# 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

$$E[\bar x]= \mu$$

• E[f(X)] != f(E[X]) in general. – The Laconic Mar 22 '19 at 20:54
• 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 $n \mu$ – Quality Mar 22 '19 at 21:01

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.