概率论与数理统计理工类第四版吴赣昌主编课后习题答案第三章.doc

leading cadres awareness of right in place, study the party Constitution and party rules, series of speeches can make proper effect. Party cadres to two lead by example, to lead by example we must change our mind, recognizing that two is important. First, learn the Communist Party Constitution Party rules, learning learning series important speech by General Secretary can enhance theory. With the third revolution the rise of rapid changes in our life are feeling, the Communists should adhere to the theory of confidence will continue to learn from the voices of the times, the times, in turn, will have new requirements for leading party cadres. Two is the most basic meaning of members all mastering the core theory and the most advanced weapons theory, complement the spirit of calcium. Secondly, the Communist Party Constitution Party rules, learning learning series important speech by General Secretary be able to firmly build the ideological foundation. Under the impact of multiple values, the two is to help cultivate independent judgment in numerous miscellaneous multiple concepts, so that the majority consensus of party members, the important magic weapon of the party with the resonance frequency. Finally, the Constitution of the Communist Party, party rules, learning learning series important speech can stand crowds, General Secretary position. Now, some grass-roots work in the the old way didnt work, hard way cannot, the new approach would not phenomenon, sometimes due to a mass of party members and cadres not understanding, does not meet. In fact, the mass convincing and identityBehind are likely to be party members and leading cadres themselves ignoring beliefs held, resulting in lack of persuasion and sense of identity. Two is in fact guaranteed party cadres work, an important prerequisite for convincing the masses. Bacon said