设二维随机变量（a,b）有分布密度：p(x,y)=3x, 0<y<x<1; p(x,y)= 0, 其它；求边际期望和协方差。 设二维随机变量（X,Y）的联合概率密度为f（X,Y）=8XY...
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\u2461\u6c42P(Y<1/8,X<1/4)\u3002P(Y<1/8,X<1/4)=\u222b(0,1/8)dy\u222b(y,1/4)f(x,y)dx=\u222b(0,1/8)dy\u222b(y,1/4) 3xdx3/1024\u3002
\u2234P(Y<1/8\u4e28X<1/4)=P(Y<1/8,X<1/4)/P(X<1/4)=3/16\u3002
\u4f9b\u53c2\u8003\u3002
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\u53c2\u8003\u8d44\u6599\u6765\u6e90\uff1a\u767e\u5ea6\u767e\u79d1-\u6982\u7387\u5bc6\u5ea6
=>EX = 3/4
EY = ∫ dx∫ y*p(x,y)dy 同样，第一个积分符号上限1，下限0，第二个积分符号上限x，下限0
=> EY = 3/8
EX EY就是所求的2个边际期望
要求协方差，有公式 Cov(X, Y) = EXY - EX*EY
所以只要再求出EXY即可
EXY = ∫ dx ∫ p(x, y) dy 第一个积分上限1，下限0，第二个积分上限x，下限0
解得EXY = 1
所以 Cov(X, Y) = 1 - 3/4 * 3/8 = 23/32
绛旓細EX = 鈭 dx 鈭 x*p(x, y) dy 绗竴涓Н鍒嗙鍙蜂笂闄1锛屼笅闄0锛岀浜屼釜绉垎绗﹀彿涓婇檺x锛屼笅闄0 =>EX = 3/4 EY = 鈭 dx鈭 y*p(x,y)dy 鍚屾牱锛岀涓涓Н鍒嗙鍙蜂笂闄1锛屼笅闄0锛岀浜屼釜绉垎绗﹀彿涓婇檺x锛屼笅闄0 => EY = 3/8 EX EY灏辨槸鎵姹傜殑2涓竟闄呮湡鏈 瑕佹眰鍗忔柟宸紝鏈夊叕寮 Cov(X, ...
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绛旓細鑱斿悎姒傜巼瀵嗗害:f(x,y)=1/(b-a)(d-c)鍦ㄧ煩褰㈠煙a<x<b,c<y<d涓,0锛 鍏朵粬 杈圭紭姒傜巼:fX(x)=1/(b-a)鍦ㄥ尯闂碼<x<b,涓,0锛 鍏朵粬 fY(y)=1/(d-c)鍦ㄥ尯闂碿<y<d涓,0锛 鍏朵粬
绛旓細8锛璁句簩缁撮殢鏈哄彉閲(X锛孻)鐨勮仈鍚堝瘑搴︿负锛歠(x锛寉)= 锛屽垯 鈭 Xi²=__?___.i=1 9. 100浠朵骇鍝侊紝鍏朵腑5浠朵笉鍚堟牸鍝侊紝5浠朵笉鍚堟牸鍝佷腑鍙堟湁3浠舵槸娆″搧锛2浠跺簾鍝併傚湪100浠朵腑浠绘剰鎶戒竴浠躲傛眰锛1锛夋娊寰楁槸搴熷搧B鐨勬鐜;2/100=1/50 锛2锛夊凡鐭ユ娊寰楃殑鏄笉鍚堟牸鍝A锛瀹冩槸搴熷搧鐨勬鐜嘝(B|A)銆
绛旓細绛旀涓猴細F锛坅锛宐锛+1-[F1锛坅锛+F2锛坆锛塢鐢变簬F锛坅锛宐锛=P{X鈮锛孻鈮}锛孎1锛坅锛=P{X鈮锛孻锛+鈭瀩锛孎2锛坆锛=P{X锛+鈭烇紝Y鈮}锛岃岋細P{X锛瀉锛孻锛瀊}=P{X锛+鈭烇紝Y锛+鈭瀩-P{X鈮锛孻锛+鈭瀩-P{X锛+鈭烇紝Y鈮}+P{X鈮锛孻鈮} 鈭碢{X锛瀉锛孻锛瀊}=1-F1锛坅锛-...
绛旓細a+0.2=0.3,鏁卆=0.1;0.3+0.4+0.1+b=1锛屾晠b=0.2.P(X=-1)=0.3,P(X=0)=0.4,P(X=2)=0.3;P(Y=1)=0.5,P(Y=3)=0.5 浠わ紱a+1/6+1/12+ +1/6+1/6+1/6+ +1/12+1/6+b=1,寰楋細a+b+1=1锛屽嵆锛歛+b=0銆傚洜涓篴>=0, b>=0,鏁呯煡閬撳繀鏈夛細a=0锛宐=0...
绛旓細1.璁句簩缁撮殢鏈哄彉閲忥紙X锛孻锛夌殑鍒嗗竷寰嬩负锛屽垯P{X+Y=0}=锛堛锛堿.0.2B.0.3C.0.5D.0.7绛旀锛欳2.璁句簩缁撮殢鏈哄彉閲忥紙X锛孻锛夌殑姒傜巼瀵嗗害涓哄垯甯告暟c=锛堛锛堿.B.C.2D.4绛旀锛欰3锛庤浜岀淮闅忔満鍙橀噺锛圶锛孻锛夌殑鍒嗗竷寰嬩负璁緋ij=P{X=i,Y=j}i,j=0,1锛屽垯涓嬪垪鍚勫紡涓敊璇殑鏄紙銆锛堿锛巔00<p01B锛...
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绛旓細F(+鈭,+鈭)=A(B+蟺/2)(C+蟺/2)=1 瑙ｅ緱锛欰=1/蟺^2,B=蟺/2,C=蟺/2 f(x,y)=dF(x,y)/dxdy=1/[蟺^2 (1+x^2)(1+y^2)]杈圭紭鍑芥暟 fx(x)=鈭玣(x,y)dy 浠庤礋鏃犵┓绉垎鍒版鏃犵┓ =1/[蟺(1+x^2)]fy(y)=鈭玣(x,y)dx 浠庤礋鏃犵┓绉垎鍒版鏃犵┓ =1/[蟺(1+y^2)]
绛旓細鏍规嵁鍏紡璁＄畻锛歅(X鈮1,Y鈮0)=P(X=-1,Y=0)+P(X=1,Y=0)+P(X=1,Y=1)+P(X=1,Y=1)=0.1+0.1+0.2+0=0.4銆侳(0,0)=P(X鈮0,Y鈮0)=P(X=-1,Y=-2)+P(X=-1,Y=0)=0.3+0.1=0.4銆
