Second Edition
413
r
= =
1 (r)(1  r)
0
1 x
x u 1 u
e(u+x/) du xr1 ex/ , (r)r
xr1 ex/ r (r)(1  r)
r
eu du =
0
since the integral is equal to (1  r) if r < 1. b. Use the transformation t = / to get
p ()d =
0
1 (r)(1  r)
r1 (  )r d =
0
1 (r)
Second Edition
915
where c = 2 / 2 . The derivative (with respect to c) of this last expression is bfn1 (bc)  afn1 (ac), and hence is equal to zero if both c = 1 (so the interval is unbiased) and bfn1 (b) = afn1 (a). From the form of the chi squared
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Solutions Manual for Statistical Inference
9.52 a. The LRT of H0 : = 0 versus H1 : = 0 is based on the statistic (x) = sup,=0 L (, 0  x) . sup,(0,) L(, 2  x)
In the denominator, 2 = (xi  x)2 /n and = x are the MLEs, while in the numerator, ^ ^ 2 0
Second Edition
913
9.46 The proof is similar to that of Theorem 9.3.5: P ( C (X) = P (X A ( ) P (X A( ) = P ( C(X) , where A and C are any competitors. The inequality follows directly from Definition 8.3.11. 9.47 Referring to (9.3.2), we want to show tha
410
Solutions Manual for Statistical Inference
the Poisson(p) pmf. Thus Y  Poisson(p). Now calculations like those in a) yield the pmf of Y , for y = 0, 1, . . ., is fY (y) = 1 (y + ) ()y!(p) p 1+p
y+
.
Again, if is a positive integer, Y negative binomi
912
a
Solutions Manual for Statistical Inference
9.39 Let a be such that  f (x) dx = /2. This value is unique for a unimodal pdf if > 0. Let be the point of symmetry and let b = 2  a. Then f (b) = f (a) and b f (x) dx = /2. a a since  f (x) dx = /2 1/
Second Edition
49
4.31 a. EY = Ecfw_E(Y X) = EnX = n . 2 n n2 + . 12 6
VarY = Var (E(Y X) + E (Var(Y X) = Var(nX) + EnX(1  X) = b. P (Y = y, X x) = c. P (y = y) = 4.32 a. The pmf of Y , for y = 0, 1, . . ., is
n y x (1  x)ny , y
y = 0, 1, . . . , n
Second Edition
911
We need to solve this numerically for the smallest n that satisfies the inequality (n  1)n 2 n1,.1 . t2 n1,.025 64 Trying different values of n we find that the smallest such n is n = 276 for which t2 n1,.025 (n  1)n = 306.0 305.5
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Solutions Manual for Statistical Inference
The log of this is t2 t2 = = t2 /2 as ,  /2t  + t 2 + 2 (t 2/ ) + 2 1  t/ 2 so the mgf converges to et /2 , the mgf of a standard normal. b. Since P (2 2 ) = for all , 2Y 2Y, 2 2Y,  2 z as . 8 In stand
Second Edition
47
fZ1 ,Z 2 (z1 , z2 ) can be factored into two densities. Therefore Z1 and Z2 are independent and Z1 gamma(r + s, 1), Z2 beta(r, s). 4.25 For X and Z independent, and Y = X + Z, fXY (x, y) = fX (x)fZ (y  x). In Example 4.5.8, fXY (x, y)
Second Edition
99
an IG n + a, (y + 1 )1 . The Bayes HPD region is of the form cfw_ : (y) k, which is b an interval since (y) is unimodal. It thus has the form cfw_ : a1 (y) a2 (y), where a1 and a2 satisfy 1 1 1 1 1 1 e a1 (y+ b ) = n+a+1 e a2 (y+ b
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Solutions Manual for Statistical Inference
9.25 The confidence interval derived by the method of Section 9.2.3 is C(y) = : y + 1 1 log y + log 1  n 2 n 2
where y = mini xi . The LRT method derives its interval from the test of H0 : = 0 versus H1 : =
Second Edition
45
4.20 a. This transformation is not onetoone because you cannot determine the sign of X2 from Y1 and Y2 . So partition the support of (X1 , X2 ) into A0 = cfw_ < x1 < , x2 = 0, A1 = cfw_ < x1 < , x2 > 0 and A2 = cfw_ < x1 < , x2 < 0
Second Edition
97
x 0 1 2 3
interval [.000,.305) [.305,.634) [.362,.762) [.695,1].
2
9.19 For FT (t) increasing in , there are unique values U (t) and L (t) such that FT (t) < 1  if and only if < U (t) and FT (t) > if and only if > L (t). Hence, 2 P
44
Solutions Manual for Statistical Inference
If V < 0, then X < Y . So for v = 1, 2, . . ., the joint pmf is fU,V (u, v) = P (U = u, V = v) = P (X = u, Y = u  v) = p(1  p)u1 p(1  p)uv1 = p2 (1  p)2uv2 .
If V = 0, then X = Y . So for v = 0, th
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Solutions Manual for Statistical Inference
9.15 Fieller's Theorem says that a 1  confidence set for = Y /X is : x2  t2 n1,/2 n1 s2 X 2  2 xy  t2 n1,/2 n1 sY X + y2 
t2
t2 n1,/2 n1
s2 Y
0 .
a. Define a = x2  ts2 , b = xy  tsY X , c = y 2
Second Edition
43
1 V 2 3 4
1 12 1 12 1 12
U 2
1 6 1 6 1 6
3
1 12 1 12 1 12
4.11 The support of the distribution of (U, V ) is cfw_(u, v) : u = 1, 2, . . . ; v = u + 1, u + 2, . . .. This is not a crossproduct set. Therefore, U and V are not independent
Second Edition
95
9.13 a. For Y = (log X)1 , the pdf of Y is fY (y) =
2
/y , y2 e
0 < y < , and
2
P (Y /2 Y ) =
1
/y e dy = e/y y2
= e1/2  e1 = .239.
b. Since fX (x) = x , 0 < x < 1, T = X is a good guess at a pivot, and it is since fT (t) =
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Solutions Manual for Statistical Inference
The LRT statistic is (x) =
a ^ ^ a0 0 n/2 
e
1 ^ 2a0 0
^ (xi 0 )2
e

