X has an uniform distribution on interval $(0,\theta]$ where $\theta$ is a positive parameter Given the estimator: $$T(X_1,X2, \ldots, X_n)=\frac{2}{n} \sum_{i=1}^n X_i$$ Find whether this estimator is consistent (i am not certain whether it is a best english equivalent of what i mean). But in symbols i mean:
Generaly speaking i know it is satisfied by $R(\theta)=\frac{\theta^2}{3n}+0$ and the fact that this expression converges to zero. However i do not understand why and where does this answer comes from.
Assuming that $X_i$ are i.i.d.
1) You can simply use the WLLN and continuous mapping theorem to deduce that $\frac{1}{n}\sum_{i=1}^n X_i \xrightarrow{p}EX = \frac{\theta}{2}$, hence $\frac{2}{n}\sum_{i=1}^n X_i \xrightarrow{p} \theta$.
2) You can use the theorem that if $\lim_{n\to \infty} MSE(\hat{\theta}_n) = 0$ then $\hat{\theta}_n$ converges in probability to $\theta$. In your case it is straightforward to show that $$ MSE(\hat{\theta}) = Var(\hat{\theta})+b^2(\hat{\theta}) = \frac{2^2}{n^2}nVar(X_i) = \frac{\theta^2}{3n} $$ which is clearly goes to $0$ as $n \to \infty$.
(Note that $EX_i = \frac{\theta}{2}$ and $Var(X_i) = \frac{\theta^2}{12}$)