I have shown that the expected value of $\displaystyle \delta (x)$ is $\displaystyle \theta+1$ How possibly can $\displaystyle \delta (x)$ be an consistent estimator? Thanks! casper
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$\displaystyle P\left(\left|\hat\theta-\theta\right|>\epsilon\right)= {1\over n} \to 0$ while $\displaystyle E\left(\hat\theta-\theta\right)^2=(n)^2{1\over n}+0\left(1-{1\over n}\right)=n\to \infty$
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