13
$\begingroup$

Suppose that $ X_{0},X_{1},\ldots,X_{n} $ are i.i.d. random variables that follow the Poisson distribution with mean $ \lambda $. How can I prove that there is no unbiased estimator of the quantity $ \dfrac{1}{\lambda} $?

$\endgroup$

migrated from mathoverflow.net Dec 16 '13 at 4:04

This question came from our site for professional mathematicians.

  • 3
    $\begingroup$ I presume you mean, "lambda?" Anyways, this isn't appropriate for MO. $\endgroup$ – Noah S Dec 16 '13 at 0:12
  • 3
    $\begingroup$ Is this for some subject? It looks like a fairly standard textbook exercise. Please check the self-study tag, and its tag wiki info and add the tag (or please give some indication how else such a question arises). Note that such questions, while welcome, place some requirements on you (and restrictions on us). What have you tried? $\endgroup$ – Glen_b Dec 16 '13 at 5:33
  • 2
    $\begingroup$ You should be able to use a similar argument to the one here. $\endgroup$ – Glen_b Dec 16 '13 at 5:36
11
$\begingroup$

Assume that $g(X_0, \ldots, X_n)$ is an unbiased estimator of $1/\lambda$, that is, $$\sum_{(x_0, \ldots, x_n) \in \mathbb{N}_0^{n+1}} g(x_0, \ldots, x_n) \frac{\lambda^{\sum_{i=0}^n x_i}}{\prod_{i=0}^n x_i!} e^{-(n + 1) \lambda} = \frac{1}{\lambda}, \quad \forall \lambda > 0.$$ Then multiplying by $\lambda e^{(n + 1) \lambda}$ and invoking the MacLaurin series of $e^{(n + 1) \lambda}$ we can write the equality as $$ \sum_{(x_0, \ldots, x_n) \in \mathbb{N}_0^{n+1}} \frac{g(x_0, \ldots, x_n)}{\prod_{i=0}^n x_i!} \lambda^{1 + \sum_{i=0}^n x_i} = 1 + (n + 1)\lambda + \frac{(n + 1)^2 \lambda^2}{2} + \ldots , \quad \forall \lambda > 0, $$ where we have an equality of two power series of which one has a constant term (the right-hand side) and the other doesn't: a contradiction. Thus no unbiased estimator exists.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.