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Given the following sentences:

Let $X_1,..., X_n$ be a random sample from a $Pois(mu)$ distribution. Consider the following estimator for $e^-mu=P(X_i=0)$: $T=e^-overlineX_n$.

The independent random variables $X_1;...;X_n$ have a geometric distribution with parameter $p$. Look at the following estimator for $p$: $S=frac1overlineX_n$.

Prove that the estimators are biased.

In my opinion both estimators are unbiased:
$E[T]=e^E[overlineX_n]=e^-mu$ that is unbiased for the parameter $e^-mu$.
$E[S]=frac1E[overlineX_n]=frac11/p=p$ that is unbiased for the parameter $p$.

Why I'm wrong in both cases? Where are my mistakes? Thanks.

You can't write$$Elefte^overlineX_nright=e^EoverlineX_n$$but we have $$Elefte^overlineX_nright=Elefte^X_1over ne^X_2over ncdots e^X_nover nright\=left(Elefte^X_1over nrightright)^n\=(exp(mu(e^1over n-1)))^n\=expBig(mu n(sqrt[n]e-1)Big)$$according to Poisson distribution which is biased.

As with the first one, we can argue similarly for the second one as following$$ES=Eleftnover X_1+cdots + X_nright\=nEleft1over X_1+cdots + X_nright\=nEleft1over Yright$$where $Y$ has negative binomial distribution with parameters $p $ and $n$. Further calculations are nasty in this case but you can refer to Geometric distribution section Parameter estimation.

The problem is that the exponential and the reciprocal aren't linear. You can't get an unbiased estimator for the standard deviation by taking the square root of an unbiased estimator for the variance, and we have the same problem here.

In order to see these examples more clearly, consider one-element samples. In the Poisson example, we get a probability of $e^-mu$ of sample value $0$ and estimate $1$, a probability of $mu e^-mu$ of sample value $1$ and estimate $e$, a probability of $fracmu^22e^mu$ of sample value $2$ and estimate $e^2$, and so on. Add them up, and the expected value of what we get is a value of the moment-generating function $sum_n=0^infty fracmu^n e^nn!e^-mu = e^mu(e-1)$. That's not the mean we wanted, and there's no simple way to correct it.

The second example, on the geometric distribution, has similar issues. The probability of getting $n$ is $(1-p)cdot p^n$, so our one-element estimate returns $frac1n$ with that probability. Sum that, and we get $-(1-p)ln(1-p)$. That's definitely not what we want - it goes to zero as $pto 1^-$.

Here $n < +infty$. We compute $Bbb E (exp(bar X_n))$ where $bar X_n$ is the mean of a i.i.d. sample $(X_1,dots,X_n)$ with distribution $textPoisson(mu)$:
beginaligned
Bbb E (exp(bar X_n)) &= Bbb E left(prod_i=1^n exp fracX_inright) \
&= prod_i=1^n Bbb E left( exp fracX_inright) \
&= Bbb E left( exp fracX_1nright)^n \
&= left(e^-musum_k=0^infty frac(e^1/n mu)^kk! right)^n \
&= expleft(n (e^1/n-1)muright) & neq exp (mu)
endaligned
One can note that the bias vanishes as $nto +infty$. For the second example, one can prove that the estimator is biased by using Jensen's inequality (see this post).