Why is this estimator biased? $X_{1},X_{2},..,X_{n}$ 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
$E[\bar x]= \mu$
 A: 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 $e^{-\mu}$
A: 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.
