is the second central moment, get $$ E\left[\Biggl(\frac{\sum_{i=1}^n X_i}{\theta} - n\Biggr)^2\right] = E\left[\Biggl(\frac{\sum_{i=1}^n X_i}{\theta} - E\biggl[\frac{\sum_{i=1}^n X_i}{\theta}\biggr]\Biggr)^2\right] = \text{Var}\left[\frac{\sum_{i=1}^n X_i}{\theta}\right] = $$ $$ =\frac{\sum_{i=1}^n \text{Var}[X_i]}{\theta^2} = \frac{n\theta}{\theta^2}=\frac{n}{\theta}. $$

Substitute it to the Cramer-Rao Bound and compare the result with variance of estimate.


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