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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 $e^{-\theta}$ ?

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

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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 $e^{-\bar{x}}$

Derivation of $\hat{\theta}$:

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