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I am trying to get my head around MLE of a function.

Say I have $X_i \sim \text{Poisson}(\theta)$ samples. I want to find the MLE of $\pi(\theta) = \exp(-\theta) = P_\theta(X = 0)$.

MLE of the Poisson's parameter is the sample means, i.e. $\bar{X}$, so is the MLE of $\pi$ be:

$$ \exp \left( -\bar{X} \right) $$

where $\bar{X}$ is the sample mean?

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If $T$ is a statistic, which is MLE for parameter $\theta$, and $f$ is a continuous one-to-one function then $f(T)$ is the MLE of $f(\theta)$. Prove this by using transformation of variables formula for probability distributions and the definition of MLE. For a quick overview see here.

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  • $\begingroup$ f does not need to be continuous or one-to-one. It holds for any f. $\endgroup$ – James Yang Jul 11 at 4:29

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