Let $X_1,X_2,…X_n$ be random sample from $U(\theta,0)$ where
$\theta<0$. Let $T_n=\min(X_1,X_2…X_n)$ the which of the following
estimators are consistent estimator of $\theta$$A=T_n$
$B=T_n-1$
$C=T_n+\frac{1}{n}$
$D=T_n-1-\frac{1}{n^2}$
$A=E(T_n)=\lim_{n \to \infty}\dfrac{n\theta}{n+1}=\theta$
$B=E(T_n-1)=\lim_{n \to \infty}\dfrac{n\theta-n-1}{n+1}=\dfrac{\theta-1-\frac{1}{n}}{1+\frac{1}{n}}=\theta-1$
$C=E(T_n+\frac{1}{n})=\lim_{n \to \infty}\dfrac{n\theta}{n+1}+\dfrac{1}{n}=\theta$
$D=E(T_n-1-\frac{1}{n^2})=\lim_{n \to \infty}\dfrac{n\theta-n-1}{n+1}-\dfrac{1}{n^2}=\dfrac{\theta-1-\frac{1}{n}}{1+\frac{1}{n}}-\dfrac{1}{n^2}=\theta-1$
$A$ and $C$ is correct can anyone confirm if it is correct or not ?
Best Answer
Since $Y_i=-\frac{1}{\theta}(X_i-\theta)\stackrel{\text{i.i.d}}\sim U(0,1)$, we have
$$Y_{(1)}=-\frac{1}{\theta}(T_n-\theta)\sim \text{Beta}(1,n)$$
Therefore,
$$\operatorname E_{\theta}(T_n)=\theta-\frac{\theta}{n+1}=\frac{n\theta}{n+1}$$
And $$\operatorname{Var}_{\theta}(T_n)=\frac{n\theta^2}{(n+1)^2(n+2)}$$
Clearly, $$\operatorname E_{\theta}(T_n)\to \theta \quad\text{ and } \quad\operatorname{Var}_{\theta}(T_n)\to 0$$
Similarly, $$\operatorname E_{\theta}\left(T_n+\frac{1}{n}\right)\to \theta \quad\text{ and } \quad\operatorname{Var}_{\theta}\left(T_n+\frac{1}{n}\right)\to 0$$
By the sufficient condition for convergence in probability, both $T_n$ and $T_n+\frac{1}{n}$ are consistent for $\theta$.