# Maximum likelihood with Poisson check please

Suppose a random sample of size n is drawn from probability model $$p_X(x;\theta) = \frac{\theta^{2x}e^{-\theta^2}}{x!}\;, \quad x=0, 1, 2, ...$$ Find a forumla for the maximum likelihood estimator, $$\hat{\theta}$$

$$L(\theta) = \prod_{i=1}^{n}\frac{\theta^{2x_i}e^{-\theta^2}}{x_i!}$$

$$L(\theta) = \frac{\theta^{2\sum_{i=1}^{n}x_i}e^{-n\theta^2}}{x_1!x_2!...x_n!}$$

$$ln[L(\theta)] = 2\sum_{i=1}^{n}x_iln(\theta)-n\theta^2 - ln(x_1!x_2!...x_n!)$$

$$T = \frac{d\big[ln[L(\theta)]\big]}{d\theta} = \frac{2\sum_{i=1}^{n}x_i}{\theta}-2n\theta$$

Let $$T=0$$

$$2n\theta = \frac{2n\bar{x}}{\theta}$$

$$\theta = \sqrt{\bar{x}}$$

This implies that $$\hat{\theta} = \sqrt{\bar{x}}$$

I have not got an answer in the back of the book for this one, does anybody see anything wrong with it?

$$T' = \frac{-2n\bar{x}}{\theta^2} - 2n$$

Which is negative for all $$\theta$$ because $$x \ge 0$$, thus the likelihood is a maximum.

## 1 Answer

Your answer is correct. You can also check the second derivative to confirm maximality.