Use Jensen's inequality to show that $e^{\frac{\sum^3_{i=1} X_i}{3}}$ is not an unbiased estimator of $e^{\theta}$

Question

My problem:

Let $X_{1}, X_{2},X_{3}$ be I.I.D. Poisson variables with common mean $\theta > 0$.

Use Jensen's inequality to show that $e^{\frac{\sum^3_{i=1}X_i}{3}}$ is not an unbiased estimator of $e^{\theta}$ and then use moment generating functions to compare this result.

My work so far:

With Jensen's inequality, I know I am trying to show that

$$E(h(x)) \geq h(E(x)).$$

I have also noted that for $e^{\frac{\sum^3_{i=1}X_i}3}$, the sum of three independent Poison variables will be an independent Poisson variable.

I also know that $E\left(e^{\frac{\sum^3_{i=1}X_i}3}\right)=e^{ \lambda(e-1)}$

From here I start to get lost. I am having quite some trouble with this one. I appreciate the help.

Answer 1

To show that your estimator is biased it is very simple. In fact, using Jensen's inequality you get that

$$\mathbb{E}\left[e^{\overline{X}_3}\right]\ne e^{\mathbb{E}[\overline{X}_3]}=e^{\theta}$$

...it is not necessary to identify if your expectation is greater or lower than $e^{\theta}$