Let $X$ and $Y$ be independent random variables taking values in the positive integers and having the same mass function. Why is it that
$$ \mathbb{P}(X<Y) = \mathbb{P}(Y<X). $$
I understand that
$$ \begin{align*} \mathbb{P}(Y>X) &= 1- \mathbb{P}(Y\le X)\\ &= 1- \mathbb{P}(X=Y) - \mathbb{P}(Y<X). \end{align*}$$
For the mass function $f(x)=2^{-x}$, $x=1,2,\dots$ I found that
$$ \mathbb{P}(X=Y) = \sum_{x=1}^\infty f(x)f(x) = \sum_{x=1}^\infty4^{-x}=\frac{1}{3}. $$
This yields that
$$ \mathbb{P}(X<Y)+\mathbb{P}(Y<X)=\frac{2}{3}. $$
So if $\mathbb{P}(X<Y)=\mathbb{P}(Y<X)$ I see that this is $\frac{1}{3}$, but I don't understand why they are the same.
not even sure they have to be independent (after all correlation is symmetric), but in the independence case:
$$\begin{align}\mathbb{P}(Y\lt X) &= \mathbb{P}(Y-X \lt 0)\\ &=\int\limits_{\mathbb{R}}\mathbb{P}(Y-X \lt 0\;|\;X=t)\cdot\phi_X\cdot\mathrm{d}t\\ &=\int\limits_{\mathbb{R}}\mathbb{P}(Y\lt t )\cdot\phi_X\cdot\mathrm{d}t\end{align}\\$$
And similarly: $\mathbb{P}(X\lt Y) = \int\limits_{\mathbb{R}}\mathbb{P}(X\lt t )\cdot\phi_Y\cdot\mathrm{d}t\\$, but since $\phi_X=\phi_Y$ (the respective densities) and $\mathbb{P}(X\lt t )=\mathbb{P}(Y\lt t )$ the two probabilities are equivalent; (someone please correct any flaws in this!) Cheers