《概率论与数理统计》习题参考答案第四章.pdf

1 5VnO6 SKY SK4 (A) 1.w, EX = 1,EY = 1.1. 2.d1 = P5 k=1PX = k = P5 k=1 C k C = 60 137, dEX = P5 k=1kPX = k = 5C = 300 137. 3.w, X?1,0,1?VO 1 6, 1 3, 1 2, dEX = 1 3. 4. EX 6= 1.y:eEX = 1,K1 = EX = R Rxf(x)dx = R1 0 x cxb= c b+2. q1 = R Rf(x)dx = R1 0 cxbdx = c b+1, g. 5.?u.E(X2) = 0.8,?(EX)2= 0.64. 6.dK E(Xn) = Z R xnf(x)dx = Z 1 0 xnxdx + Z 2 1 xn(2 x)dx = 2(2n+1 1) (n + 1)(n + 2) 7.dK (1)PX = 1 = 0.4,PX = 2 = 0.2,PX = 3 = 0.4, PY = 1 = 0.3,PY = 0 = 0.4,PY = 1 = 0.3 EX = 2,EY = 0 (2)E(Z) = E ? Y X ? = X i X j yj xi pij= 1 15 (3)E(Z) = E(X Y )2= X i X j (xi yj)2pij= 5 8.duX,Yp,dk E(X Y ) = EX EY = ?Z x fX(x)dx ? ?Z y fY(y)dy ? = 2 3 6 = 4 9. DX = EX2 (EX)2= 1.7 10.duf(x)3R,dEX = R Rxf(x)dx = 0, ? DX =E(X2) (EX)2= Z R x2f(x)dx = Z 1 1 x2 1 x2 dx = 2 Z 1 0 x2 1 x2dx = 2 Z 2 0 sin2 cos cosdx = 2 Z 2 0 1 cos2 2 d = 1 2(C“ x = sin) 11.duf(x)3R,dEX = R Rxf(x)dx = 0, ? DX = E(X2) (EX)2= Z R x2f(x)dx = Z + 0 x2 1 2e xdx = Z + 0 x2exdx = 2 12.?3g?u)?gX,u)?Vp,KXlp?0 1 (0 =D(XY ) = E(X2Y 2) E(XY )2 = E(X2)E(Y 2) (EX EY )2 m=E(X2) (EX)2E(Y 2) (EY )2 + (EX)2E(Y2) (EY )2 +(EY )2E(X2) (XY )2 = E(X2)E(Y 2) (EX EY )2 16.dK D(X + Y ) =DX + DY + 2cov(X,Y ) = DX + DY + 2XY DXDY = 85 D(X Y ) =DX + DY 2cov(X,Y ) = DX + DY 2XY DXDY = 37 Cov(X + Y,X Y ) =E(X + Y )(X Y ) E(X + Y ) E(X Y ) =E(X2) (EX)2 E(Y 2) (EY )2 = DX DY = 11 17.dK XY= Cov(X,Y ) DXDY= D(X + Y ) DX DY DXDY= 0.6 18.duf(x,y)u?:%,dkxyf(x,y),xf(x,y),yf(x,y)Oux,x,y?, dk Cov(X,Y ) = E(XY ) EX EY = 0 0 0 = 0 19.dK = Cov(,) DD= Cov(aX + b,cY + d) pD(aX + b)pD(cY + d)= acCov(X,Y ) a2DXc2DY= ac |ac|XY 20.w, F(x,y) = FX(x) FY(y),(x,y) R2, X,Yp. (B) 1. (1) ; (2) (D(X + Y ) = D(X Y ) Cov(X,Y ) = 0); (3) (dD(X + Y ) = D(X Y )Cov(X,Y ) = 0,l?X,Y); (4) ; (5) (dX + Y= nCov(X,Y ) = Cov(X,n X) = Cov(X,X) = DX = XY DXDY = XY DXpD(n X) = XY DXDX, l?XY= 1); (6) (Cov(X + Y,X Y ) = DX DY = 0). 2. (1) 1 6; w, f(x)R?,dEX = 0,?E(X2) = 2 R1 0 x2(1 x)dx = 1 64, d, DX = E(X2) (EX)2= 1 6. (2) 1 2,5; (w, = p100p(1 p) q 100 ?p+1p 2 ?2 = 5,?