# condition on $a_1,\ldots,a_n$ so that $a_1X_1+\ldots+a_nX_n$ is an unbiased estimator of the mean

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

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 $\mu\neq0$):
$$\sum_{i=1}^n a_i=1$$
• Is this because $E(X_i)=\mu$ for $i=1,\ldots, n$? – Slugger Sep 2 '13 at 22:51
• 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'? – Slugger Sep 2 '13 at 22:55
• But it must be true for all $\mu$ to be an unbiased estimator. But a population has only one mean right? – Slugger Sep 2 '13 at 22:59