# Method of moments: unbiased estimator for small samples

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..

• 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. – Did Dec 13 '14 at 7:50
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.