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Let $X_1,\ldots, X_n$ be a random sample from a population with mean $\mu$. What condition must be imposed on $a_1,\ldots,a_n$ such that $$a_1X_1+\ldots+a_nX_n$$ is an unbiased estimator of $\mu$.

I am new to statistics and am not sure how to handle this problem. The definition for an unbiased estimator that I am using is:

$\hat{\Theta}$ is an unbiased estimator of $\theta$ (or $\mu$ in this case) if $E(\hat{\Theta})=\theta$ for all $\theta$.

Here $\hat{\Theta}=a_1X_1+\ldots+a_nX_n$. Now my intuition would say that $a_i=\frac{1}{n}$ for $i=1,\ldots n$ definitely makes $\hat{\Theta}=a_1X_1+\ldots+a_nX_n$ an unbiased estimator. Maybe also any combination satisfying $\sum\limits_{i=1}^{n}a_n=1$ but I am not sure about this one. However I am not sure how to prove this using the definition of the expected value. I am pretty sure this question should not be too difficult but I'm new to stats, so any help would be greatly appreciated!! Thanks in advance

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In order to have unbiased estimator, you should have $\mathbb{E}(\hat{\Theta})=\mu$. Now you should have $\mathbb{E}(a_1X_1+\ldots+a_nX_n)=\mu$. By linear property of expectation you get: $$ \mathbb{E}(a_1X_1+\ldots+a_nX_n)=a_1\mathbb{E}(X_1)+\ldots+a_n\mathbb{E}(X_n)=\sum_{i=1}^n a_i\mu. $$ Unbiased condition means that $\sum_{i=1}^n a_i\mu=\mu$ which is (for $\mu\neq0$):

$$ \sum_{i=1}^n a_i=1 $$

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  • $\begingroup$ Is this because $E(X_i)=\mu$ for $i=1,\ldots, n$? $\endgroup$ – Slugger Sep 2 '13 at 22:51
  • $\begingroup$ Exactly. That is the reason. $\endgroup$ – Arash Sep 2 '13 at 22:52
  • $\begingroup$ Allright thanks so much that makes a lot of sense! There is one more part that is confusing to me that you might be able to explain. The definition of unbiased estimator requires $\mathbb{E}(\hat{\Theta})=\mu$ for all $\mu$, with emphasis on the for all. Should I interpret that as 'whatever the mean might be in this case'? $\endgroup$ – Slugger Sep 2 '13 at 22:55
  • $\begingroup$ Yes, an unbiased estimator should not touch the mean value of the original random variable. $\endgroup$ – Arash Sep 2 '13 at 22:57
  • $\begingroup$ But it must be true for all $\mu$ to be an unbiased estimator. But a population has only one mean right? $\endgroup$ – Slugger Sep 2 '13 at 22:59

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