Math 5080-2, F 2015. Assignment 1. Chapter 6, Exercises 2, 4, 9, 10, 13, 14(b),16(a).

2. (a) For 0 < y < 1, FY (y) = P (X1/4 ≤ y) = P (X ≤ y4) = FX(y4) = y4,so fY (y) = 4y3.

(b) For e−1 < w < 1, FW (w) = P (e−X ≤ w) = P (X ≥ − lnw) = 1 + lnw,so fW (w) = 1/w.

(c) For 0 < z < 1 − e−1, FZ(x) = P (1 − e−X ≤ z) = P (e−X ≥ 1 − z) =P (X ≤ − ln(1− z)) = − ln(1− z), so fZ(z) = 1/(1− z).

(d) For 0 < u < 1/4, FU (u) = P (X(1 −X) ≤ u) = P ((X − 1/2)2 ≥ −u +1/4) = P |X − 1/2| ≥

√1/4− u) = 2(1−

√1/4− u), so fU (u) = (1/4− u)−1/2.

4. (a) FX(x) = 1− e−(x/θ)β for x > 0, so FY (y) = P ((X/θ)β ≤ y) = P (X ≤θy1/β) = FX(θy1/β) = 1− e−y for y > 0, hence fY (y) = e−y for y > 0.

(b) FY (w) = P (ln(X) ≤ w) = P (X ≤ ew) = FX(ew) = 1 − e−eβw/θβ for w

real, so fW (w) = e−eβw/θβ (eβw/θβ)β for w real.

(c) FZ(z) = P ((lnX)2 ≤ z) = P (−√z ≤ lnX ≤

√z) = P (e−

√z ≤ X ≤

e√z) = FX(e

√z)−FX(e−

√z) = e−(e

−√z/θ)β − e−(e

√z/θ)β for z > 0, and fZ(z) is

the derivative of this.

9. (a) A = (0, 1), B = (0, 1). fY (y) = fX(y4)|(y4)′| = 4y3 for y ∈ B.(b) A = (0, 1), B = (e−1, 1). fW (w) = fX(− lnw)|(− lnw)′| = 1/w for

w ∈ B.(c) A = (0, 1), B = (0, 1 − e−1). fZ(z) = fX(− ln(1 − z))|(− ln(1 − z))′| =

1/(1− z) for z ∈ B.(d) A1 = (0, 1/2), A2 = (1/2, 1), B = (0, 1/4). Now, if u ∈ (0, 1/4), u =

x(1− x) implies

x =1±√1− 4u

2

with the minus sign if x ∈ A1 and the plus sign if x ∈ A2. So

fU (u) = fX

(1−√1− 4u

2

)∣∣∣∣(1−√1− 4u

2

)′∣∣∣∣+ fX

(1 +√1− 4u

2

)∣∣∣∣(1 +√1− 4u

2

)′∣∣∣∣= 2(1− 4u)−1/2 = (1/4− u)−1/2

if u ∈ B.

10. (a) for x > 0, fY (y) = fX(−x)|(−x)′| + fX(x)|(x)′| = e−x, which is anexponential density.

(b) P (W = 0) = 1/2 and P (W = 1) = 1/2 by symmetry of fX . SoP (W ≤ w) = 0 for w < 0, = 1/2 for 0 ≤ w < 1, and = 1 for 1 ≤ w.

13. y = u(x) = x2 so x = ±√y if 0 < y < 4 and x =√y if 4 < y < 16.

Therefore,

fY (y) =

{(y/24)(1/(2

√y))2 if 0 < y < 4

(y/24)(1/(2√y)) if 4 < y < 16

=

{√y/24 if 0 < y < 4√y/48 if 4 < y < 16.

1

14. (b) X = V and Y = V/U , so

J = det

(0 1

−v/u2 1/u

)= v/u2,

so fU,V (u, v) = f(v, v/u)v/u2 = 4(v/u2)e−2v(1+1/u) for v > 0 and v/u > 0, orequivalently, u > 0 and v > 0.

16. (a) X1 = V and X2 = U/V , so

J = det

(0 1

1/v −u/v2)

= −1/v,

so fU,V (u, v) = f(v)f(u/v)(1/v) = (1/v2)(1/(u/v)2)(1/v) = 1/(u2v) for v ≥ 1and u/v ≥ 1, or equivalently, u ≥ v ≥ 1.

