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$X_{1},X_{2},..,X_{n}$ are iid $\sim Poisson(\mu)$

than the MLE for $\theta=e^{-\mu}$ is $\hat \theta =e^{-\bar x}$

Why is this considered to be biased for $\theta$?

Is $E[\hat \theta]$ not $\theta$ ?

as

$E[\bar x]= \mu$

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    $\begingroup$ E[f(X)] != f(E[X]) in general. $\endgroup$ – The Laconic Mar 22 at 20:54
  • $\begingroup$ I know that general result but I am still confused how to do it here since the sum of poisson is also poisson with mean $n \mu$ $\endgroup$ – Quality Mar 22 at 21:01
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Recall that the moment generating function of $X \sim \operatorname{Poisson}(\mu)$ is $$ E[e^{s X}] = \exp(\mu(e^s - 1)) $$ for all $s \in \mathbb{R}$

Proof. We just compute: $$ \begin{aligned} E[e^{s X}] &= \sum_{k=0}^\infty e^{s k} e^{-\mu} \frac{\mu^k}{k!} \\ &= e^{-\mu} \sum_{k=0}^\infty \frac{\left(\mu e^s\right)^k}{k!} \\ &= e^{-\mu} e^{\mu e^s} \\ &= \exp(\mu(e^s - 1)). \end{aligned} $$ We will use this result with $s = -1/n$.

If $X_1, \ldots, X_n \sim \operatorname{Poisson}(\mu)$ are i.i.d. and $$ \widehat{\theta} = \exp\left(-\frac{1}{n} \sum_{i=1}^n X_i\right) = \prod_{i=1}^n \exp\left(-\frac{X_i}{n}\right), $$ then, using independence, $$ \begin{aligned} E[\widehat{\theta}] &= \prod_{i=1}^n E\left[e^{-X_i / n}\right] \\ &= \prod_{i=1}^n \exp(\mu(e^{-1/n} - 1)) \\ &= \exp(n \mu(e^{-1/n} - 1)) \\ &\neq e^{-\mu}, \end{aligned} $$ so $\widehat{\theta}$ is not an unbiased estimator of $e^{-\mu}$

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As an aside from finding the exact expectation, you can use Jensen's inequality, which says that for a random variable $X$ and a convex function $g$,

$$E\left[g(X)\right]\ge g\left(E\left[X\right]\right)$$

, provided the expectations exist.

Verify that $g(x)=e^{-x}$ is a convex function from the fact that $g''>0$.

The equality does not hold because $g$ is not an affine function or a constant function.

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