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This is the problem 2.11 from Lehman book "Theory of point estimation" 2-nd edition.

Construct a sequence $\{\delta_{n}\}$ of estimators of $g(\theta)$, satisfying

$$ \sqrt{n}[\delta_{n} - g(\theta)]\stackrel{d}{\to}\mathcal{N}[0,v(\theta)], \; v>0, $$

but for which the bias $b_{n}(\theta) = E[\delta_{n}] - g(\theta)$ does not tend to zero.

By another words, asymptotic normality does not guaranty that $\{\delta_{n}\}$ is unbiased or even that its bias tends to zero (p 439).

Lars
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  • Could you show your own efforts or present a solution you think might work? Although good job on referencing your reference material. – Zach466920 May 02 '15 at 16:03
  • If I had I would have written it :) now I am confused...

    ok, may be for beginning someone has an example:

    (a) $k_{n}[\delta_{n} - g(\theta)] \stackrel{d}{\to} H $, where $E[H] = 0$ -- this is asymptotic unbiasedness.

    (b) $\delta_{n} \stackrel{p}{\to} g(\theta)$ -- consistency.

    (a) plus (b) does not imply that $b_{n} {\to} 0$ as $n{\to} \infty$

     – Lars May 02 '15 at 16:18
  • In addition, I want to say that I have no doubts that this sequence exists... YES, it looks strange, BUT convergence in distribution does not imply the convergence of mean values. For example: let $X_{1}$, $X_{2}$, ... be $U(-1, 1)$ and set

    $$ Y_{n} = X_{n}, ; for ; |X_{n}| \leq 1- \frac{1}{n} $$

    and $$ Y_{n} = n, ; otherwise. $$

    In this case $E[X_{n}]\not\to E[Y]$

     – Lars May 02 '15 at 16:36
  • erm... mistake in the end of the previous comment:

    In this case $Y_{n} \to Y=U(-1, 1)$, but $E[Y_{n}]\not\to E[Y]$.

    (A. Gut p 183)

     – Lars May 02 '15 at 16:43
  • You should probably put this in your question under a subtitle called my own attempt. – Zach466920 May 02 '15 at 16:47

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