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Doing some exercises I stumbled upon this tricky one:

Suppose we have an independent random sample $(X_1, ... , X_n)$ with $X_i \sim Poisson(\lambda)$. Define $\theta = e^{-\lambda}$.

Let $$ \delta_X = \begin{cases} 1 \quad if \quad X = 0 \\ 0 \quad if \quad X\geq 1 \end{cases} $$

Show that $T_n = n^{-1} \sum_{i = 1} ^ n \delta_{X_i}$ for $n = 1$ is the unique estimator with minimum variance for $\theta$. If $n \geq 1$ is $T_n$ an unbiased and efficient estimator?

Let's stick to the easiest case $n = 1$. Showing that the estimator is unbiased is trivial. I know that if I prove that $T_1$ is efficient I also have the uniqueness so I tried to compute the variance and compare it to the lower bound given by the Cramer - Rao theorem. I inverted the functional and obtained $$ \lambda = -log{\ \theta} $$ I substituted it into the likelihood function but then computing the Fisher information is quite tricky even in the easiest case.

Is there another approach to work it out? Does the same approach hold with $n \geq 1$?

Thank you for your attention, I'd be glad if you could help me!

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    $\begingroup$ Note that $\delta_X \sim \text{Bernoulli}(\theta)$; reframing the problem that way may make it simpler to solve. $\endgroup$ – jbowman Jan 9 at 0:04
  • $\begingroup$ Yes, it did. Thank you very much! I'll write down the details tomorrow. $\endgroup$ – Simone Vernengo Jan 9 at 0:16
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    $\begingroup$ You might want to write up your findings as an answer and accept it (that's OK), as that way people will a) know it's been answered and b) be able to look up the answer if they have the same or a similar problem. $\endgroup$ – jbowman Jan 9 at 1:27
  • $\begingroup$ stats.stackexchange.com/a/436402/119261 $\endgroup$ – StubbornAtom Jan 9 at 6:13

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