0
$\begingroup$

Some students asked me why variances get added despite two random variables are subtracted. The example looked like this $$y(n)=x(n)-x(n-1)$$ where x(n) is assumed to be iid. We look for the expected value of $y$ and its variance.

I see why the variance has to be added. It is quite intuitive, but I had a hard time explaining it to them. Numerical examples worked, but is there a quick way how one could proof it? Some research on the internet showed me only that it is defined that way.

$\endgroup$
2
  • 1
    $\begingroup$ I suppose variables vary positively and negatively either side of the mean - you don't know which in any given trial. Addition or subtraction of variables is then essentially the same thing, you may be adding a negative or subtracting a positive. $\endgroup$
    – Paul
    Sep 17, 2019 at 9:00
  • $\begingroup$ @Paul thats similar how I explained it. Once you see it it makes perfect sense. To some of them, however, it did not. I hoped I could proof it somehow more algebraic. $\endgroup$
    – Mr.Sh4nnon
    Sep 17, 2019 at 9:02

1 Answer 1

2
$\begingroup$

$$ \mathop{\mathrm{Var}}(x-y)= \mathbb E(x-y)^2 - \Big(\mathbb E(x-y)\Big)^2=\\ \mathbb Ex^2-2(\mathbb Exy)+\mathbb Ey^2-(\mathbb Ex)^2+2(\mathbb Ex)(\mathbb Ey)-(\mathbb Ey)^2\\ $$

Since $x$ and $y$ are independent $\mathbb Exy=(\mathbb Ex)(\mathbb Ey)$. Finally: $$ \mathop{\mathrm{Var}}(x-y) = \mathbb Ex^2 -(\mathbb Ex)^2 + \mathbb Ey^2 -(\mathbb Ey)^2 = \mathop{\mathrm{Var}} x+ \mathop{\mathrm{Var}}y $$

$\endgroup$
2
  • $\begingroup$ Thank you very much. I feel stupid not to have thought about it. $\endgroup$
    – Mr.Sh4nnon
    Sep 17, 2019 at 12:24
  • 1
    $\begingroup$ You can also show students, that variance of $x$ and $x'=-x$ is the same (just pdf is mirrored). So subtracting to variables is pretty much the same as adding. $\endgroup$ Sep 17, 2019 at 12:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .