# Comparing efficiency between estimators

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 $$o(1), n\to\infty$$.

• 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).
– whuber
Jul 11 at 19:20
• @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 $B$. Jul 11 at 19:29
• But why use $\hat\theta_2$ at all, which is systematically biased? Just replace it with $\hat\theta_2-B.$
– whuber
Jul 11 at 19:33
• 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} Jul 11 at 20:40

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 $$V_1$$.
• 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!
• 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 $V/n$ Jul 12 at 2:12
• 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 $o_p(1)$ Jul 13 at 7:41