$X_{1},X_{2},..,X_{n}$ 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 $\theta$ ?


$E[\bar x]= \mu$

  • 7
    $\begingroup$ E[f(X)] != f(E[X]) in general. $\endgroup$ Mar 22, 2019 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, 2019 at 21:01

2 Answers 2


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 $e^{-\mu}$


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.