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Does consistency imply asymptotically unbiasedness? I know the statement doesn't work in the other direction. My guess is it does, although it obviously does not imply unbiasedness.

I think it wouldn't be too hard if one digs into measure theory and makes use of convergence in measure. But I have a gut feeling that this could be proved with only elementary probability theory.

But can anybody give me a proof? Thanks!

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Let $Y_n$ take the value $0$ with probability $(n-1)/n$ and take the value $n$ with probability $1/n$. Then $(Y_n)$ is a consistent sequence of estimators for zero but is not asymptotically unbiased: the expected value of $Y_n$ is $1$ for all $n$.

If we assume a uniform upper bound on the variance, $\mathrm{Var}(Y_n-X)\leq \mathrm{Var}(Y_n)+\mathrm{Var}(X)<C$ for all $n$, then consistency implies asymptotic unbiasedness. Let $Y_1,Y_2,\dots$ be a consistent sequence of estimators for a random variable $X$. This means that for all $\epsilon>0$ the probability of the event $|Y_n-X|>\epsilon$ tends to zero as $n\to\infty$. We want to show that $Y_1,Y_2,\dots$ is asymptotically unbiased: that the expected value of $Y_n-X$ tends to zero as $n\to\infty$. This follow from Cauchy-Schwarz for example: $$\mathbb E[|Y_n-X|]\leq \epsilon+\sqrt{\mathbb E [|Y_n-X|^2]\mathbb{P}[|Y_n-X|\geq\epsilon]}\to 0.$$

It is also useful to note that by Chebyshev's inequality, if the variance tends to zero then asymptotic unbiasedness implies consistency.

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  • $\begingroup$ Can you provide more details about how to get the inequality of $E(|Y_n-X|)$? Thanks. $\endgroup$
    – Lucas
    Feb 13 at 23:26
  • $\begingroup$ @Lucas here is some more details: $\mathbb E |Y_n - X | =\mathbb E \left[ | Y_n - X | {\bf 1} \{|Y_n - X | < \epsilon \} \right] + \mathbb E \left[ | Y_n - X | {\bf 1} \{|Y_n - X | \ge \epsilon \} \right]$. Use Cauchy-Schwarz on the second term and the first term is obviously at most $\epsilon$. $\endgroup$
    – James
    Jul 22 at 5:13

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