统计推断答案(打印版)

1、Solutions Manual forStatistical Inference, Second EditionGeorge Casella University of FloridaRoger L. BergerNorth Carolina State University Damaris SantanaUniversity of Florida Second Edition3-13c. (i) h(x) = 1 I(x), c() = 0, w() = , w() = ,x 0x()12t1(x) = log(x), t2(x) = x.(ii) A line.xd. (i) h(x)

2、= C exp(x4)I(x), c() = exp(4) 1, -2, 5 a. In Exercise 3.34(a) w1() = 1and for a n(e, e ), w1() = 1 .2eb. EX = = , then = . Therefore h(x) = 1 I(x),x 0x 0, w1() = , w2() = , t1(x) = log(x), t2(x) = x.c. From (b) then (1, . . . , n, 1, . . . , n) = (1, . . . , n, 1 , . . . , n )3.37 The pdf (

3、1 )f ( (x) ) is symmetric about because, for any 0,o1 f .(+) . =1 f . . =1 f . . =1 f .() . . Thus, by Exercise 2.26b, is the median.3.38 P (X x) = P (Z + z + ) = P (Z z) by Theorem .39 First take = 0 and = 1.a. The pdf is symmetric about 0, so 0 must be the median. Verifying this, write 1 1

4、11.1 . .1.P (Z 0) =0 1+z2 dz =tan(z).=.02 0= 2 .b. P (Z 1) = 1 tan1(z). = 1 = 1 . By symmetry this is also equal to P (Z 1).1 . 24 .424Writing z = (x )/ establishes P (X ) = 1and P (X + ) = 1 .3.40 Let X f (x) have mean and variance 2. Let Z = X . ThenEZ =. 1 .E(X ) = 0andVarZ = Var. X . 1 .=2Var(X

5、) =. 1 .22VarX =2= 1.Then compute the pdf of Z, fZ (z) = fx(z + ) = fx(z + ) and use fZ (z) as the standardpdf.3.41 a. This is a special case of Exercise 3.42a.b. This is a special case of Exercise 3.42b.3.42 a. Let 1 2. Let X1 f (x 1) and X2 f (x 2). Let F (z) be the cdf corresponding tof (z) and l

6、et Z f (z).ThenF (x | 1)=P (X1 x)=P (Z + 1 x)=P (Z x 1)=F (x 1)F (x 2)=P (Z x 2)=P (Z + 2 x)=P (X2 x)=F (x | 2).3-14Solutions Manual for Statistical InferenceThe inequality is because x 2 x 1, and F is nondecreasing. To get strict inequality for some x, let (a, b be an interval of length 1 2 with P

7、(a 0.Let x = a + 1. ThenF (x | 1)=F (x 1)=F (a + 1 1)=F (a) 2. Let X1 f (x/1) and X2 f (x/2). Let F (z) be the cdf corresponding tof (z) and let Z f (z). Then, for x 0,F (x | 1)=P (X1 x)=P (1Z x)=P (Z x/1)=F (x/1)F (x/2)=P (Z x/2)=P (2Z x)=P (X2 x)=F (x | 2).The inequality is because x/2 x/1 (becaus

8、e x 0 and 1 2 0), and F is nondecreasing. For x 0, F (x | 1) = P (X1 x) = 0 = P (X2 x) = F (x | 2). Toget strict inequality for some x, let (a, b be an interval such that a 0, b/a = 1/2 andP (a 0. Let x = a1. ThenF (x | 1)=F (x/1)=F (a1/1)=F (a) 0, by Theorem 2.1.3. For 1 2,FY (y|1) = 1 FX. 1 . 1y.