it well: practical men can handle individual matters, but looking at the whole operation globally, but only man can do to knowledge. Giving up two effective, must first raise awareness of leading cadres of party members, as party members learn real responsibility to know to music , good changes, in parallel with the ground, do not forget to also answer the antenna, draw catches on meteorology, which sit between heaven and Earth which melds together the dreams of practicing Communist. Members cadres should in two learn a do in the based post do contribution in all members in the carried out learn Constitution Party rules, and learn series speech, do qualified members learning education, this is following party of mass line education practice activities and three strict three real topic education zhihou, deepening party education of and once important practice, is promoted three strict three real topic education from key minority to all members expand, and from concentrated education to regular education extends of important initiatives. Vast numbers of party members and cadres in the course of two, should be第三章 多维随机变量及其分布3.1 二维随机变量及其分布习题1设(X,Y)的分布律为求a.解答：由分布律性质,可知解得a=习题2(1)2. (1)PaXb,Yc;解答：PaXb,Yc=F(b,c)-F(a,c).(2)P0Yb;解答P0a,Yb.解答：PXa,Yb=F(+,b)-F(a,b).习题3(1)设二维离散型随机变量的联合分布如下表：试求：(1)P1/2X3/2,0Y4解答：P1/2X2/3,0Y4=PX=1,Y=1+PX=1,Y=2+PX=1,Y=3=1/4+0+0=1/4.(2)P1X2,3Y4;解答：P1X2,3Y4=PX=1,Y=3+PX=1,Y=4+PX=2,Y=3+PX=2,Y=4=0+1/16+0+1/4=5/16.(3)F(2,3).解答：F(2,3)=P(1,1)+P(1,2)+P(1,3)+P(2,1)+P(2,2)+P(2,3)=1/4+0+0+1/16+1/4+0=9/16.习题4设X,Y为随机变量，且PX0,Y0=3/7,PX0=PY0=4/7,求PmaxX,Y0,PminX,YY=1，且由正态分布图形的对称性，知PXY=PXY,故PXY=1/2.习题7设随机变量(X,Y)的概率密度为 (1)确定常数k;(2)求PX1,Y3;(3)求PX1,有F(x,y)=PX1,Yy=4最后，设x1,0y1,有F(x,y)=PX1,Yy=4函数F(x,y)在平面各区域的表达式(见课后答案)习题9设二维随机变量(X,Y)的概率密度为 求边缘概率密度 解答： = 习题10设(X,Y)在曲线,y=x所围成的区域G里服从均匀分布，求联合分布密度和边缘分布密度.解答：区域G的面积,由题设知(X,Y)的联合分布密度为 从而即 即 3.2 条件分布与随机变量的独立性习题1二维随机变量(X,Y)的分布律为(1)求Y的边缘分布律；(2)求PY=0X=0,PY=1X=0;(3)判定X与Y是否独立？