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Could someone clarify if I'm interpreting this question correctly?

As the sample size increases and approaches infinity, then the expected value of the estimator would approach $0.$ The estimator is biased because $θ$ is not equal to $θ/n.$

But, if $n = 1,$ then wouldn't the expected value${} = θ$? Is this how the statistician's estimator would become unbiased?

Clarification appreciated!

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We don't have to take limit of $n$ to $\infty$, currently we have

$$\mathbb{E}[\hat{\theta}]=\frac{\theta}{n}$$

We want to find $k$ such that

$$\mathbb{E}[k\hat{\theta}]=\frac{k\theta}{n}=\theta$$

Can you solve for $k$?

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  • $\begingroup$ So that would solve for k = n? $\endgroup$ – MathsHelp May 8 '18 at 17:49
  • $\begingroup$ yup, can you an unbiased estimator $\tilde{\theta}$ such that $\mathbb{E}[\tilde{\theta}]=\theta$ now? $\endgroup$ – Siong Thye Goh May 8 '18 at 17:51
  • $\begingroup$ Wouldn't that just be any number =/= 0? Unless k represents something specifically? $\endgroup$ – MathsHelp May 8 '18 at 17:54
  • $\begingroup$ i thought you have conclude that $k=n$, that is $\mathbb{E}[k\hat{\theta}]=\theta$. can you find $\tilde{\theta}$ such that $\mathbb{E}[\tilde{\theta}]=\theta$? $\endgroup$ – Siong Thye Goh May 8 '18 at 17:56
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    $\begingroup$ $$V(n \hat{\theta})=n^2 V(\hat{\theta})=\frac{n^3\sigma^2}{n-1}$$ $\endgroup$ – Siong Thye Goh May 8 '18 at 19:12

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