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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 $Pr(X\geq Y)\geq\frac{1}{2}.$

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$$\begin{align} \mathbb{P} \left( X \geq Y \right) & = \int_y \mathbb{P} \left( X \geq y | Y =y \right) f_Y(y) dy = \int_y \mathbb{P} \left( X \geq y \right) f_Y(y) dy = \int_y \left( 1 - \mathbb{P} \left( X \leq y \right) \right) f_Y(y) dy\\ & = 1 - \int_y \mathbb{P} \left( X \leq y \right) f_Y(y) dy = 1 - \int_y F_X(y) f_Y(y) dy \geq 1 - \int_y F_Y(y) f_Y(y) dy\\ & = 1 - \int_y F_Y(y) d \left( F_Y(y) \right) = 1 - \int_y d \left( \frac{F_Y^2(y)}{2}\right) = 1 - \lim_{y \rightarrow \infty} \frac{F_Y^2(y)}{2} + \lim_{y \rightarrow -\infty} \frac{F_Y^2(y)}{2}\\ & = 1 - \frac12 + 0 = \frac12 \end{align}$$ Hence, $$\mathbb{P} \left( X \geq Y \right) \geq \frac12$$

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