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Let TnT_nTn​​ be an estimator, based on a sample x1,x2,…,xnx_1,x_2,\ldots,x_nx1​,x2​,…,xn​​, of the parameter θ\thetaθ​. Then TnT_nTn​​ is a
Question

Let TnT_n​ be an estimator, based on a sample x1,x2,,xnx_1,x_2,\ldots,x_n​, of the parameter θ\theta​. Then TnT_n​ is a consistent estimator of θ:\theta:​​

A.

​​​​​P(Tnθ>ϵ)=0, ϵ>0P(T_n-\theta>\epsilon)=0,\;\forall\epsilon>0​​

B.

​​​​​​​​​​​​​​​​​​​P(Tnθ<ϵ)=0P(|T_n-\theta|<\epsilon)=0​​

C.

limnP(Tnθ>ϵ)=0, ϵ>0 \lim_{n\to\infty}P(|T_n-\theta|>\epsilon)=0,\;\forall\epsilon>0​​

D.

limnP(Tnθ<ϵ)=0, ϵ>0 \lim_{n\to\infty}P(|T_n-\theta|<\epsilon)=0,\;\forall\epsilon>0​​

Correct option is C

Given:
TnT_n​ is an estimator of the parameter θ.\theta.​​
Formula Used:
An estimator TnT_n​ is consistent if
limnP(Tnθ>ϵ)=0,ϵ>0.\lim_{n\to\infty}P(|T_n-\theta|>\epsilon)=0,\quad\forall\epsilon>0.​​
Solution:
By the definition of consistency, the probability that the estimation error exceeds any positive ϵ \epsilon​ tends to zero as n.\to\infty.​​
Hence,
limnP(Tnθ>ϵ)=0,ϵ>0.\lim_{n\to\infty}P(|T_n-\theta|>\epsilon)=0,\quad\forall\epsilon>0.​​
The correct answer is (c).

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