=?p = 1 p=p = 1 2 ?); (3) 1; 1 = E(X 1)(X2) = E(X2) 3EX + 2 = DX + (EX)2 3EX + 2 = + 2 3 + 2 = = 1; (4) 8 9; PY = 1 = PX 0 = Z 2 0 1 3dx = 2 3,PY = 0 = PX = 0 = 0, PY = 1 = 1 PY = 1 PY = 0 = 1 3,EY = 1 3,F(Y 2) = 1,DY = 8 9 3 (5) 46; duX1,X2,X3p,dDY = D(X1 2X2+ 3X3) = DX1+ 4DX2+ 9DX3= (60)2 12 + 4 22+ 9 3 = 46; (6) N(0,5); duX,Yp,dk EZ = E(X 2Y + 7) = EX 2EY + 7 = 0,DZ = DX + 4DY = 5 (7) 0; duXij,i,j = 1,2,np,d EY = ? ? ? ? ? ? ? ? ? ? 222 222 . . . . . . . . . 222 ? ? ? ? ? ? ? ? ? ? = 0 (8) 0.02; dK, X2,Y 2 ? L1SK4(B)-2(8) X2 Y 2 01PX2 00.180.220.4 10.320.280.6 PY20.50.51 d, Cov(X2,Y 2) = E(X2Y2) E(X2)E(Y2) = 0.28 0.6 0.5 = 0.02. (9) 0; duCov(X,Y ) = E(XY ) EX EY = 0.12 0.6 0.2 = 0,dXY= 0. (10) 0.9; Cov(Y,Z) = Cov(Y,X 0.4) = Cov(X,Y ), Y Z= Cov(Y,Z) DYDZ= Cov(X,Y ) DYDX= XY= 0.9 (11) 6; duEX = EY = 0,dk E(X + Y )2=E(X2) + E(Y 2) + 2E(XY ) = 2 + 2 + 2(E(XY ) EX EY ) = 4 + 2Cov(X,Y ) =4 + 2XY DXDY = 4 + 2 0.5 pE(X2) (EX)2pE(Y 2) (EY )2 = 6 3. E(Y ) =E ? 1 X ? = Z + 0 1 x x a2 e x 2a2dx = 1 a Z + 0 e 1 2( x a) 2d?x a ? ? -y = x a ? = 1 a Z + 0 e 1 2y 2dy =1 2a Z + e 1 2y 2dy =1 2a ?ZZ R2 e x2+y2 2 dxdy ?1 2 = 2 2a 4. (1) F(x) = 0,x a = Z + a f(x)dx qA = X a,B = Y ap, P(A B) = 3 4, ?P(AB) = P(A)P(B) 0 12 = P ?X 1 12 ? = 1 (12 ) ET = 25(12 ) 21(10 ) 5 5 d ET = 0, 25(12 ) = 21(10 ),?(x) = 1 2e x2 2,x R,l?k = 11 1 2 ln 25 21. 11.dK EX =10 0.85+ 5 C1 5 0.2 0.8 4 + 0 C0 5 0.2 2 0.83 +(2) (1 0.85 C1 5 0.2 0.8 4 C0 5 0.2 2 0.83) = 5.2 12.?mY ,K Y = g(X) = 5 X,0 X 5 25 X,5 3 = Z 3 1 2 cos x 2 dx = sin x 2 ? ? ? ? 3 = 1 2 = Y B ? 4, 1 2 ? E(Y 2) = (EY )2 + DY = 22+ 1 = 5 15. (1)b?gX,K PX = 0 = 1 20,PX = 1 = 9 20,PX = 2 = 9 20,PX = 3 = 1 20 , EX = 1.5. (2) P = 1 20 0 + 9 20 1 6 + 9 20 2 6 + 1 20 3 6 = 1 4 16. (1)df(x,y)x,y?5 PX r2 u?:,dCov(X,Y ) = E(XY )EX EY = 0 0 0 = 0. 6 (2). 