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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} $?

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    $\begingroup$ I presume you mean, "lambda?" Anyways, this isn't appropriate for MO. $\endgroup$
    – Noah S
    Dec 16, 2013 at 0:12
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    $\begingroup$ 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
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    $\begingroup$ 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

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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.

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

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