1 ^ (xi )2 2^ a^
=
1 ^ 2a0 0
n/2
en/2 e

1 ^ 2a0 0
^ (xi 0 )2
The rejection region of a size test is cfw_x : (x) c , and a 1  confid
42
Solutions Manual for Statistical Inference
The complete definition of FXY is x 0 or y 0 02 x y/8 + y 2 x/4 0 < x < 2 and 0 < y < 1 2 x and 0 < y < 1 FXY (x, y) = y/2 + y 2 /2 . 2 x /8 + x/4 0 < x < 2 and 1 y 1 2 x and 1 y d. The function z = g(x) = 9/
Second Edition
93
The sign of the derivative is given by the expression in square brackets, a parabola. It is easy to see that for 0, the parabola changes sign from positive to negative. Since this is the sign change of the derivative, the function must
Chapter 4
Multiple Random Variables
4.1 Since the distribution is uniform, the easiest way to calculate these probabilities is as the ratio of areas, the total area being 4. a. The circle x2 + y 2 1 has area , so P (X 2 + Y 2 1) = . 4 2 b. The area below
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Solutions Manual for Statistical Inference
therefore 0 is the MLE. The LRT statistic is ^2 X ^2 0
m/2 n/2
Y ^2
m/2
(^0 ) 2
(m+n)/2
,
and the test is: Reject H0 if (x, y) < k, where k is chosen to give the test size . 2 2 2 b. Under H0 , Yi2 /(0 X ) 2
Second Edition
317
3.50 The proof is similar to that of part a) of Theorem 3.6.8. For X negative binomial(r, p), Eg(X)
=
x=0
g(x)
r+x1 r p (1  p)x x r+y2 r p (1  p)y1 y1 y r+y1 (set y = x + 1)
=
y=1
g(y  1)
=
y=1
g(y  1)
r+y1 r p (1  p)y1
Chapter 9
Interval Estimation
9.1 Denote A = cfw_x : L(x) and B = cfw_x : U (x) . Then A B = cfw_x : L(x) U (x) and 1 P cfw_A B = P cfw_L(X) or U (X) P cfw_L(X) or L(X) = 1, since L(x) U (x). Therefore, P (AB) = P (A)+P (B)P (AB) = 11 +12 1 = 11 2
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Solutions Manual for Statistical Inference
To evaluate the second term, let u = obtain
t
x 1+x2 ,
dv = xex
2
/2
dx, v = ex
2
/2
, du =
1x2 (1+x2 )2 ,
to
x2 x2 /2 e dx = 1 + x2 =
2 x (ex /2 ) 2 1+x 2 t et /2 + 2 1+t

t t
2 1  x2 (ex /2 )d