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Suppose that $\hat \theta_1, \hat \theta_2$ are two estimators of $\theta$. Furthermore, assume that \begin{align} \sqrt{n}(\hat \theta_1-\theta)\overset{d}{\to}N(0,V_1)\\ \sqrt{n}(\hat \theta_2-\theta-B)\overset{d}{\to}N(0,V_2), \end{align} where $V_{1}=\lim_{n\to\infty} V_{1,n}$, $V_{2}=\lim_{n\to\infty} V_{2,n}$ and $B:=B_n$ are all known (including $V_{1,n},V_{2,n}$).

Does it make sense to talk about relative efficiency between $\hat \theta_1$ and $\hat \theta_2$ by comparing $V_1$ and $V_2$ (or even $V_{1,n}$ and $V_{2,n}$) in the presence of term $B$?

For example, if I show that $V_{1,n}/n-V_{2,n}/n-B<0$, then can I say that $\hat\theta_1$ is relatively more efficient that $\hat\theta_2$?

*Based on the answer and a comment, I clarify that the term $B$ is relevant in the expression for $\hat \theta_2$. Hence it depends on $n$. $B$ itself is $o(1), n\to\infty$.

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    $\begingroup$ If you know $B$ you would just subtract it from $\hat\theta_2.$ If you don't know $B,$ how can you hope to compare the estimators? Usually, ARE is used to assess relative costs of using estimators (in terms of sample size requirements). That suggests comparing $V_1^2/n$ to $B^2 + V_2^2/n$ -- for which you must know $B$ (or have a sufficiently accurate guess of it). $\endgroup$
    – whuber
    Jul 11 at 19:20
  • $\begingroup$ @whuber I've added more details to the question (roughly, any term there is known). I'm not too confident to compare the two estimators because for $\hat \theta_2$ there is the term $B$. $\endgroup$ Jul 11 at 19:29
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    $\begingroup$ But why use $\hat\theta_2$ at all, which is systematically biased? Just replace it with $\hat\theta_2-B.$ $\endgroup$
    – whuber
    Jul 11 at 19:33
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    $\begingroup$ Possibly you wanted to express it like $$\begin{align} \sqrt{n}(\hat \theta_1-\theta)\overset{d}{\to}N(0,V_1)\\ \sqrt{n}(\hat \theta_2-\theta)\overset{d}{\to}N(B,V_2), \end{align}$$ $\endgroup$ Jul 11 at 20:40

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This is a reasonable question in settings where you 'know' $B$ in theory but you can't estimate it effectively, or if you have a bound rather than an actual value or whatever. For example, I've had settings a bit like this where there's a slightly misspecified model and the bias in an estimator is bounded above by the log likelihood ratio between the assumed and true models. A sensible comparison is of the mean squared errors, so $B^2+V_2$ vs $V_1$.

There are extensions of the Cramer-Rao inequality for MSE. Asymptotic efficiency results such as the local asymptotic minimax theorem carry over -- the local asymptotic minimax theorem bounds any 'bowl-shaped symmetric loss', which includes MSE.

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  • $\begingroup$ Because $\hat\theta_i-\theta \sim \mathcal{N}(0, V_i/n) + B\mathcal{I}(i=2),$ the mean squared errors are not $B^2+V_2$ and $V_1:$ they are $B^2+V_2/n$ and $V_1/n.$ That makes a huge difference in the comparison! $\endgroup$
    – whuber
    Jul 11 at 21:18
  • $\begingroup$ Yes, that's right. When I did it, the bias was on the same scale as the standard deviation, so it was $B/\sqrt{n}$ vs $V/n$ $\endgroup$ Jul 12 at 2:12
  • $\begingroup$ So, just to be clear, you are analyzing a situation where the bias vanishes asymptotically, right? $\endgroup$
    – whuber
    Jul 12 at 13:39
  • $\begingroup$ Depends what you mean -- it doesn't vanish relative to the standard error (which is what I'd mean by 'asymptotically unbiased'), but it's $O(n^{-1/2})$ and a fortiori $o_p(1)$ $\endgroup$ Jul 13 at 7:41

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