When we want to find an estimator of a parameter of a distribution function using the method of moments (order 1), is it true that the estimator found is unbiased ? I am pretty sure that this is not the case but I am a little bit confused. For instance, if we take a sample of i.i.d $X_1,\ldots,X_n$ with law
\begin{gather*} f(x,\theta) = \theta(\theta+1)x^{\theta-1}(1-x)1_{]0,1[}(x), \ \theta>0, \end{gather*}

then by using the method of moments of order 1, we find $\hat{\theta}_{MOM} = \frac{2\bar{X}}{1-\bar{X}}$. I succeeded to prove that it is asymptotically unbiased, but is it also true for small sample (saying by definition of this method) ? I need this, because I want to prove that it is an inefficient estimator (not asymptotically!).

On wikipedia, it is written that an unbiased estimator $T$ of $\theta$ is efficient if
\begin{gather*} e(T) = \frac{1/I(\theta)}{Var(T)} = 1. \end{gather*}

If it is biased estimator, I think that this is
\begin{gather*} e(T) = \frac{(g'(\theta))^2/I(\theta)}{Var(T)} = 1, \end{gather*} by the Cramér-Rao lower bound theorem, where $E_{\theta}[T] = g(\theta)$, but too difficult to compute in this case..

Thank you for your help.

  • $\begingroup$ How is the accepted answer addressing your question? There is no "single observation $x$" here since the question invokes a i.i.d. sample of size $n$ and, presumably, $n\ne1$. Please explain. $\endgroup$
    – Did
    Dec 13, 2014 at 7:50

1 Answer 1


What is the expectation of your $\hat \theta_{MOM}$ for a single observation $x$? It is clearly not $\theta$, so it cannot be an unbiased estimator.

  • $\begingroup$ Yes... Shame on me... Thank you. So, is my definition of efficiency correct? How to prove the innefficiency of this biased estimator? $\endgroup$
    – MCrassus
    Nov 14, 2014 at 18:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.