Does efficiency imply unbiased and consistency? If I can prove that for an estimator $\hat{k}( \theta)$ I can write: 
$$\frac{\partial l(X_1, \dots , X_n)}{\partial \theta} = a(n, \theta)(\hat{\theta} - \theta)$$ 
Am i sure that the estimator is unbiased? and consistent?
NB:


*

*$l$: is the log likelihood

*$X_1$ is generated from a regular model

*$\hat{\theta}$ is the estimator for $\theta$

*$a(\cdot,\cdot)$ is a function of $n$ and $\theta$ (without any particular meaning i guess)

 A: Estimators that are asymptotically efficient are not necessarily unbiased but they are asymptotically unbiased and consistent. An estimator that is efficient for a finite sample is unbiased. Since efficient estimators achieve the Cramer-Rao lower bound on the variance and that bound goes to 0 as the sample size goes to infinity efficient estimators are consistent.
A: In an attempt to... un-unanswer this question:
If one can write the derivative of the log-likelihood as the OP states, then this estimator equals the true value always, irrespective of the realized sample (what a dream, hey?). This is because we choose the estimator so as to make this derivative zero:
$$\hat \theta : \frac{\partial l(\hat \theta \mid X_1, \dots , X_n)}{\partial \theta} =0$$
So, if $$\frac{\partial l(\hat \theta \mid X_1, \dots , X_n)}{\partial \theta} =a(n, \theta) \cdot (\hat{\theta} - \theta) =0 \Rightarrow \hat \theta = \theta$$
(the case $a(n, \theta) =0$ is trivial). In such a case the estimator does not really need any of the usual properties, obviously.
