I read one example in my notes, but I couldn't find out how the answer in my notes is derived.
If $x_1,...,x_n$ are realizations of a random variable distributed with the following PDF:
$f(z; \theta)= ( \theta +1)x^ \theta$ if $0<x<1$, and $f(z; \theta)=0$ otherwise. How do we reach the Method of moments estimator for $\theta$:
$$ \frac {2 \bar{x}-1}{1- \bar{x}} $$
Who can describe how we get this?
First find $E(X)=\int_0^1(\theta+1)x^{\theta}.xdx=(\theta+1)\int_0^1x^{\theta+1}dx=(\theta+1)\dfrac{x^{\theta+2}}{\theta+2}]_{x=0}^{x=1}=\dfrac{\theta+1}{\theta+2}$. Now find the first order mean of the sample i.e. $\bar{X}$.
Now set these two equal, and solve for $\theta$. This gives you the method of moments estimate of $\theta$.
$\dfrac{\theta+1}{\theta+2}=\bar{X}\implies\theta+1=\theta \bar{X}+2\bar{X}\implies\theta=\dfrac{2\bar{X}-1}{1-\bar{X}}$