概率論復(fù)習(xí)題解答(第4章)

第4XX–202pi求E(X)，E(X)，E(3X．2=EX)=20.30.3=ii12p=40.4+00.3+40.3=EX)=xi2i1EXEX)=30.2=謝謝閱讀感謝閱讀1XXiiP{X1/i,6感謝閱讀8X88感謝閱讀EX)謝謝閱讀iEX)pp感謝閱讀12nn精品文檔放心下載精品文檔放心下載感謝閱讀XX~B,pEXnp精品文檔放心下載11Y,則Y,感謝閱讀記A，BpA∪=p+p-pp1212EYnp+p-pp)1212將nnXn．

X~BEX)謝謝閱讀精品文檔放心下載精品文檔放心下載設(shè)X~P()P{XP{X．謝謝閱讀（XPXk=kk!ek=又PX5X6,5e6e6)=XXk}6k22,k4,,問(wèn)X(()(()(感謝閱讀6661，而k1kk1k12k22kk2k1k1k11X.kk1謝謝閱讀1xe,xx/3f(x)9其它.0)=xf(x)dxxxedxx11x/3精品文檔放心下載2ex/3dx0990感謝閱讀謝謝閱讀．251,xF(x)x2其它.0Xf(x)F(x)謝謝閱讀當(dāng)x>5f(x)2xx33x5時(shí)，f(x))=xf(x)dxx|-5xx35)=10.X謝謝閱讀f(x)42x(1x),0x精品文檔放心下載0其它.求X．)=xf(x)dx12x)5-0X．3x)，1x，223f(x)x)

22,0x，，.0313E(X)xf(x)dxxx)dxxx)0.謝謝閱讀221202設(shè)X~B(4,p)E(sinX)．精品文檔放心下載2XP{Xk}Cnk，k=，謝謝閱讀pp)kk

nX4sinX以2XE(sin)1Cpp)31Cpp)142p)精品文檔放心下載1133244VW是VWkV，感謝閱讀2k>W1,0vaVf(v)a其它a1k11E2f(v)dxkv精品文檔放心下載2(v3)|2a0aa330,YYX01203/289/283/2813/143/14021/2800求YX–Y．)=謝謝閱讀謝謝閱讀感謝閱讀24xy,0xyxy1設(shè)隨機(jī)變量f(x,y)感謝閱讀謝謝閱讀)yx24x211xyx1D00121x24x22224x3x4x36x45)謝謝閱讀1102550

0)y24xydxdy24y22/511y精品文檔放心下載Dxy24xydxdy24x2001ydydx2123031x00D8421(36x45x6).謝謝閱讀3530X精品文檔放心下載X10111314pi0.20.30.30.10.1YX),求．謝謝閱讀=］×××××=感謝閱讀Y)=］2=1000[(12-10)）））謝謝閱讀22222）26-6XP{Xk}p)k1p,k,感謝閱讀其中0<p<1．

p則謝謝閱讀E(X)(kP{X(kqk1p)pkqpdqkdqk1k1k1k1k1pqp()1/pdd1kdqdq1qk0E(X)(kP{X(kqk1p)k(kqk1kq精品文檔放心下載222k1]k1k1k1k1k21/ppqq

pqk(kd2d2

qq1/ppq(k)1/pk1k22k0k1d212()1/p1/p2q/p1/p

21qq)23)=)=p謝謝閱讀211,|x|Xf(x)1x2．

其它精品文檔放心下載dxxf(x)dxx0111x12Xxf(x)dxx211221x12x2t[/2,/2]/2/sin2tcost21cos2t1/2dt220X，)具有)=9，)=4，D(X4)．XY1/6D(XY)，XYCov(X,Y)D(X))Cov(X,Y)XYD(X))謝謝閱讀D(XY)D(X))2Cov(X,Y)942感謝閱讀D(XD(X)9D(Y)2Cov(X,3Y)96(精品文檔放心下載在題13Cov，))．XY精品文檔放心下載X0+12感謝閱讀2222Y0+12感謝閱讀2222X)］=精品文檔放心下載222Y］謝謝閱讀222CovXY))謝謝閱讀=Cov)D(X))(9/2845/112-5謝閱讀XY謝在題14Cov，X+．精品文檔放心下載XY222E(X)E(Y),E(XY),Cov(X,Y)E(XY)E(X)E(Y)5精品文檔放心下載E(X2)24x11x13E(Y2)0052141D(X)E(X2)E(X))

