已知二维随机变量（X,Y）的概率密度为f(X,Y)=21/4x^2y,(x^2<y<1).0(其它)。求概率P(X<Y)的求解详细过程！ 设随机变量X-N(0 1)求Y=e^x概率密度？
\u5df2\u77e5\u4e8c\u7ef4\u968f\u673a\u53d8\u91cf\uff08X,Y\uff09\u7684\u6982\u7387\u5bc6\u5ea6\u4e3af(X,Y)=21/4x^2y,(x^2<=y<=1),0(\u5176\u5b83)\u3002\u89c1\u4e0b\u56fe\uff1a
\u76f4\u63a5\u7528\u516c\u5f0f\u6cd5\uff0c\u7b80\u5355\u5feb\u6377\uff0c\u7b54\u6848\u5982\u56fe\u6240\u793a
画出图像就很明显，所要求的积分区域就是y=x²、y=1与y=x所围成的较大的那个区域。
则P(X<Y) = ∫(0,1)∫(-√y,y)f(x,y)dxdy
=7/4 ∫(0,1)y(4)+y(5/2)dy
= 17/20
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绛旓細fx(x)=鈭(0->2) x^2+1/3xy dy=2x^2+2/3x 0<=x<=1 鍏朵粬0
绛旓細fx(x)=鈭(0->2) x^2+1/3xy dy=2x^2+2/3x 0
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绛旓細P=鈭(0-->1)e^(-y)dy鈭(0-->1-y)2e^(-2x)dx =鈭(0-->1)e^(-y)(1-e^(2(1-y))dy =鈭(0-->1)(e^(-y)-e^2e^y)dy=(1-e)(1+e^2)
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绛旓細杩欏氨鏄痻銆亂鐨勫彇鍊艰寖鍥淬傝姹侾(X<Y)灏辨槸姹俧(x,y)鍦ㄥ煎煙閲屼笖X<Y閮ㄥ垎鐨勪簩閲嶇Н鍒嗐傜敾鍑哄浘鍍忓氨寰堟槑鏄撅紝鎵瑕佹眰鐨勭Н鍒嗗尯鍩熷氨鏄痽=x²銆亂=1涓巠=x鎵鍥存垚鐨勮緝澶х殑閭ｄ釜鍖哄煙銆傚垯P(X<Y) = 鈭(0,1)鈭(-鈭歽,y)f(x,y)dxdy =7/4 鈭(0,1)y(4)+y(5/2)dy = 17/20 ...
绛旓細1锛塮(x,y)=1,0<x<1,0<y<1,=0,鍏朵粬 锛2锛夌浉浜掔嫭绔嬨俧X(x)=3x^2 fY(y)=4y^3 f(x,y)=fX(x)*fY(y) 鎵浠ワ細鐙珛
绛旓細鍖栫疮娆＄Н鍒嗭紝鍏堝y绉垎銆傚乏杈箉绉垎绾 0鍒x锛x绉垎绾0鍒1/2銆傚彸杈箉绉垎绾0鍒1-x,x绉垎绾1/2鍒1銆
绛旓細1. a = 1 2. e^(-x) ,-$<x<+ 3. 1-e^[(1/2)*x].
