Suppose $X_1,....,X_n$ are $iid$ random variables and for each of them $Variance(X_i)= \sigma^2$. $a_1...a_n$ are also real numbers and $\sum_{i=1}^n a_i = 1$ If $S = \sum_{i=1}^na_iX_i$, prove $var(S)$ is minimum if $a_i = 1/n, \ i=1,...,n$

So i did following:

$Var(S) = \sum_{i=1}^na_i^2Var(X_i) + 2\sum_{i<j}a_ia_jcov(X_i, X_j)$

I'm not sure how to take this from here. Any ideas?

| cite | improve this question | | | | |
  • 3
    $\begingroup$ What is $cov(X_i, X_j)$ if $X_i$ and $X_j$ are independent? $\endgroup$ – jbowman Dec 26 '17 at 19:51
  • $\begingroup$ Are the a$_i$s all nonnegative. $\endgroup$ – Michael R. Chernick Dec 26 '17 at 23:21
  • $\begingroup$ @Michael Good question--but it doesn't matter! $\endgroup$ – whuber Jan 16 '19 at 13:06

Since $X_i$'s are independent, $\operatorname{Cov}(X_i,X_j)=0$ if $i\neq j$. It follows that $$\operatorname{Var}(S)=\operatorname{Var}\left(\sum_{i=1}^na_iX_i\right)=\sigma^2\sum_{i=1}^na_i^2.$$ We can bypass the lengthy Lagrangian method by using Cauchy-Schwarz inequality: $$1=\left(\sum_{i=1}^na_i\right)^2=\left(\sum_{i=1}^n\frac{a_i\sigma}{\sigma}\right)^2\leq\left(\sum_{i=1}^na_i^2\sigma^2\right)\left(\sum_{i=1}^n\frac{1}{\sigma^2}\right)=\frac{n\operatorname{Var}(S)}{\sigma^2}.$$ Now if $a_i=1/n$, then $$\operatorname{Var}(S)=\sigma^2\sum_{i=1}^n\frac{1}{n^2}=\frac{\sigma^2}{n},$$ which is exactly the lower bound given by Cauchy-Schwarz.

| cite | improve this answer | | | | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.