X and Y random variables. How can I interpret $\sqrt{E(Y-X)^{2}}$ as the distance between X and Y?

Question

X and Y random variables. How can I interpret $\sqrt{E(Y-X)^{2}}$ as the distance between X and Y by showing (prooving) that i) and ii) below works:

1. if $\sqrt{E(Y-X)^{2}}$ = 0 then $P(Y=X)=1$

2. The triangle inequality works: $\sqrt{E(Y-X)^{2}} \leq \sqrt{E(Y^{2})} +\sqrt{E(X^{2})}$

So, I have to use i) and ii) to show that $\sqrt{E(Y-X)^{2}}$ can be interpretated as the distance between X and Y.

I am completely lost here.

Any help?

Answer 1

A distance function satisfies

1. $d(x,y)=0$ implies $x=y$
2. $d(x,y) \leq d(x,c)+d(c,y)$.

Here $d(x,y)=\sqrt{\mathbb{E}(X-Y)^2}$ and $c=0$. In words, if the distance function is zero, the points are the same (distance from a point to itself is zero, right?). The triangle inequality just means that the distance from $x$ to $y$ is less than the sum of the distances from $x$ to $c$ and the distance from $c$ to $y$ where $c$ is some intermediate point. They ask you to prove $\sqrt{\mathbb{E}(\cdot)}$ satisfies these two properties, so that it can be interpreted as a distance function.

They seemed to have omitted symmetry though...