解答：(1)由(X,Y)的分布律知，Y只取0及1两个值.PY=0=PX=0,Y=0+PX=1,Y=0=7/15+7/30=0.7PY=1= (2)Py=0x=0=Px=0,y=0Px=0=23,Py=1x=0=13.(3)已知Px=0,y=0=715,由(1)知Py=0=0.7,类似可得Px=0=0.7.因为Px=0,y=0Px=0Py=0,所以x与y不独立.习题2将某一医药公司9月份和8份的青霉素针剂的订货单分别记为X与Y. 据以往积累的资料知X和Y的联合分布律为(1)求边缘分布律;(2)求8月份的订单数为51时，9月份订单数的条件分布律.解答：(1)边缘分布律为Xpk0.180.150.350.120.20对应X的值,将每行的概率相加,可得PX=i.对应Y的值(最上边的一行),将每列的概率相加，可得PY=j.Ypk0.280.280.220.090.13(2)当Y=51时,X的条件分布律为PX=kY=51=PX=k,y=51PY=51=pk,510.28,k=51,52,53,54,55.列表如下:kPX=kY=516/287/285/285/285/28习题3已知(X,Y)的分布律如下表所示，试求：(1)在Y=1的条件下,X的条件分布律;(2)在X=2的条件下,Y的条件分布律.XY0120121/41/8001/301/601/8解答：由联合分布律得关于X,Y的两个边缘分布律为X012pk3/81/37/24Y012pk5/1211/241/8故(1)在Y=1条件下，X的条件分布律为X(Y=1)012pk3/118/110(2)在X=2的条件下，Y的条件分布律为Y(X=2)012pk4/703/7习题4已知(X,Y)的概率密度函数为f(x,y)=3x,0x1,0yx0,其它,求：(1)边缘概率密度函数；(2)条件概率密度函数.解答：(1)fX(x)=-+f(x,y)dy=3x2,0x10,其它,fY(y)=-+f(x,y)dx=32(1-y2),0y10,其它.(2)对y(0,1),fXY(xy)=f(x,y)fY(y)=2x1-y2,yx1,0,其它,对x(0,1),fYX(yx)=f(x,y)fX(x)=1x,0yX=05x52(5-y)125dydx=13.习题7设随机变量X与Y都服从N(0,1)分布，且X与Y相互独立，求(X,Y)的联合概率密度函数.解答：由题意知，随机变量X,Y的概率密度函数分别是fX(x)=12e-x22，fY(y)=12e-y22因为X与Y相互独立，所以(X,Y)的联合概率密度函数是f(x,y)=12e-12(x+y)2.习题8设随机变量X的概率密度f(x)=12e-x(-x0,各有PXa,Xa=PXaPXa,而事件XaXa,故由上式有PXa=PXaPXa,PXa(1-PXa)=0PXa=0或1=PXa(a0)但当a0时，两者均不成立，出现矛盾，故X与X不独立.习题9设X和Y是两个相互独立的随机变量，X在(0,1)上服从均匀分布，Y的概率密度为fY(y)=12e-y2,y00,y0,(1)求X与Y的联合概率密度；(2)设有a的二次方程a2+2Xa+Y=0,求它有实根的概率.解答：(1)由题设易知fX(x)=1,0x10,其它,又X,Y相互独立，故X与Y的联合概率密度为f(x,y)=fX(x)fY(y)=12e-y2,0x00,其它;(2)因a有实根=判别式2=4X2-4Y0=X2Y,故如图所示得到：Pa有实根=PX2Y=x2yf(x,y)dxdy=01dx0x212e-y2dy=-01e-x22dx=1-1e-x22dx-0e-x22dx=1-212-1e-x22dx-12-0e-x22dx=1-2(1)-(0),又(1)=0.8413,(0)=0.5,于是(1)-(0)=0.3413,所以Pa有实根=1-2(1)-(0)1-2.510.3413=0.1433.3.3 二维随机变量函数的分布习题1设随机变量X和Y相互独立，且都等可能地取1,2,3为值，求随机变量U=maxX,Y和V=minX,Y的联合分布.解答：由于UV,可见PU=i,V=j=0(ij),于是，随机变量U和V的联合概率分布为V概率U12311/92/92/9201/92/93001/9习题2设(X,Y)的分布律为XY-112-121/101/53/101/51/101/10试求：(1)Z=X+Y;(2)Z=XY;(3)Z=X/Y;(4)Z=maxX,Y的分布律.