9、1 FX. 1 . 2y.= FY (y|2)for all y, since FX (x|) is stochastically increasing and if 1 2, FX (x|2) FX (x|1) for all x. Similarly, FY (y|1) = 1 FX ( 1 |1) 2, FX (x|2) 2 and 1, 2 0 then 1 1 . Therefore1FX (x| 1 ) FX (x| 1 ) for all x and FX (x| 1 ) 2/9 = EX2/b2.Thus EX2/b2 is a better bound. But for b

10、= 2, E|X|/b = 1/Thus E|X|/b is a better bound.2 0.b.MX (t)=etxfX (x)dxa aetxfX (x)dxeta fX (x)dx=etaP (X a),where we use the fact that etx is decreasing in x for t 0.c. h(t, x) must be nonnegative.2123.46 For X uniform(0, 1), = 1and 2 = 1 , thus. 1k 1k . 2k . 1 12k k) = 1 P2 12 X 2 + 12=0k 3,For X e

11、xponential(), = and 2 = 2, thusP (|X | k) = 1 P ( k X + k) =. 1 + e(k+1) ek1k 1e(k+1)k 1.From Example 3.6.2, Chebychevs Inequality gives the bound P (|X | k) 1/k2.Comparison of probabilitiesku(0, 1)exactexp()exactChebychev.1.942.926100.5.711.61741.423.135821.44300.0651.33200.0498.25400.006

12、74.06251000.01So we see that Chebychevs Inequality is quite conservative.3.472P (|Z| t)=2P (Z t)=21tex2 /2dx2. 2 1+x2=ex /2dxt1+x2. 2 . 12x22 =ex /2dx+ ex /2dx. .t1+x2t1+x23-16Solutions Manual for Statistical Inference222To evaluate the second term, let u = x , dv = xex /2dx, v = ex /2, du = 1x , to

13、obtain x221+x2 x 2.1 x2(1+x2 )22t1 + x2ex /2dx=1 + x2(ex /2). .tt(1 + x2)2(ex /2)dx t21 x22Therefore,=1 + t2et /2 + t(1 + x2)2ex /2dx. 2 t t2 /2. 2 .1 1 x2 .x2 /2P (Z t)= 1 + t2 e+1 + x2 + (1 + x2)2edx2t. 2t=et /2 +. 2 .2. ex /2dx2 1 + t2. 2 t t2 /2.t(1 + x2)2 1 + t2 e3.48 For the negative binomialP

14、 (X = x + 1) =.r + x + 1 1.x + 1pr (1 p)x+1 =. r + x .x + 1(1 p)P (X = x).For the hypergeometric (M x)(kx+x+1)(x+1)if x k, x M , x M (N k)P (X=x)MP (X = x + 1) =( MN x+1)(kx 1)N(k)if x = M (N k) 10otherwise.3.49 a. 1 1x/E(g(X)(X ) =0g(x)(x ) () x edx.Let u = g(x), du = g(x), dv = (x )x1ex/ , v = xex

15、/ . Then 1. x/ . x/.Eg(X)(X ) = ()g(x)x e.+ .00g(x)x edx .Assuming g(x) to be differentiable, E|Xg(X)| u.4.12 One interpretation of “a stick is broken at random into three pieces” is this. Suppose the length of the stick is 1. Let X and Y denote the two points where the stick is broken. Let X and Y

16、both have uniform(0, 1) distributions, and assume X and Y are independent. Then the joint distribution of X and Y is uniform on the unit square. In order for the three pieces to form a triangle, the sum of the lengths of any two pieces must be greater than the length of the third. This will be true

17、if and only if the length of each piece is less than 1/2. To calculate the probability of this, we need to identify the sample points (x, y) such that the length of each piece is less than 1/2. If y x, this will be true if x 1/2, y x 1/2 and 1 y y, each piece will have length less than 1/2 if y 1/2,

18、 x y 1/2 and 1 x 0, then X Y . So for v = 1, 2, . . ., the joint pmf isfU,V (u, v)=P (U = u, V = v)=P (Y = u, X = u + v)=p(1 p)u+v1p(1 p)u1=p2(1 p)2u+v2.Second Edition4-154.45 a. We will compute the marginal of X. The calculation for Y is similar. Start with1fXY (x, y)=2X Y ,122. 1 . xX .2. xX . . y

