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