解答：与一维离散型随机变量函数的分布律的计算类型，本质上是利用事件及其概率的运算法则.注意，Z的相同值的概率要合并.概率1/101/53/101/51/101/10(X,Y)X+YXYX/Ymaxx,Y(-1,-1)(-1,1)(-1,2)(2,-1)(2,1)(2,2)-1-2-2241-1-1/2-于是(1)X+Y-20134pi1/101/51/21/101/10(2)XY-20134pi1/21/51/101/101/10(3)X/Y-2-1-1/212pi1/51/53/101/51/10(4)maxX,Y-112pi1/101/57/10习题3设二维随机向量(X,Y)服从矩形区域D=(x,y0x2,0y1的均匀分布，且U=0,XY1,XY,V=0,X2Y1,X2Y,求U与V的联合概率分布.解答：依题(U,V)的概率分布为PU=0,V=0=PXY,X2Y=PXY=01dxx112dy=14,PU=0,V=1=PXY,X2Y=0,PU=1,V=0=PXY,X2Y=PYX2Y=01dyy2y12dx=14,PU=1,V=1=1-PU=0,V=0-PU=0,V=1-PU=1,V=0=1/2,即UV01011/401/41/2习题4设(X,Y)的联合分布密度为f(x,y)=12e-x2+y22,Z=X2+Y2,求Z的分布密度.解答：FZ(z)=PZz=PX2+Y2z.当z0时，FZ(z)=P()=0;当z0时，FZ(z)=PX2+Y2z2=x2+y2z2f(x,y)dxdy=12x2+y2z2e-x2+y22dxdy=1202d0ze-22d=0ze-22d=1-e-z22.故Z的分布函数为FZ(z)=1-e-z22,z00,z00,z0.习题5设随机变量(X,Y)的概率密度为f(x,y)=12(x+y)e-(x+y),x0,y00,其它,(1)问X和Y是否相互独立？(2)求Z=X+Y的概率密度.解答：(1)fX(x)=-+f(x,y)dy=0+12(x+y)e-(x+y)dy,x00,x0under2line令x+y=tx+12te-tdt=12(x+1)e-x,x00,x0,由对称性知fY(y)=12(y+1)e-y,y00,y0,显然f(x,y)fX(x)fY(y),x0,y0,所以X与Y不独立.(2)用卷积公式求fZ(z)=-+f(x,z-x)dx.当x0z-x0即x0x0时，fZ(z)=0z12xe-xdx=12z2e-z.于是，Z=X+Y的概率密度为fZ(z)=12z2e-z,z00,z0.习题6设随机变量X,Y相互独立，若X服从(0,1)上的均匀分布，Y服从参数1的指数分布，求随机变量Z=X+Y的概率密度.解答：据题意，X,Y的概率密度分布为fX(x)=1,0x10,其它,fY(y)=e-y,y00,y0,由卷积公式得Z=X+Y的概率密度为fZ(z)=-+fX(x)fY(z-x)dx=-+fX(z-y)fY(y)dy=0+fX(z-y)e-ydy.由0z-y1得z-1y0时，fZ(z)=0+fX(z-y)e-ydy=max(0,z-1)ze-ydy=e-max(0,z-1)-e-z,即fZ(z)=0,z01-e-z,01.习题7设随机变量(X,Y)的概率密度为f(x,y)=be-(x+y),0x1,0y+,0,其它.（1）试确定常数b；（2）求边缘概率密度fX(x),fY(y)；（3）求函数U=maxX,Y的分布函数.解答：（1）由-+-+f(x,y)dxdy=1，确定常数b.01dx0+be-xe-ydy=b(1-e-1)=1，所以b=11-e-1，从而f(x,y)=11-e-1e-(x+y),0x1,0y+,0,其它.（2）由边缘概率密度的定义得fX(x)=0+11-e-1e-(x+y)dy=e-x1-e-x,0x1,0,其它，fY(x)=0111-e-1e-(x+y)dx=e-y,0y+,0,其它（3）因为f(x,y)=fX(x)fY(y)，所以X与Y独立，故FU(u)=PmaxX,Yu=PXu,Yu=FX(u)FY(u)，其中FX(x)=0xe-t1-e-1dt=1-e-x1-e-1,0x1，所以FX(x)=0,x0,1-e-x1-e-1,0x1,1,x1.同理FY(y)=0ye-tdt=1-e-y,0y+,0,y0，因此FU(u)=0,u0,(1-e-u)21-e-1,0u00,x0,2(y)=e-y,y00,y0,其中0,0,试求系统L的寿命Z的概率密度.