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Suppose $X_i$ is $iid$ $Poisson(\theta)$

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

I already found the MLE for the $\theta$. how do you then find the MLE of $e^{-\theta}(= P(Xi = 0))$ ?

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By the functional invariance property of the maximum likelihood estimator, the maximum likelihood estimator of $e^{-\theta}$ is just $\color{blue}{e^{-\widehat{\theta}_{MLE}}}$, where $\widehat{\theta}_{MLE}$ is the maximum likelihood estimator of $\theta$.

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  • $\begingroup$ so then would that be, the MLE is: $e^{-\hat \theta} = e^{-1/n \sum X_i}$? Since I calculate MLE of $\theta$ is $\hat \theta = 1/n \sum X_i$ $\endgroup$ – ISuckAtMathPleaseHELPME Mar 6 at 15:36
  • $\begingroup$ Yes, that would be it. $\endgroup$ – Minus One-Twelfth Mar 6 at 17:19

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