设随机变量X，Y相互独立，且服从同一分布，试证明 P{a＜min(X，Y)≤b}=[p(X＞a)]2-[p(X＞b)]2．
【答案】：因为X与Y独立同分布，故P{a＜min(X，Y)≤b}=P{min(X，Y)≤b}P{min(X，Y)≤a}
=1-P{min(X，Y)＞b}-[1-P{min(X，Y)＞a}]
=P{min(X，Y)＞a}-P}min(X，Y)＞b}
=P(X＞a，Y＞a)-P(X＞b，Y＞b)
=P(X＞a)P(Y＞a)-P(X＞b)P(Y＞b)
=[P(X＞a)]2-[P(X＞b)]2．
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