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I would like to know why the sum of variances of independent variables assuming a normal distribution is $\sigma_{x+y}^2=\sigma_x^2+\sigma_y^2$ instead of $\sigma_{x+y}=\sigma_x+\sigma_y$?

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    $\begingroup$ $5^2=3^2+4^2$ but $5 \neq 3+4.$ To be closer to your question, the math just works out that the variances add, but not the standard deviations. $\endgroup$ – coffeemath Sep 28 '15 at 15:00
  • $\begingroup$ When $X$ and $Y$ are independent random variables, $\operatorname{Var}(X+Y)=\operatorname{Var}(X)+\operatorname{Var}(Y)$. Here $\operatorname{Var}(X)=\sigma_X^2$, $\operatorname{Var}(Y)=\sigma_Y^2$ and $\operatorname{Var}(X+Y)=\sigma_{X+Y}^2$ $\endgroup$ – Zoran Loncarevic Sep 28 '15 at 15:05
  • $\begingroup$ Hmm but why the standart deviations don't add? Is there any deduction that leads to $SD_{x+y}^2=SD_x^2+SD_y^2$? $\endgroup$ – Roger Danilo Figlie Sep 28 '15 at 16:53
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$\sigma$ refers to the square root of the variance, not to the variance itself. You have to square it again to get the original variances back.

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