# Efficient Estimator from Insufficient Statistic

Suppose that I have a statistic $$T(X)$$, and I know for sure that it is not sufficient to estimate a parameter $$\theta$$.

Is it still possible to have an estimator $$\hat\theta(T(X))$$ that is efficient (under convex loss), or is there a theorem (something like a reverse Rao-Blackwell) that says this is impossible?

You may answer the question under the efficiency definition of attaining CRLB for unbiased estimators or mean squared error averaged over the real line, or if it will help some other performance measure that is more amenable towards answering the question.

• As far as I understand, efficiency comes into play after consistency. If we have a consistent estimator, then we are interested to know how rapidly that estimator converges to the parameter of interest (efficiency). So I think you can also ask if there is a consistent estimator which is not sufficient. – StubbornAtom Nov 1 '18 at 16:39
• @StubbornAtom I was thinking of maximum likelihood estimates of parameters from non exponential family. They are not sufficient but asymptotically consistent and perhaps efficient. But once infinite data involved maybe things are different. – Cagdas Ozgenc Nov 1 '18 at 16:46
• Not quite true. The $U(0,\theta)$ family is non-exponential. Yet for a sample of size $n$ drawn from this distribution, $\hat\theta_{\text{MLE}}=\max_{1\le i\le n} X_i$ is a sufficient statistic as well as a consistent estimator for $\theta$. – StubbornAtom Nov 1 '18 at 16:51
• @StubbornAtom That's the changing support case (which is an exception to Pitman Koopman theorem) . You missed the point of my comment. Basically it doesn't always apply. – Cagdas Ozgenc Nov 1 '18 at 17:03
• Thinking about it: For a sample of size $n$ drawn from Cauchy$(\theta,1)$ distribution, some sample quantiles like median are consistent estimators of $\theta$. However, the only set of sufficient statistic is the sample itself or the full set of order statistics. But I think I read that the sample median is inefficient in this case. (Related: stats.stackexchange.com/questions/373526/…). – StubbornAtom Nov 2 '18 at 7:14

Since [under assumptions of its existence] a minimal sufficient statistic $$S_n$$ is a function of a sample $$(X_1,\ldots,X_n)$$,$$S_n=S_n(X_1,\ldots,X_n)$$ an efficient estimator $$\hat{\theta}(S)$$ can be written as$$\hat{\theta}(S(X_1,\ldots,X_n))$$which makes the question difficult to understand.