2

Math 5080-2, F 2015. Assignment 2. Chapter 6, Exercises 18, 25, 27, 28, 29,31.

18. (a) X = T and Y = S − T , so

J = det

(0 11 −1

)= −1,

and fS,T (s, t) = f(t, s− t) = e−(s−t), where 0 < t < s− t <∞, or s > 2t > 0.(b) fT (t) =

∫∞2te−(s−t) ds = ete−2t = e−t for t > 0.

(c) fS(s) =∫ s/20

e−(s−t) dt = e−s(es/2 − 1) = e−s/2 − e−s for s > 0.

25. If X is POI(λ), then MX(t) =∑∞n=0 e

tne−λλn/n! = e−λ exp(λet) =

eλ(et−1).(a) Therefore,

e25(et−1) =MX1

(t)[e5(et−1)]3,

hence MX1(t) = e10(et−1), so X1 is POI(10).

(b) W = X1 +X2 is POI(15), as can also be shown by means of MGFs.

27. Misprint: Note that Y1 should be Yi.(a) ln

∏ni=1 Yi =

∑ni=1 lnYi is N(µ1 + · · ·+ µn, σ

21 + · · ·+ σ2

n), so∏ni=1 Yi is

LOGN(µ1 + · · ·+ µn, σ21 + · · ·+ σ2

n)(b) ln

∏ni=1 Y

ai =

∑ni=1 lnY

ai = a

∑ni=1 lnYi is N(a(µ1 + · · · + µn), a

2(σ21 +

· · ·+ σ2n)), so

∏ni=1 Y

ai is LOGN(a(µ1 + · · ·+ µn), a

2(σ21 + · · ·+ σ2

n)).(c) ln(Y1/Y2) = ln(Y1)−ln(Y2) is N(µ1−µ2, σ

21+σ

22), so Y1/Y2 is LOGN(µ1−

µ2, σ21 + σ2

2).(d) If Y is LOGN(µ, σ2), then E[Y ] is the MGF of N(µ, σ2) at t = 1, hence

it is eµ+(1/2)σ2

. By part (a), the answer is eµ1+···+µn+(1/2)(σ21+···+σ

2n).

28. (a) Now F (x) = x2 for 0 ≤ x ≤ 1, so by Eqs. (6.5.5) and (6.5.6),

g1(y1) = 2[1− y21 ]2y1 = 4y1(1− y21), g2(y2) = 2[y22 ]2y2 = 4y32 ,

for 0 < y1 < 1 and 0 < y2 < 1, resp.(b) By Eq. (6.5.4),

g12(y1, y2) = 2 · 2y1 · 2y2 = 8y1y2, 0 < y1 < y2 < 1.

(c) Let R = Y2 − Y1 and S = Y1. Then Y1 = S and Y2 = R+ S, so

fR,S(r, s) = g12(s, r + s) = 8r(r + s), r > 0, s > 0, r + s < 1.

Integrating out s we get fR(r) =∫ 1−r0

8r(r + s) ds = 8r2(1− r) + 4r(1− r)2 =r(1− r)[8r + 4(1− r)] = 4r(1− r2), 0 < r < 1.

29. (a) f(y1, . . . , yn) = n!/(y1 · · · yn)2, 1 ≤ y1 < y2 < · · · < yn <∞.(b) g1(y1) = n(1/y1)

n−1(1/y21), 1 < y1 <∞.(c) gn(yn) = n(1− 1/yn)

n−1(1/y2n), 1 < yn <∞.

1

(d) See p. 220. g12(y1, y2) = 2/(y21y22), 1 < y1 < y2 < ∞. Let R = Yn − Y1

and S = Y1. Then Y1 = S and Yn = R+ S and

fR,S(r, s) = g1n(s, r + s) = 2/(s2(r + s)2), r > 0, s > 1,

so the marginal density of R is

fR(r) =

∫ ∞1

2

s2(r + s)2ds =

2

r3

(r(2 + r)

1 + r− 2 log(1 + r)

), r > 0.

(I used Mathematica to evaluate the integral.)(e) From (6.5.3), gr(yr) =

n!(r−1)!2 (1−1/yr)

r−1(1/y2r)(1/yr)r−1, 1 < yr <∞.

