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$\newcommand{\Var}{\operatorname{Var}}$ $\newcommand{\Cov}{\operatorname{Cov}}$ I wonder how come that the variance $\Var(sX)=s^2\Var(X)$, but $\Var\left(\sum_{i=1}^s X_i\right)=\sum_{i=1}^s \Var(X_i)+\sum_{i=1,j=1 ,i\neq j}^{s}2\Cov(X_i,X_j)=\sum_{i=1}^s \Var(X_i)=s\Var(X)$ for $s$ uncorrelated random variables with equal distribution. What is the main reason for that?

Thank you.

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    @ChrisEagle: Thanks, I meant to write it.2012-10-25

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$\newcommand{\Var}{\operatorname{Var}}$ $\newcommand{\Cov}{\operatorname{Cov}}$ Assuming that the $X_i$'s are a random sample, it is because $X_i$ and $X_j$ are independent for $i\neq j$. On the other hand, $\Var(sX)=\sum_{i=1}^s\sum_{j=1}^s\Cov(X,X)=s^2\Cov(X,X)=s^2\Var(X)$.

In general, for independent $A$ and $B$, we have $\Var(aA+bB)=a^2\Var(A)+b^2\Var(B)$.

You can read a derivation here: http://en.wikipedia.org/wiki/Variance#Properties.

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    In this case independence suffices but the weaker hypothesis of uncorrelatedness is enough.2012-10-25