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Let $\vec{X}=(X_1,X_2,\ldots,X_n)$ be a random sample of some random variable $X$ whose distribution $F$ depends on some real-valued parameter $\theta_F$. Let $\hat\theta(\vec{X})$ be an unbiased estimator of $\theta_F$. Let $(\vec{x}_n)$ be a sequence of observations of $\vec{X}$. Is it true that the sequence $(\hat\theta(\vec{x}_n))$ will have $\theta_F$ as a cluster point?

My guess is that it should be true, but I am not sure how to prove this from the fact that the estimator is unbiased. Any tips?

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What if the sequence of observations is a constant sequence? For example let theta(a,a,...,a) be different from the actual parameter. Now consider the sequence of observations (a,a,...,a),(a,a,...,a),(a,a,...,a),... –  Amr Nov 7 '12 at 20:07
Now the sequence theta(a,...,a), theta(a,...,a),.... will not have theta (the actual parameter) as a cluster point. –  Amr Nov 7 '12 at 20:08
You are right about that. Hmm...I am trying to get a feel for what unbiased means in terms of actual data. I guess I will have to think harder. –  echoone Nov 7 '12 at 20:43

1 Answer 1

No. In particular, if the possible range of $\theta_F$ is compact, then you can create an unbiased estimator $\tilde\theta$ which only takes values on the boundary, and so no interior point will be a cluster point of any sequence of estimates even if you put additional constraints (e.g. independent samples) to deal with Amr's concern. If $\theta_F$ takes values on the entire real line you can build an unbiased estimator which only takes integer values by considering nested compact sets.

In addition, a biased estimator can have the true value be a cluster point. For example, if $X_i$ are normal with known variance $\sigma^2$ and we are trying to estimate the mean, we can use the biased estimator $\tilde\theta =\frac1n \sum x_i + 1$, which is clearly biased but has every point on the real line as a cluster point if we have an infinite sequence of independent random samples, since the sampling distribution of $\tilde \theta$ has support on the entire real line.

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