21P(ABC)=P(A)P(B)P(C)不一定推出A,BC相互独立 令9={1,2,3,4,5,6,7,8},且在每一点的概率都为1/8.考虑事件B1={1,2,3,4},B2 B3={1,5,6,7}则P(B1)=P(B2)=P(B3)=1/2,B1B2B={1},P(B1B2B3) k=P(B1)P(B2)P(B3)然而B2与B3之间显然不独立 22X2与y2独立但X与Y不独立-1 如果随机变量X与Y独立,那么X2与Y2一定独立.反之不然 考虑一个二维随机向量(X,Y),其概率分布为 P(X=i,Y=j),i,j=-1,0,1 其中p11=p-1,1=1/32,p-1,-1=p1,-1=p1.0=p0.1=3/32,p-1.0=p0.-1 5/32,p0=8/32.容易验证X2,Y2独立,但X,Y不独立 23X2与y2独立但X与Y不独立2 设随机向量(X,Y)的联合密度函数为 f(e, y) (1+xy),|xl<1,ly<1 0, 其他. 由于fx(x)=/y(y)=是,所以f(x,y)≠fx(x)y(y)可见x与Y不相互独立但是 fx2(x)=(4)-1/2,fy2(y)=(4y)-1/2, 而且 fx2y2(x,y)=(4Vxy)-1. 可见对一切x,y有 ∫x2(x)·fy2(y)=f(xy(x,y) 故X2与Y2独立21 P(ABC)=P(A)P(B)P(C)ؽíÑA,B,CƒpÕá -Ω = {1, 2, 3, 4, 5, 6, 7, 8}§…3z:VÇя1/8. į‡B1 = {1, 2, 3, 4}, B2 = B3 = {1, 5, 6, 7}. KP(B1) = P(B2) = P(B3) = 1/2§B1B2B3 = {1}§P(B1B2B3) = 1 8 = P(B1)P(B2)P(B3). , B2†B3ƒmw,ØÕá. 22 X2†Y 2ÕáX†Y ØÕá-1 XJ‘ÅCþX†Y Õá§@oX2†Y 2½Õá. ‡ƒØ,. ć‘‘Å•þ(X, Y )§ÙVǩُ pi,j := P(X = i, Y = j), i, j = −1, 0, 1, Ù¥p1,1 = p−1,1 = 1/32§p−1,−1 = p1,−1 = p1,0 = p0,1 = 3/32§p−1,0 = p0,−1 = 5/32§p0,0 = 8/32. N´yX2 , Y 2Õá§X, Y ØÕá. 23 X2†Y 2ÕáX†Y ØÕá-2 ‘Å•þ(X, Y )éÜÝ¼ê f(x, y) = ( 1 4 (1 + xy), |x| < 1§|y| < 1, 0, Ù¦. dufX(x) = fY (y) = 1 2§¤±f(x, y) 6= fX(x)fY (y). Œ„X†Y ؃pÕá. ´ fX2 (x) = (4x) −1/2 , fY 2 (y) = (4y) −1/2 , … fX2,Y 2 (x, y) = (4√ xy) −1 . Œ„éƒx§yk fX2 (x) · fY 2 (y) = f(X2,Y 2) (x, y). X2†Y 2Õá. 11