19、Y . yY .exp 2(12)2XX+YYand compute1 1 2 2z+z2 )Y dz,fX (x)=fXY (x, y)dy = e 2(12 ) ( 2X Y ,12where we make the substitution z = yY , dy = Y dz, = xX . Now the part of theYXexponent involving 2 can be removed from the integral, and we complete the square in zto get2e 2(12 ) 1 22 22 2fX (x)=2X,12e 2(1

20、2 ) (z2z+ ) dz222 22 1 2e /2(1 )e /2(1 ) =e 2(12 ) (z)dz.2X ,12The integrand is the kernel of normal pdf with 2 = (1 2), and = , so it integratesto 2,12. Also note that e /2(1 )e /2(1 ) = e /2. Thus,222 2222e /2 1 1 xX 22 2 . .fX (x) =Xthe pdf of n(X , 2 ).b.2X,122,1= eX,2XfY |X (y|x) 1 . xX 2xXyYyY

21、 2 . X . 2.X .Y .+.Y .e12X Y 12=2(12 ) 1 1 2 22 (xX )X2X e.1 1. xX 22xX 2xXyYyY 2 . 2(12 )X . (1 ).X . 2.X .Y .+.Y .=2Y ,1e2 2 . .2 .11 2(12 ).2 . xX2xXyY+ yY=2Y ,1e2XXYY2 22 (12 .(yY ). X (xX ).11YY=21 2 e,Y , which is the pdf of n.(Y (Y /X )(x X ) , Y ,1 2.c. The mean is easy to check,E(aX + bY )

22、= aEX + bEY = aX + bY ,4-16Solutions Manual for Statistical Inferenceas is the variance,XVar(aX + bY ) = a2VarX + b2VarY + 2abCov(X, Y ) = a22Y+ b22+ 2abX Y .aTo show that aX + bY is normal we have to do a bivariate transform. One possibility is U = aX + bY , V = Y , then get fU,V (u, v) and show th

23、at fU (u) is normal. We will do this in the standard case. Make the indicated transformation and write x = 1 (u bv), y = v and=.obtaina.Then.|J | = .1/ab/a .1 01. 1112 2 .2 2(12 ) . a (ubv)a (ubv)+vfUV (u, v) =,e.2a12Now factor the exponent to get a square in u. The result is 1. b2 + 2ab + a2 . .u2.

24、 b + a . 2. 2(12)a2b2 + 2ab + a2 2b2 + 2ab + a2uv + v.Note that this is joint bivariate normal form since U = V = 0, 2 = 1, 2 = a2 + b2 + 2abandthus =Cov(U, V )U VE(aXY + bY 2)=U Vvua + b=,a2 + b2 + ab2 2222222a + ab + b (1 ) = 1 (1 )a (1 )a ua2 + b2 + 2ab = a2 + b2 + 2ab =2where a,12 = U ,12. We ca

25、n then writefUV (u, v) =1exp ., 1. u2uv,2 2 v2 .+,22U V12212UU VV4.46 a.which is in the exact form of a bivariate normal distribution. Thus, by part a), U is normal.EX=aX EZ1 + bX EZ2 + EcX=aX 0 + bX 0 + cX=cXVarX=a2 VarZ1 + b2 VarZ2 + VarcX=a2+ b2X XXXEY=aY 0 + bY 0 + cY=cYVarY=a2 VarZ1 + b2 VarZ2

26、+ VarcY=a2+ b2Y YYYCov(X, Y )=EXY EX EY=E(aX aY Z2 + bX bY Z2 + cX cY + aX bY Z1Z2 + aX cY Z1 + bX aY Z2Z112+ bX cY Z2 + cX aY Z1 + cX bY Z2) cX cY =aX aY + bX bY ,since EZ2 = EZ2 = 1, and expectations of other terms are all zero.12b. Simply plug the expressions for aX , bX , etc. into the equalitie

