Is there any way to make sense out of this formula intuitively?

I rederived it algebraically (took me a while...), which made me happy because I used to be incapable of doing that kind of stuff, but I still have no intuition for it.

Geometric intuition would be best, but anything is very welcome.

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    $\begingroup$ Here's a trivial example that may or may not help with intuition: $Var(X + X) = Var(X) + Var(X) + 2Cov(X, X)=Var(X) + Var(X) + 2Var(X)=4Var(X)=Var(2X)$. $\endgroup$ – lmo Jun 27 '16 at 16:19
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    $\begingroup$ Geometric intuition is straightforward in vector representation of (centered) variables in subject space. There, vector length is st. deviation and angle b/w vectors is Pearson $r$ (see). Then, it appears that the st. dev. of the sum X+Y is the longer axis of the parallelogram, and by its trigonometric law you get the formula you ask about. See the 1st pic with formulas here. $\endgroup$ – ttnphns Jun 27 '16 at 16:40
  • $\begingroup$ See also a not the same, but a related question. It is about correlation between X and X+Y (when X and Y are uncorrelated, as a particular case). $\endgroup$ – ttnphns Jun 27 '16 at 16:58
  • $\begingroup$ As for algebraic way, the hint is $(a+b)^2=a^2+b^2+2ab$, from school. One can start from here. $\endgroup$ – ttnphns Jun 27 '16 at 20:29
  • $\begingroup$ @ttnphns, thank you for the links. I remember just enough Linear Algebra to make sense of them. $\endgroup$ – jeremy radcliff Jun 28 '16 at 8:05
  1. More mathematical way

The variance of two variables, X and Y, can explained by separating it into four sections: how X varies independently, how Y varies independently, how X varies due to changes in Y, and how Y varies due to changes in X. Essentially, all changes in X and Y have to be caused by one of these four things.

That gives us:

$Var(X) + Var(Y) + Cov(X,Y) + Cov(Y,X)$

which can be simplified, because $Cov(X,Y) = Cov(Y,X)$ into:

$Var(X+Y) = Var(X) + Var(Y) + 2Cov(X,Y)$

  1. Not mathematical way

This is a vast oversimplification and is not mathematically correct, but it helped me with remembering.

Consider the much simpler algebraic equation:


Also consider that:




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    $\begingroup$ Thank you, that's very helpful, especially the first part $\endgroup$ – jeremy radcliff Jun 28 '16 at 8:04

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