Please refer to the question in image
I have tried to find $ E(x) $ but i ended up with $\overline x $ = $\frac{\theta + 1}{\theta} $ which statisfies no option , i also tried to find $ E(x-1)^2 $ but then it gives $\frac{\sum (x-1)^2}{n}$= $\frac{\theta + 2}{\theta^2}$.
Please suggest the correct method.
Best Answer
Since the first order (population) moment $1=E\left(\frac{1}{n}\sum\limits_{i=1}^n X_i\right)$ is independent of $\theta$, we can consider the second order raw moment $\frac{1}{\theta}+1=E\left(\frac{1}{n}\sum\limits_{i=1}^n X_i^2\right)$.
By method of moments,
$$\frac{1}{n}\sum_{i=1}^n X_i^2=\frac{1}{\theta}+1$$
So a valid method of moments estimator of $\theta$ is simply $$\hat\theta(X_1,\ldots,X_n) =\frac{1}{\frac{1}{n}\sum\limits_{i=1}^n X_i^2-1}$$
Since $E\left[\frac{1}{n}\sum\limits_{i=1}^n (X_i-1)^2\right]=\frac{1}{\theta}$, we again have by method of moments
$$\frac{1}{n}\sum\limits_{i=1}^n (X_i-1)^2=\frac{1}{\theta}$$
Thus giving the estimator $$\hat\theta'(X_1,\ldots,X_n)=\frac{n}{\sum\limits_{i=1}^n (X_i-1)^2}$$
Here we equated sample variance with population variance, i.e. considering central moments instead of raw moments. Looking at the options, this seems to be the convention followed in the question.