Suppose $X$ and $Y$ are independent random variables over $\mathbb R$ with distributions $P_X(x)$ and $P_Y(x)$, respectively, and $x \in \mathbb R$. Is $$f(P_X,P_Y) \equiv 2 \max\{ P( X < Y ), P( Y < X ) \} - 1 $$ a valid distance measure for one dimensional distributions? If so, does it have a name?
The idea is that if $P_X$ has little or no overlap with $P_Y$, then either $P( X < Y )$ or $P( Y < X )$, the probability of one r.v. being greater than another, will be close to 1. If they overlap strongly, then $\max\{ P( X < Y ), P( Y < X )\}$ should be close to $\frac{1}{2}$.
The assumption is that $$\max\{ P( X < Y ), P( Y < X )\} \geqslant \frac{1}{2}.$$ This should be true since $P( Y < X ) = 1 - P( X < Y )$, and $\max\{ P( X < Y ), 1 - P( X < Y ) \}$ has to be at least one half for $0 \leqslant P( X < Y ) \leqslant 1$.
Then $f$ as defined above should range from zero to one, depending on the extent of overlap of $P_X(x)$ and $P_Y(x)$. I don't know how to prove this, so I'm posting it as a question here to see if there's a flaw in my argument or not :)
there are $X,Y$ so that $f(P_X,P_Y)=0$ while $P_X\neq P_Y.$
A random variable is $Z$ is even if $P_Z(x)=P_Z(-x).$ If $Z$ is even then, $P_Z(x)=P_{-Z}(x).$
If both $X,Y$ are even, continuous and independent, then:
$$P(X<Y)=P(-X<-Y)=P(X>Y)$$
So, $f(P_X,P_Y)=0.$
It is not hard to show there are such $X,Y$ with $P_X\neq P_Y.$
But a metric should not be zero between unequal points in your space.