解答：设Z=minX,Y,则F(z)=PZz=Pmin(X,Y)z=1-Pmin(X,Y)z=1-PXz,Yz=1-1PXz1-PYz=1-1-F1z1-F2z由于F1(z)=0ze-xdx=1-e-z,z00,z0,F2(z)=1-e-z,z00,z0,故F(z)=1-e-(+)z,z00,z00,z0.习题9设随机变量X,Y相互独立，且服从同一分布，试证明：Paa2-PXb2.解答：设minX,Y=Z,则Paz=1-PXz,Yz=1-PXzPYz=1-PXz2,代入得Pab2-(1-PXa2)=PXa2-PXb2.证毕.复习总结与总习题解答习题1在一箱子中装有12只开关，其中2只是次品，在其中取两次，每次任取一只，考虑两种试验：(1)放回抽样；(2)不放回抽样.我们定义随机变量X,Y如下：X=0,若第一次取出的是正品1,若第一次取出的是次品, Y=0,若第二次取出的是正品1,若第二次取出的是次品,试分别就(1),(2)两种情况，写出X和Y的联合分布律.解答：(1)有放回抽样，(X,Y)分布律如下：PX=0,Y=0=10101212=2536; PX=1,Y=0=2101212=536,PX=0,Y=1=1021212=536, PX=1,Y=1=221212=136,(2)不放回抽样，(X,Y)的分布律如下：PX=0,Y=0=1091211=4566, PX=0,Y=1=1021211=1066,PX=1,Y=0=2101211=1066, PX=1,Y=1=211211=166,YX010145/6610/6610/661/66习题2假设随机变量Y服从参数为1的指数分布，随机变量Xk=0,若Yk1,若Yk(k=1,2),求(X1,X2)的联合分布率与边缘分布率.解答：因为Y服从参数为1的指数分布，X1=0,若Y11,若Y1, 所以有 PX1=1=PY1=1+e-ydy=e-1, PX1=0=1-e-1,同理PX2=1=PY2=2+e-ydy=e-2,PX2=0=1-e-2,因为PX1=1,X2=1=PY2=e-2，PX1=1,X2=0=PX1=1-PX1=1,X2=1=e-1-e-2,PX1=0,X2=0=PY1=1-e-1,PX1=0,X2=1=PX1=0-PX1=0,X2=0=0,故(X1,X2)联合分布率与边缘分布率如下表所示：X1slashX201PX1=i01-e-101-e-11e-1-e-2e-2e-1PX2=j1-e-2e-2习题3在元旦茶话会上，每人发给一袋水果，内装3只橘子，2只苹果，3只香蕉. 今从袋中随机抽出4只，以X记橘子数，Y记苹果数，求(X,Y)的联合分布.解答：X可取值为0,1,2,3,Y可取值0,1,2.PX=0,Y=0=P=0,PX=0,Y=1=C30C21C33/C84=2/70,PX=0,Y=2=C30C22C32/C84=3/70,PX=1,Y=0=C31C20C33/C84=3/70,PX=1,Y=1=C31C21C32/C84=18/70,PX=1,Y=2=C31C22C31/C84=9/70,PX=2,Y=0=C32C20C32/C84=9/70,PX=2,Y=1=C32C21C31/C84=18/70,PX=2,Y=2=C32C22C30/C84=3/70,PX=3,Y=0=C33C20C31/C84=3/70,PX=3,Y=1=C33C21C30/C84=2/70,PX=3,Y=2=P=0,所以，(X,Y)的联合分布如下：XY012301203/709/703/702/7018/7018/702/703/709/703/700习题4设随机变量X与Y相互独立，下表列出了二维随机变量(X,Y)的联合分布律及关于X与Y的边缘分布律中的部分数值，试将其余数值填入表中的空白处：XYy1y2y3pix11/8x21/8pj1/61解答：由题设X与Y相互独立，即有pij=pipj(i=1,2;j=1,2,3), p1-p21=p11=16-18=124,又由独立性，有p11=p1p1=p116故p1=14.从而p13=14-124-18, 又由p12=p1p2, 即18=14p2.从而p2=12. 类似的有p3=13,p13=14,p2=34.将上述数值填入表中有XYy1y2y3pix11/241/81/121/4x21/83/81/43/4pj1/61/21/31习题5设随机变量(X,Y)的联合分布如下表：求：(1)a值； (2)(X,Y)的联合分布函数F(x,y)； (3)(X,Y)关于X,Y的边缘分布函数FX(x)与FY(y).