likelihood of poisson distribution

Question

If $(x_1,\cdots, x_n)$ is a sample from a Poisson(θ) distribution, where $θ ∈ (0,∞)$ is unknown, then determine the $MLE$ of $θ$.

my attempt:

so the probability density of poisson is

$$p_{\theta}(x) = \frac{\theta^x e^{-\theta}}{x!}$$

the likelihood is given by

$L(\theta | \bar{x}) = \prod_{i=1}^{n} \frac{\theta^{x_i}e^{-\theta}}{x_i!} = e^{-n\theta} \left(\prod_{i=1}^{n} \frac{1}{x_i!} \right) \theta^{\sum_{i=1}^{n}x_i}$

How do I do the factorial ?

For some reason in the solution their answer is just $e^{-n\theta} \theta^{\sum_{i=1}^{n}x_i}$ ?

Answer 1

The factorial part is a (positive) constant (with respect to the parameter $\theta$), so when maximising the likelihood with respect to $\theta$, it can be ignored. (It is still technically part of the likelihood though.)

It's like if you wanted to find $\theta$ to maximise a function $5f(\theta)$, it's the same as just finding the maximiser of $f(\theta)$ (since $5$ is a positive constant).

(Remember, when we find maximum likelihood estimators, we maximise the likelihood or log-likelihood with respect to the parameter $\theta$, and treat the $x_i$ as constants.)