31. (a) g1(y1) = n[1− (1− e−y1)]n−1e−y1 = ne−ny1 , y1 > 0.(b) gn(yn) = n[1− e−yn ]n−1e−yn , yn > 0.(c) g1n(y1, yn) = n(n − 1)e−y1 [e−y1 − e−yn ]n−2e−yn , 0 < y1 < yn. Let

R = Yn − Y1 and S = Y1. Then Y1 = S and Yn = R+ S and

fR,S(r, s) = g1n(s, r + s) = n(n− 1)e−s[e−s − e−(r+s)]n−2e−(r+s)

= n(n− 1)e−nse−r(1− e−r)n−2

for r > 0 and s > 0. Integrate out s > 0 to get fR(r) = (n− 1)e−r(1− e−r)n−2,r > 0.

(d) The joint pdf of Y1, . . . , Yn is n!e−(y1+···+yn), 0 < y1 < · · · < yn. Wemust integrate over yn > . . . > yr+1 (in that order), where yr+1 > yr. We getn(n− 1) · · · (n− r + 1)e−(y1+···+yr)−(n−r)yr , 0 < y1 < · · · < yr.

2

Assignment 3, F 2015. Chapter 7, Exercises 1, 2, 3.

1. (a) FX1:n(y) = P (X1:n ≤ y) = 1 − P (X1:n > y) = 1 − P (X1 > y)n =

1− (1− F (y))n = 1− y−n for y ≥ 1.(b) This converges to 1 if y ≥ 1, 0 if y < 1. Therefore X1:n converges in

distribution to the constant 1.(c) Zn = Xn

1:n has cdf Fn(z) = FX1:n(z1/n) = 1− z−1 for z ≥ 1.

2. (a) Xn:n has cdf F (x)n = (1 + e−x)−n → 0 as n→∞, so Xn:n does notconverge in distribution.

(b) Xn:n − lnn has cdf F (y + lnn)n = (1 + e−y−lnn)−n = (1 + e−y/n)−n →e−e−y

for y real. This is a cdf, so the answer is yes.

3. (a) X1:n has cdf G(y) = 1 − (1 − F (y))n = 1 − y−2n → 1 for y > 1 andG(y) = 0 for y ≤ 1. Hence X1:n converges in distribution to the constant 1.

(b) Xn:n has cdf G(y) = F (y)n = (1 − y−2)n → 0 for y > 1, so Xn:n doesnot converge in distribution.

(c) n−1/2Xn:n has cdf Fn(y) = F (n1/2y)n = (1 − (n1/2y)−2)n = (1 −y−2/n)n → e−y−2

for y > 0. Since this defines a cdf, convergence in distri-bution is established.

1

Assignment 4. Chapter 7, Exercises 7, 15, 16, 17.

7. The WEI(1,2) distribution has pdf f(x) = 2xe−x2

for x > 0, meanµ = Γ( 3

2 ) = 12Γ( 1

2 ) = 12

√π and variance σ2 = Γ(2)− Γ( 3

2 )2 = 1− π/4.(a) By the central limit theorem, this holds with a = µ − 1.96σ/

√n and

b = µ+ 1.96σ/√n, so if n = 35 we have a = 0.7328 and b = 1.0397.

(b) For n odd, X(n+1)/2:n is approximatelyN(x1/2, c2/n), where c2 = 1/(4f(x1/2)2).

Now F (x1/2) = 12 , so since F (x) = 1− e−x2

, it follows that x1/2 =√

ln 2. Also,c2 = 1/(4 ln 2). Now a = x1/2−1.96c/

√n and b = x1/2 +1.96c/

√n, so if n = 35

we have a = 0.6336 and b = 1.0315.

15. (a) µ =∫ B0θB−θwθdw = Bθ/(θ + 1) and σ2 =

∫ B0θB−θwθ+1dw −

B2θ2/(θ + 1)2 = B2[θ/(θ + 2) − θ2/(θ + 1)2] = B2θ/[(θ + 1)2(θ + 2)]. If θ = 3and B = 80 we have µ = 60 and σ2 = 240. By the central limit theorem,the probability is question is approximately 1−Φ((6025−nµ)/

√nσ2), which if

n = 100 gives 1− Φ(.1614) = 0.4359.(b) Fn(y) = 1 − (1 − F (y))n → 1 if y > 0 and → 0 if y ≤ 0. This gives

convergence in distribution to 0, therefore convergence in probability.(c) Fn(y) = F (y)n → 0 if y < B and → 1 if y ≥ B.(d) (Wn:n/B)n has cdf Fn(y) = F (By1/n)n, where F (w) = (w/B)θ, so

Fn(y)→ yθ for 0 < y < 1.(e) It is asymptotically N(x1/2, c

2/n), where c2 = 1/(4f(x1/2)2). Setting

F (w) = 12 we find that x1/2 = B/21/θ and c2 = B2/(θ222/θ).

