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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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Your second statement is only true if the variables are uncorrelated. In your first statement, the summed variables are maximally correlated: they're all equal. – Chris Eagle Oct 25 '12 at 16:18
Take $s=2$. $$\text{var}(sX) = \text{var}(X+X) = \text{var}(X)+\text{var}(X)+2\text{cov}(X,X) = 4\text{var}(X)=s^2\text{var}(X)$$ – Dilip Sarwate Oct 25 '12 at 16:19
@ChrisEagle: Thanks, I meant to write it. – Ben Benli Oct 25 '12 at 16:21
up vote 1 down vote accepted

$\newcommand{\Var}{\operatorname{Var}}$ $\newcommand{\Cov}{\operatorname{Cov}}$ 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:

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In this case independence suffices but the weaker hypothesis of uncorrelatedness is enough. – Michael Hardy Oct 25 '12 at 16:30

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