\begin{aligned} &X_{1}, X_{2}, \ldots, X_{n} \text { are n samples from a population with density }\\ &f(x)=\left\{\begin{array}{ll} (\theta+1) x^{\theta} & 0<x<1 ; \quad \theta>0 \\ 0 \text { otherwise } \end{array}\right. \end{aligned}
I have found that method of moment for this is $\frac{2 \sum_{i=1}^{n} x_{i}-n}{n-\sum_{i=1}^{n} x_{i}}$ and MLE is $\frac{-\left(n+\sum_{i=1}^{n} \ln \left(x_{i}\right)\right)}{\sum_{i=1}^{n} \ln \left(x_{i}\right)}$ .
Now I have this question ; Suppose $\mathrm{n}=2,$ and $X_{1}=.125, X_{2}=.375,$ what is your (moment) esti- mated value of $\theta$ ?
The value turns out to be negative but parmeter $\theta > 0$ so what is method of moment estimate in this case ?
The fact that the MoM estimate can give an estimation that is not acceptable, out of the parameter support, is one of the reason why it is not always a good estimator, moreover if you compare it with the ML one.
Usually your MoM estimator can be expressed in a more suitable way
$$\hat{\theta}_{MM}=\frac{2\overline{X}_n-1}{1-\overline{X}_n}$$
EDIT: in this case, both the estimators are evidently out of the specified parameter support.
Actually your $f(x)$ is a beta distribution, $Beta(\theta+1;1)$ thus it is well defined for any $\theta>-1$.
Your exercise states $\theta>0$ and states a particular pair of observation to get you in trouble... thus the solution is the following
$$\hat{\theta}_{MM}=\max\left\{\frac{2\overline{X}_n-1}{1-\overline{X}_n};0^+\right\}$$
Actually, setting $\theta>0$ or $\theta\geq 0$ the only change is that if $\theta=0$ $X\sim U(0;1)$ thus I will write
$$\hat{\theta}_{MM}=\max\left\{\frac{2\overline{X}_n-1}{1-\overline{X}_n};0\right\}$$
Same story for MLE