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} $?

  • 3
    $\begingroup$ I presume you mean, "lambda?" Anyways, this isn't appropriate for MO. $\endgroup$
    – Noah S
    Dec 16, 2013 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, 2013 at 5:33
  • 2
    $\begingroup$ You should be able to use a similar argument to the one here. $\endgroup$
    – Glen_b
    Dec 16, 2013 at 5:36

1 Answer 1


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.

  • $\begingroup$ How could you obtain the first equality? $\endgroup$
    – TrungDung
    Nov 19, 2020 at 21:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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