estimatorsmaximum likelihoodpoint-estimation
Suppose $X_1, X_2,…,X_n$ is a random sample from a $\text{Poisson} (\theta)$ distribution with probability mass function:
$$P(X=x)=\frac{\theta^ {x}e^{-\theta}}{x!}, x=1,2,…; 0<\theta$$
What is the maximum likelihood estimator for: $e^{-\theta}= P(X = 0)$?
I already found the MLE for the $\theta$. How do you then find the MLE of $P(X = 0)$ which is equal to $e^{-\theta}$ ?
Let $X_1,\dots,X_n$ be a random sample from density $f(x_i)=(\theta/x_i^2)\,I_{(\theta,\infty)}(x_i)$, for $\theta>0$. Since $I_{(\theta,\infty)}(x_i)=I_{(0, x_i)}(\theta)$, writing $x=(x_1,\dots,x_n)$, the likelihood function is $$ L_x(\theta) = \frac{\theta^n}{\prod_{i=1}^n x_i^2} I_{(0,x_{(1)})}(\theta) \, , \qquad (*) $$ in which $x_{(1)}=\min\{x_1,\dots,x_n\}$. The way the density is defined implies that there is no MLE in the usual sense, because the candidate $x_{(1)}\neq\arg\max_\theta L_x(\theta)$. In fact, $L_x(x_{(1)})=0$. If, for each $\theta>0$, we change the version of the density in just one point, and this doesn't change the family of sampling distributions, doing $f(x_i)=(\theta/x_i^2)\,I_{[\theta,\infty)}(x_i)$, then it's true that $\hat{\theta}_{\mathrm{MLE}}=X_{(1)}$.
This is not a serious difficulty, but it's a curious case in which the particular versions of the sampling densities chosen for the problem change the answer. In the second edition of DeGroot's "Probability and Statistics" there is a similar example starting on page 343.
Also, since $\mathrm{E}_\theta[X_i]=\infty$, for every $\theta>0$, asking for an MLE of this quantity doesn't make sense.
This is not intended as a full answer but as a series of hints. First, look at your likelihood.
We have
$$L(\theta) = \prod_{i=1}^n 1 \quad I\left( \theta \leq x_i \leq \theta + 1 \right)$$
where $I( x \in A)$ is the indicator function, returning $1$ if indeed $x \in A$ and zero otherwise. This is how the parameter $\theta$ enters our likelihood here.
Note here that for the likelihood to be non-zero, all the $x$s have to be in that interval. But then, this will happen if and only if the minimum is greater than $\theta$ and the maximum is smaller than $\theta+1$, would you agree? These conditions will give you an interval in which $\theta$ can lie.
But since this is an interval and the likelihood is constant in that interval, the mle is not unique. So if you are looking at this problem by yourself, you have the freedom of choice. That is, you can take a boundary point or a point in the interior, it does not matter. However, only one of these three estimators fullfils the condition. I leave it to you to find which one.
Best Answer
Invariance principle : The maximum likelihood estimator of the transform is the transform of the maximum likelihood estimator.