解答：(1)because由分布律的性质可知ijPij=1, 故14+14+16+a=1,a=13.(2)因F(x,y)=PXx,Yy当x1或y-1时，F(x,y)=0;当1x2,-1y0时，F(x,y)=PX=1,Y=-1=1/4;当x2,-1y0时，F(x,y)=PX=1,Y=-1+PX=2,Y=-1=5/12;当1x0时，F(x,y)=PX=1,Y=-1+PX=1,Y=0=1/2;当x2,y0时，F(x,y)=PX=1,Y=-1+PX=2,Y=-1+PX=1,Y=0+PX=2,Y=0=1;综上所述，得(X,Y)联合分布函数为F(x,y)=0,x1或y-11/4,1x2,-1y05/12,x2,-1y01/2,1x2,y01,x2,y0.(3)由FX(x)=PXx,Y+=xixj=1+pij, 得(X,Y)关于X的边缘分布函数为：FX(x)=0,x114+14,1x214+14+16+13,x2=0,x11/2,1x21,x2,同理，由FY(y)=PX+,Yy=yiyi=1+Pij, 得(X,Y)关于Y的边缘分布函数为FY(y)=0,y-12/12,-1y01,y0.习题6设随机变量(X,Y)的联合概率密度为f(x,y)=c(R-x2+y2),x2+y2R0,x2+y2R,求：(1)常数c; (2)PX2+Y2r2(rR).解答：(1)因为1=-+-+f(x,y)dydx=x2+y2Rc(R-x2+y)dxdy=020Rc(R-)dd=cR33,所以有c=3R3.(2)PX2+Y2r2=x2+y2r23R3R-x2+y2dxdy=020r3R3(R-)dd=3r2R2(1-2r3R).习题7设f(x,y)=1,0x2,max(0,x-1)ymin(1,x)0,其它, 求fX(x)和fY(y).解答：max(0,x-1)=0,x1x-1,x1, min(1,x)=x,x11,x1,所以，f(x,y)有意义的区域(如图)可分为0x1,0yx,1x2,1-xy1,即f(x,y)=1,0x1,0yx1,1x2,x-1y1,0,其它 所以fX(x)=0xdy=x,0x0,y00,其它,(1)确定常数c; (2)求X,Y的边缘概率密度函数；(3)求联合分布函数F(x,y); (4)求PYX;(5)求条件概率密度函数fXY(xy); (6)求PX2Y00,x0=2e-2x,x00,x0,fY(y)=-+f(x,y)dx=0+2e-2xe-ydx,y00,其它=e-y,y00,y0.(3)F(x,y)=-x-yf(u,v)dvdu=0x0y2e-2ue-vdvdu,x0,y00,其它=(1-e-2x)(1-e-y),x0,y00,其它.(4)PYX=0+dx0x2e-2xe-ydy=0+2e-2x(1-e-x)dx=13.(5)当y0时，fXY(xy)=f(x,y)fY(y)=2e-2xe-ye-y,x00,x0=2e-2x,x00,x0.(6)PX2Y1=PX2,Y1PY1=F(2,1)01e-ydy=(1-e-1)(1-e-4)1-e-1=1-e-4.习题10设随机变量X以概率1取值为0, 而Y是任意的随机变量，证明X与Y相互独立.解答：因为X的分布函数为F(x)=0,当x0时1,当x0时, 设Y的分布函数为FY(y),(X,Y)的分布函数为F(x,y)，则当x0时，对任意y, 有F(x,y)=PXx,Yy=P(Xx)(Yy)=P(Yy)=P=0=FX(x)FY(y);当x0时，对任意y, 有F(x,y)=PXx,Yy=P(Xx)(Yy)=PS(Yy)=PYy=Fy(y)=FX(x)FY(y),依定义，由F(x,y)=FX(x)FY(y)知，X与Y独立.习题11设连续型随机变量(X,Y)的两个分量X和Y相互独立，且服从同一分布，试证PXY=1/2.解答：因为X,Y独立，所以f(x,y)=fX(x)fY(y).PXY=xyf(x,y)dxdy=xyfX(x)fY(y)dxdy=-+fY(y)-yfX(x)dxdy=-+fY(y)FY(y)dy=-+FY(y)dFY(y)=F2(y)2-+=12,也可以利用对称性来证，因为X,Y独立同分布，所以有PXY=PYX,而PXY+PXY=1, 故PXY=1/12.习题12设二维随机变量(X,Y)的联合分布律为若X与Y相互独立，求参数a,b,c的值.