A useful criterion for first order stochastic dominance of two random variables $X$ and $Y$, denoted $X\le Y$, is to check whether the densities (or pmf if discrete) cross at most at one point. Specifically, if density $f$ of $X$ starts above the density $g$ of $Y$, then they cross and afterwards $f$ stays below $g$. (If they do not cross at all, it is trivial)
I have no trouble showing that this is true when at least one of the random variables has finite support. However, if I assume that both $X$ and $Y$ have nonzero densities on, say, $(0,\infty)$, I am at a loss, my argument breaks down. How does one prove this result for infinite supports? Cheers.
Let $F(x)=P(X\le x)$ and $G(x)=P(Y\le x)$. Assume that both $X$ and $Y$ are continuous random variables, so there exist functions $f(x)$ and $g(x)$ such that $F(x)=\int_{-\infty}^x f(t)\,dt$ and $G(x)=\int_{-\infty}^x g(t)\,dt$. We are given that there is a certain crossover point, $x^*$, such that $$ x<x^*\implies f(x)>g(x)\\ x>x^*\implies f(x)<g(x) $$
In order for $X\preceq Y $ to occur, it must be true that $F(x)\ge G(x)$ for all $x\in \mathbb R$. We prove this in two cases.
If $x\le x^*$, then $$F(x)=\int_{-\infty}^xf(t)\,dt>\int_{\infty}^x g(t)\,dt=G(x).$$
If $x>x^*$, then $$F(x)=1-\int_{x}^\infty f(t)\,dt>1-\int_x^\infty g(t)=G(x).$$
In either case, $F(x)<G(x)$, as desired.