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?

  • 3
    $\begingroup$ What is $cov(X_i, X_j)$ if $X_i$ and $X_j$ are independent? $\endgroup$
    – jbowman
    Dec 26, 2017 at 19:51
  • $\begingroup$ Are the a$_i$s all nonnegative. $\endgroup$ Dec 26, 2017 at 23:21
  • $\begingroup$ @Michael Good question--but it doesn't matter! $\endgroup$
    – whuber
    Jan 16, 2019 at 13:06

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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