解答：关于X的边缘分布为Xx1x2x3pka+1/9b+1/9c+1/3关于Y的边缘分布为Yy1y2pk1/9+a+c4/9+b由于X与Y独立，则有p22=p2p2 得 b=(b+19)(b+49) p12=p1p2 得 19=(a+19)(b+49) 由式得b=29, 代入式得a=118. 由分布律的性质，有a+b+c+19+19+13=1,代入a=118,b=29, 得c=16.易验证，所求a,b,c的值，对任意的i和j均满足pij=pipj.因此，所求a,b,c的值为a=118,b=29,c=16.习题13已知随机变量X1和X2的概率分布为且PX1X2=0=1.(1)求X1和X2的联合分布律； (2)问X1和X2是否独立？解答：(1)本题是已知了X1与X2的边缘分布律，再根据条件PX1X2=0=1, 求出联合分布. 列表如下：X2X1-101PX2=j011/401/401/201/21/2PX1=i1/41/21/41由已知PX1X2=0=1,即等价于PX1X20=0,可知PX1=1,X2=1=0,PX1=-1,X2=1=0.再由p1=p-11+p11+p01, 得p01=12, p-10=p-1=p-11=14,p10=p1-p11=14,从而得p00=0.(2)由于p-10=14p-1p0=1412=18, 所以知X1与X2不独立.习题14设(X,Y)的联合密度函数为f(x,y)=1R2,x2+y2R20,其它,(1)求X与Y的边缘概率密度；(2)求条件概率密度，并问X与Y是否独立？解答：(1)当xR时，fX(x)=-+f(x,y)dy=-+0dy=0;当-RxR时，fX(x)=-+f(x,y)dy=1R2-R2-x2R2-x2dy=2R2R2-x2.于是fX(x)=2R2-x2R2,-RxR0,其它.由于X和Y的地位平等，同法可得Y的边缘概率密度是：fY(y)=2R2-y2R2,-RyR0,其它.(2)fXY(xy)=f(x,y)fY(y)注意在y处x值位于xR2-y2这个范围内，f(x,y)才有非零值，故在此范围内，有fXY(xy)=1R22R2R2-y2=12R2-y2,即Y=y时X的条件概率密度为fXY(xy)=12R2-y2,xR2-y20,其它.同法可得X=x时Y的条件概率密度为fYX(yx)=12R2-x2,yR2-x20,其它.由于条件概率密度与边缘概率密度不相等，所以X与Y不独立.习题15设(X,Y)的分布律如下表所示求：(1)Z=X+Y; (2)Z=maxX,Y的分布律.解答：与一维离散型随机变量函数的分布律的计算类似，本质上是利用事件及其概率的运算法则. 注意，Z的相同值的概率要合并.概率(X,Y)X+YXYX/YmaxX,Y 1/102/103/102/101/101/10 (-1,-1)(-1,1)(-1,2)(2,-1)(2,1)(2,2)-1-2-2241-1-1/2-221-于是(1)X+Y-20134pi1/102/105/101/101/10 (2)maxX,Y-112pi1/102/107/10习题16设(X,Y)的概率密度为f(x,y)=1,0x1,0y2(1-x)0,其他，求Z=X+Y的概率密度.解答：先求Z的分布函数Fz(z)，再求概率密度 fz(z)=dFz(z)dz.如右图所示.当z0时，Fz(z)=PX+Yz=0;当0z1时，Fz(z)=PX+Yz=x+yzf(x,y)dxdy=0zdx0z-x1dy=0z(z-x)dx=z2-12x20z=12z2；当1z2时，Fz(z)=02-zdx0z-xdy+2-z1dx02(1-x)dy=z(2-z)-12(2-z)2+(z-1)2；当z2时，Df(x,y)dxdy=01dx02(1-x)dy=1.综上所述Fz(z)=0,z012z2,0z1z(2-z)-12(2-z)2+(z-1)2,1z21,z2，故fz(z)=z,0z12-z,1z0,y00,其它,求随机变量Z=X+2Y的分布函数.解答：按定义FZ(Z)=Px+2yz, 当z0时，FZ(Z)=x+2yzf(x,y)dxdy=x+2yz0dxdy=0. 当z0时，FZ(Z)=x+2yzf(x,y)dxdy=0zdx0(z-x)/22e-(x+2y)dy=0ze-x(1-ex-z)dx=0z(e-x-e-z)dx=-e-x0z-ze-z=1-e-z-ze-z, 故分布函数为 FZ(Z)=0,z01-e-z-ze-z,z0.习题18设随机变量X与Y相互独立，其概率密度函数分别为fX(x)=1,0x10,其它, fY(y)=Ae-y,y00,y0,求：(1)常数A; (2)随机变量