27、s in a) and simplify.c. Let D = aX bY aY bX = ,12X Y and solve for Z1 and Z2,bY (XcX ) bX (Y cY )Y (X X )+X (Y Y )Z1=DY (X X )+X (Y Y ),2(1+)X YZ2=,2(1.) X YSecond Edition4-17Then the Jacobian is. z1z1 .bY bXJ =x1 y .=DD.aX bY=aY bX 1 1 z2z2aYaXD2 D2= D =,xyDD12X Yand we have that X )+X (yY )2(Y (xX

28、 )+X (yY )211111 (Y (xe efX,Y (x, y)= 222(1+)2 222(1)2 2 ,X YX Y212X Y=(2X Y ,1 2)1 exp.1. x X .2.2(1 2)X2x X . y Y . y Y .2X+YY, x , y ,a bivariate normal pdf.d. Another solution isaX=X bX = ,(1 2)X aY=Y bY=0cX=XcY=Y .There are an infinite number of solutions. Write bX = ,2 a2 ,bY = ,2 a2 , andsubs

29、titute bX ,bY into aX aY = X Y . We getXXYY . . . .aX aY +2 a22 a2= X Y .XXYYSquare both sides and simplify to get(1 2)2 2= 2 a2 2X Y aX aY + 2 a2 .X YX YY XThis is an ellipse for = 1, a line for = 1. In either case there are an infinite number of points satisfying the equations.4.47 a. By definitio

30、n of Z, for z 0) + P (X z and XY 0)=P (X z and Y 0) + P (X z and Y 0)(since z 0)=P (X z)P (Y 0) + P (X z)P (Y 0)(independence)=P (X z)P (Y 0)(symmetry of Xand Y )=P (X z)(P (Y 0)=P (X z).By a similar argument, for z 0, we get P (Z z) = P (X z), and hence, P (Z z) =P (X z). Thus, Z X n(0, 1).b. By de

31、finition of Z, Z 0 either (i)X 0 or (ii)X 0 and Y 0. So Z andY always have the same sign, hence they cannot be bivariate normal.4-18Solutions Manual for Statistical Inference4.49 a.fX (x)= (af1(x)g1(y) + (1 a)f2(x)g2(y)dy=af1(x) g1(y)dy + (1 a)f2(x) g2(y)dy=af1(x) + (1 a)f2(x).fY (y)= (af1(x)g1(y) +

32、 (1 a)f2(x)g2(y)dx=ag1(y) f1(x)dx + (1 a)g2(y) f2(x)dx=ag1(y) + (1 a)g2(y).b. () If X and Y are independent then f (x, y) = fX (x)fY (y). Then,f (x, y) fX (x)fY (y)=af1(x)g1(y) + (1 a)f2(x)g2(y) af1(x) + (1 a)f2(x)ag1(y) + (1 a)g2(y)=a(1 a)f1(x)g1(y) f1(x)g2(y) f2(x)g1(y) + f2(x)g2(y)=a(1 a)f1(x) f2

33、(x)g1(y) g2(y)=0.Thus f1(x) f2(x)g1(y) g2(y) = 0 since 0 a 1.() if f1(x) f2(x)g1(y) g2(y) = 0 thenf1(x)g1(y) + f2(x)g2(y) = f1(x)g2(y) + f2(x)g1(y).ThereforefX (x)fY (y)=a2f1(x)g1(y) + a(1 a)f1(x)g2(y) + a(1 a)f2(x)g1(y) + (1 a)2f2(x)g2(y)=a2f1(x)g1(y) + a(1 a)f1(x)g2(y) + f2(x)g1(y) + (1 a)2f2(x)g2(y)=a2f1(x)g1(y) + a(1 a)f1(x)g1(y) + f2(x)g2(y) + (1 a)2f2(x)g2(y)=af1(x)g1(y