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Exercise: Suppose that $X\sim\operatorname{Exp}(\theta)$, show that no unbiased estimator for $\theta$ exists. Hint: use the fact that $X$ is a complete and sufficient statistic for $\theta$.

I have the following definitions to work with:

Definition 1: An estimator $d(X)$ is unbiased for $\theta$ if $\operatorname{E}_\theta(d(X)) = \theta.$

Definition 2: Statistic $T$ is sufficient for $\theta\in\Omega$ if the conditional distribution of $X$ given $T$ is known (does not depend on $\theta$).

Definition 3: Statistic $T$ is complete for $\theta\in\Omega$ if for any Borel function $f$ we have that $\operatorname{E}_\theta f(T) = 0$ for all $P_\theta\in\Omega$ implies that $f(T) = 0$.

What I've tried: I'm not sure how to show that no unbiased estimator exists by using completeness or sufficiency, but I think it's possible in fact use a proof by contradiction, as is done here. If $d(X)$ is an unbiased estimator we have that $$\operatorname{E}_\theta d(X) = \int d(x)\theta e^{-\theta x}dx = \theta.$$ Unfortunately I don't know how to proceed from here. I've thought of using sufficiency and completeness to show that the above equation is not possible, but I haven't succeeded.

Question: How do I show that there exists no unbiased estimator for $\theta$?

Thanks in advance!

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So you have

$$\int d(x) e^{-\theta x} dx = 1.$$

If you are allowed to exchange derivative (w.r.t. $\theta$) and integral you obtain

$$-\int d(x) x e^{-\theta x} dx = 0,$$

so $x d(x) = 0$ (and then $d(x) = 0$ a.e.) using completeness. (Which gives the contradiction.)

This exchange is allowed for exponential families, see e.g. Lehmann/Romano (2005), Theorem 2.7.1 therein, but I do not know whether you "may" use this tool.

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  • $\begingroup$ Thanks for your answer! I'm not sure whether I'm allowed to use this. Could elaborate a bit more on what you're doing? $\endgroup$ – titusAdam Dec 14 '17 at 15:36
  • $\begingroup$ Oh, there was a mistake in my solution. (I was taking the derivative w.r.t. $\theta$, but I was writing down the derivative w.r.t. $x$). It is now corrected. Regarding the exchange this really is explained nicely in the cited reference. One somehow needs to apply the dominated convergence theorem (as always in these circumstances), and here one uses some specific properties of the exponential function to justify it. $\endgroup$ – Mathias Vetter Dec 14 '17 at 17:19
  • $\begingroup$ Thanks again! One more question: do you use the fact that $X$ is a sufficient statistic in this approach? Is it necessary for $X$ to be sufficient when using this approach? $\endgroup$ – titusAdam Dec 20 '17 at 17:14
  • $\begingroup$ I didn't use it. So maybe there is some other solution which avoids the exchange of integral and derivative. $\endgroup$ – Mathias Vetter Jan 3 '18 at 10:40
  • $\begingroup$ @titusAdam $X$ is trivially sufficient here. $\endgroup$ – StubbornAtom Jan 14 at 15:59

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