19.dK, T = X + Y , fT(t) = Z R fX(x)fY(t x)dx = ( 25te5t,t 0 0,t 2Y G,K PX = 0,Y = 0 = PX Y = SG1 SG = 1 4,PX = 0,Y = 1 = P = 0, PX = 1,Y = 0 = PY 2Y = SG3 SG = 1 2, PX = 0 = 1 4,PX = 1 = 3 4,PY = 0 = 1 2,PY = 1 = 1 2 (2) Cov(X,Y ) = E(XY ) EX EY = 1 2 3 4 1 2 = 1 8, DX = 3 4 1 4 = 3 16,DY = 1 2 1 2 = 1 4, XY= Cov(X,Y ) DXDY= 1 3 22.dK PX = 1,Y = 1 = P(AB),PX = 1,Y = 1 = P(AB), PX = 1,Y = 1 = P(AB),PX = 1,Y = 1 = P(A B), EX = P(A) P(A),EY = P(B) P(B), E(XY ) = P(AB) + P(A B) P(AB) P(AB) = 1 2P(AB) 2P(AB) eA,Bp,KA,Bp, A,Bp, A,Bp,u, E(XY ) = EX EY , XY= 0. eXY= 0,KE(XY ) = EX EY ,K1 2P(A) 2P(B) + 4P(AB) = (2P(A) 1)(2P(B) 1),u , P(AB) = P(A)P(B),=A,Bp. 23. (1) f1(x) = Z R f(x,y)dy = 1 2 ?Z R 1(x,y)dy + Z R 2(x,y)dy ? = 1 2(1X + 2X) = 1 2e x2 2 n, f2(y) = 1 2e y2 2 7 (2)d(1), EX = EY = 0,DX = DY = 1,dk XY= Cov(X,Y ) DXDY= E(XY ) EX EY DXDY= E(XY ) = ZZ R2 f(x,y)dxdy =1 2 ?ZZ R2 1(x,y)dxdy + ZZ R2 2(x,y)dxdy ? = 1 2(E(X1Y1) + E(X2Y 2) =1 2(EX1 EY1+ X1Y1 p DX1 p DY1+ EX2 EY2+ X2Y2 p DX2 p DY2) =1 2 ? 0 0 + 1 3 11 + 0 0 + ? 1 3 ? 11 ? = 0 (X1,Y1) 1(x,y),(X2,Y2) 2(x,y). 24.PD = (x,y)|1 x y 1,0 x 1,K f(x,y) = ( 2,(x,y) D 0,(x,y) / D ,fX(x) = ( 2x,0 x 1 0, ,fY(y) = ( 2y,0 y 1 0, d EX = EY = 2 3,E(X 2) = E(Y2) = 1 2,DX = DY = 1 18 E(XY ) = Z 1 0 dx Z 1 1x 2xydy = Z 1 0 xy2 ? ? ? ? 1 1x dx = Z 1 0 x(2x x2)dx = 2 3 1 4 = 5 12 D(X + Y ) = DX + DY + 2Cov(X,Y ) = DX + DY + 2(E(XY ) EX EY ) = 1 18 25. (1)dK PX = 1,Y = 1 = PU 1 = 1 4,PX = 1,Y = 1 = P = 0, PX = 1,Y = 1 = P1 1 = 1 4 (2)d(1) X + Y 202 1 4 1 2 1 4 ! ,E(X + Y ) = 0,E(X + Y )2= 2,D(X + Y ) = 2 26. (1)duA,B?duP(AB) = P(A)P(B),u, A,B?du = 0. (2)?CX,Y X = ( 1,Au) 0,Au) ,Y = ( 1,Bu) 0,Bu) K EX = P(A),EY = P(B),DX = P(A)P(A),DY = P(B)P(B) E(XY ) = 1 X i=0 1 X j=0 ijPX = i,Y = j = PX = 1,Y = 1 = P(AB) u | = ? ? ? ? E(XY ) EX EY DXDY ? ? ? ? = |XY| 1 27. (1)dK, P(AB) = P(A)P(B|A) = 1 12,P(B) = P(AB) P(A|B) = 1 6, dk PX = 1,Y = 1 = P(AB) = 1 12,PX = 1,Y = 0 = P(AB) = P(A) P(AB) = 1 6, PX = 0,Y = 1 = P(AB) = P(B) P(AB) = 1 12,PX = 0,Y = 0 = 1 1 12 1 6 1 12 = 2 3 8 (2)d(1) EX = P(A) = 1 4,DX = P(A)P(A) = 3 16, EY = P(B) = 1 6,DY = P(B)P(B) = 5 36, E(XY ) = 1 X i=0 1 X j=0 ijPX = i,Y = j = PX = 1,Y = 1 = P(AB) = 1 12, XY= Cov(X,Y ) DXDY= E(XY ) EX EY DXDY= 15 15 28. (1)dK Yi= Xi X = Xi 1 n n X i=1 Xi=