Let $X$ and $Y$ be two independent random variable with c.d.f. $F_X$ and $F_Y$. if $F_X(z)\leq F_Y(z)$ for any $z$ how can show $Pr(X\geq Y)\geq\frac{1}{2}.$
prove $Pr(X\geq Y)\geq\frac{1}{2}$
2 Answers
$\begin{align} \mathbb{P} \left( X \geq Y \right) & = \int_y \mathbb{P} \left( X \geq y | Y =y \right) f_Y(y) dy = \int_y \mathbb{P} \left( X \geq y \right) f_Y(y) dy = \int_y \left( 1 - \mathbb{P} \left( X \leq y \right) \right) f_Y(y) dy\\ & = 1 - \int_y \mathbb{P} \left( X \leq y \right) f_Y(y) dy = 1 - \int_y F_X(y) f_Y(y) dy \geq 1 - \int_y F_Y(y) f_Y(y) dy\\ & = 1 - \int_y F_Y(y) d \left( F_Y(y) \right) = 1 - \int_y d \left( \frac{F_Y^2(y)}{2}\right) = 1 - \lim_{y \rightarrow \infty} \frac{F_Y^2(y)}{2} + \lim_{y \rightarrow -\infty} \frac{F_Y^2(y)}{2}\\ & = 1 - \frac12 + 0 = \frac12 \end{align}$ Hence, $\mathbb{P} \left( X \geq Y \right) \geq \frac12$
When $X$ is a continuous r.v. one may proceed as follows. Since $F_X(X)\sim$ Uniform$[0,1]$, $P(X\geqslant Y) = \mathbb E[P(X\geqslant Y \mid X)] \stackrel{indp}{=} \mathbb E[P(X\geqslant Y)] = \mathbb E[F_Y(X)] \geqslant \mathbb E[F_X(X)] = \frac12$
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0@Zubzub by the probability integral transfom https://en.wikipedia.org/wiki/Probability_integral_transform – 2018-04-11