$X, Y$ are two independent random variables with distribution functions $F_X, F_Y$ respectively, and $F_X\leq F_Y$.
Show that $P(X\leq Y)\le \frac{1}{2}$ if both $X$ and $Y$ have probability density functions.
What if both $X$ and $Y$ have no pdfs?
This problem is quite strange so I find nowhere to start.
Thanks in advance.
Best Answer
Let $f_X, f_Y$ be the pdfs for $X$ and $Y$, respectively. By independence, the pdf for $(X,Y)$ satisfies $f(s,t)=f_X(s)f_Y(t)$ for all $(s,t) \in \mathbb R^2$. Therefore:
$$P(X\le Y) = \int_{-\infty}^\infty\int_{s}^{\infty} f_Y(t)f_X(s)dtds$$ $$=\int_{-\infty}^\infty P(Y \ge s)f_X(s)ds = \int_{-\infty}^\infty (1 - F_Y(s))f_X(s)ds$$ $$\le \int_{-\infty}^\infty (1 - F_X(s))f_X(s)ds = 1 -\int_{-\infty}^\infty F_X(s)f_X(s)ds$$ $$ = 1 - A$$
We have:
$$A = \int_{-\infty}^\infty\int_{-\infty}^{s} f_X(t)f_X(s)dtds =\int_{-\infty}^\infty \left(1 - \int_{s}^{\infty} f_X(t)dt\right)f_X(s)ds$$ $$= 1 - \int_{-\infty}^\infty\int_{s}^{\infty} f_X(t)f_X(s)dtds$$ $$1 - \int_{-\infty}^\infty\int_{-\infty}^{t} f_X(s)f_X(t)dsdt$$
$$=1 -A$$
So $A = \frac{1}{2}$ and the result follows. Notice that $A = E[F_X(X)]$.
With no pdfs the result is false, consider the case where both $X$ and $Y$ are both equal to a constant. Then $F_X = F_Y$ and $P(X\le Y) = 1$.