52525謝謝閱讀XYCov(X,Y)2D(X))3D(XY)D(X))2Cov(X,Y)275,Y1

,x2y2f(x,y)其它.0X和YX和YE(X)x/dydx2x1x11x212/011x21E(Y)y/1dydx01x211x2E(XY)1x2xy/dydx0，111x2Cov，X和Y.精品文檔放心下載XYf(x)21xXf(x,y)dy1/dy,1x1，1x11x221x2，其他，其他f(y)Yf(x,y)dx21y21y2，1y11/dx,1y11y2，其他，其他當(dāng)x+y≤1f(x,yf22X()fX和Y的Y,Y1/|y|2x,0x1f(x,y)其它.0yX和YX和Yf(x,y:0<x<|yx謝謝閱讀y2xE(X)xf(x,xdydx12112x2,2x02x23D00,E(Y)(x,112x2x20DE(XY)(x,O0Cov，yx2xxyDCov(x,y)D(x)D(y)112x2x200X與Y關(guān).fX(x)2x1f(x,y)dy2x2他其f1y11,2

y0y22421y(y)f(x,y)dx

12dx,0y

2Y24y2其他0<-2<y<2f(x,y)fX(x)fY(y)X和Y的.mnm,n,.精品文檔放心下載xQYyn(xY),0Yx

QQ(Y)mx,Yx0<xYfY(y)1yey,0y0xx11yyE(Q)Q(Yf)dymyn(xy)e.dyedy謝謝閱讀Y0xyyydexx(mn)ydenx

mxde00xemnnxeyyyy(mn)yexmxexx感謝閱讀000x(mn)xemnenxenxmxexyxx0x(mn)emnnxx精品文檔放心下載dE(Q)1xx令(mn)en(mn)en0精品文檔放心下載dxnnx則e,xmnmnd2E(Q)mnx又e0dxn當(dāng)x時(shí)，E(Q)取最大值mn謝謝閱讀X感謝閱讀酬mnr精品文檔放心下載Y則ke,k

rXXrYPYkYXrrie,krZ,則Z與YYmynrryYZgYrmrirEgYg(kk)g(kk)g(r)P(Yr)感謝閱讀rr1k0k0ekmrkne謝謝閱讀r1kimrk0irkmnenremree精品文檔放心下載r1kr1kr1kkk0k0k0k0mnnrr2ker1kemr感謝閱讀k0k0.精品文檔放心下載0時(shí)EgY此時(shí)r1503333X精品文檔放心下載謝謝閱讀XX精品文檔放心下載CkC3kP{Xk}，33C36即X01231991piE(X)019913123.2設(shè)A{X{X{X，精品文檔放心下載{X精品文檔放心下載P(Xk}P{AXk}精品文檔放心下載3k0=Xk}Xk}k13366k0k01131

=E(X).66241xcos,Xf(x)220xX獨(dú)立重復(fù)觀察4其他用YY2于3Y~Bp=PX>f(x)dx1xx13/3/32222/3謝謝閱讀所以p，Y精品文檔放心下載)=U感謝閱讀若U1若U1X;Y.若U1若U1X和YD(XY)．謝謝閱讀f(u)U1,2u24其他XYU-1且UUXYU≤-1且U，感謝閱讀謝謝閱讀11du，1244XYUdu，謝謝閱讀111

142XYU>-1且UU>所以X和YX1Y2111du，4410X和YX–11pi–1Y1pi，

，

CovX+YCov精品文檔放心下載感謝閱讀感謝閱讀XD(X)E(X)a,D(X)b(b，精品文檔放心下載XaE(Y))1．Y感謝閱讀bXP{Xx}piiE(X)E(X)xpa,i,i,ii1記YXag(X),則g(xxa1pp)pibbbxabi1i1i

iiii

i1E(Y)EXa0,)DXaD(X)1精品文檔放心下載bbb若X精品文檔放心下載由EX存在知xfxdx絕對(duì)收斂。感謝閱讀則Xafxdx1xfxdxafxdxbb1xfxdxa

bxfxdx因?yàn)榻^對(duì)收斂，所以1xfxdxa絕對(duì)收斂b即EY存在，且EYEEXa感謝閱讀Xa11EXa0感謝閱讀bbbDYDDXaDX1Xa11感謝閱讀bbbXP{Xx}pk,(k)，X，k2f(x)(xx)2pk在xE(X)為．k1k2(xdf(x)x)p0，dxk1kk則xxpp0，kkkk1k1xxppE(X)x0xE(X)kkk謝謝閱讀k1k1d2f(x)又10xE(X)f(x)(xx)p2dx2kkk1E(X2p

f(E(X))(x

kD(X)k1kX與YXX12i2/31/3記U,Y),V,Y)，謝謝閱讀求U與VCovXY12X12(X)））））ijU1222V1112V12U102，

，

CovX謝謝閱讀精品文檔放心下載1/2,1x0

f(x)1/4,0x2X令YX,F(x,y)Cov．謝謝閱讀E(X)xf(x)dx1/2xdx1/4xdx1/40210E(Y)E(X)xf(x)dxx/2dxx精品文檔放心下載2/4dx5/602222ｘ103/4dx7/802x3fｘ10則：Cov(X,Y)E(X3)E(X)E(X2)謝謝閱讀7/8(1/4)(5/6)

2/3謝謝閱讀A和0<)<0<)<，P(AB)P(P(B)P(A)P(B)P(A)P(B)A和BA和B精品文檔放心下載1．感謝閱讀