5
$\begingroup$

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.

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ 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
    Commented Jun 27, 2016 at 16:19
  • 1
    $\begingroup$ 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
    Commented Jun 27, 2016 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
    Commented Jun 27, 2016 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
    Commented Jun 27, 2016 at 20:29
  • $\begingroup$ @ttnphns, thank you for the links. I remember just enough Linear Algebra to make sense of them. $\endgroup$
    – jeremy radcliff
    Commented Jun 28, 2016 at 8:05

1 Answer 1

Reset to default
4
$\begingroup$
  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:

$(X+Y)^2=X^2+Y^2+2XY$

Also consider that:

$Var(X)=\sigma^2(X)$

So:

$[\sigma(X+Y)]^2=[\sigma(X)]^2+[\sigma(Y)]^2+2[\sigma(X)\sigma(Y)]$

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ Thank you, that's very helpful, especially the first part $\endgroup$
    – jeremy radcliff
    Commented Jun 28, 2016 at 8:04

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.