# What is the maximum likelihood estimator for $e^{-\theta} = P(X_i = 0)$?

Suppose $$X_1, X_2,...,X_n$$ is a random sample from a $$\text{Poisson} (\theta)$$ distribution with probability mass function:

$$P(X=x)=\frac{\theta^ {x}e^{-\theta}}{x!}, x=1,2,...; 0<\theta$$

What is the maximum likelihood estimator for: $$e^{-\theta}= P(X = 0)$$?

I already found the MLE for the $$\theta$$. How do you then find the MLE of $$P(X = 0)$$ which is equal to $$e^{-\theta}$$ ?

Invariance principle : The maximum likelihood estimator of the transform is the transform of the maximum likelihood estimator.

Invariance property of MLE: if $$\hat{\theta}$$ is the MLE of $$\theta$$, then for any function $$f(\theta)$$, the MLE of $$f(\theta)$$ is $$f(\hat{\theta})$$.

The MLE for the Poisson parameter is the sample mean (derivation done below).

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

The MLE of a function of this parameter is a function of the sample mean:

$$f(\hat{\theta}) = f(\bar{x})$$

In our case the Maximum Likelihood Estimator of $$e^{-\theta}$$ is $$e^{-\bar{x}}$$