(f) It converges to x1/2 = B/21/θ.

(g) n1/θW1:n/B has cdf Fn(y) = 1−(1−F (Bn−1/θy))n = 1−(1−yθ/n)n →1− e−yθ for y > 0, which is WEI(1, θ).

16. (a) Xn → µ in probability by the weak law of large numbers. Since e−µ

is continuous in µ, Yn = e−Xn → e−µ in probability.(b) By the theorem, Yn is asymtotically N(e−µ, µe−2µ/n), since [(e−µ)′]2 =

[−e−µ]2 = e−2µ.(c) This is true for the same reason that part (a) holds.

17. x1/2 = µ by symmetry, and c2 = 12 (1 − 1

2 )/f(µ)2 = 14 [(2πσ2)−1/2]−2 =

πσ2/2, so X̃n is asymptotically N(µ, πσ2/(2n)).

1

Assignment 5, F 2015. Chapter 8, Exercises 13, 15.

13. (a) Z is N(0, 1/16) so P (Z < 12 ) = P (Z/(1/4) < 2) = Φ(2) = 0.9772.

(b) Z1 − Z2 is N(0, 2) so P (Z1 − Z2 < 2) = P ((Z1 − Z2)/√

2 <√

2) =Φ(√

2) = Φ(1.4142) ≈ 0.9213.(c) Z1 + Z2 is N(0, 2), so answer is the same as in (b).(d) This is P (χ2(16) < 32) ≈ 0.988 from page 607.(e) This is P (χ2(15) < 25) = 0.950 from page 607.

15. (a) N(0, 2σ2).(b) N(3µ, 5σ2).(c) (X1 − X2)/(σ

√2) is N(0, 1), so dividing by Sz, the sample standard

deviation of the Z sample, we get t(k − 1).(d) χ2(1).(e) For the same reason as in (c), t(k − 1).(f) χ2(2).(g) This is the difference between two independent GAM(2, 1/2) random

variables, which must be unknown.(h) t(1).(i) F (1, 1).(j) By (h), Z1/|Z2| is t(1), and this is the same except for an independent

factor of ±1 with probability 1/2 each. Since t(1) is symmetric, the answer mustbe t(1), which is equivalent to the Cauchy distribution.

(k) This is the ratio of two independent nonstandard normal random vari-ables, which must be unknown.

(l)√n(X − µ)/σ is N(0, 1), and the sum is χ2(k), so the result is t(k).

(m) The first term is χ2(n) and the second is χ2(k − 1), so the result isχ2(n+ k − 1).

(n) The first term is N(µ/σ2, 1/(nσ2)) and the second is N(0, 1/k), so theresult is N(µ/σ2, 1/(nσ2) + 1/k).

(o)√k[Z ] is N(0, 1), so answer is χ2(1).

(p) This is (S2x/σ

2)/S2z , which is F (n− 1, k − 1).

1

Assignment 6. Chapter 9, Exercises 1–6.

1. (a) µ′1 = θ/(1 + θ) = X, so θ̂ = X/(1−X).

(b) µ′1 = (θ + 1)/θ = X, so θ̂ = 1/(X − 1).

(c) X is GAM(1/θ, 2), so µ′1 = 2/θ = X, and θ̂ = 2/X.

2. (a) NB(3, p). µ′1 = 3/p = X, so p̂ = 3/X.(b) GAM(2, κ). µ′1 = 2κ = X, so κ̂ = X/2.

(c) WEI(θ, 1/2). µ′1 = θΓ(1 + 1/(1/2)) = 2θ = X, so θ̂ = X/2.(d) DE(θ, η). µ′1 = η = X and µ′2 = σ2 + µ2 = 2θ2 + η2 = M ′2, so η̂ = X

and θ̂ =

√(1/2)(M ′2 −X

2) = σ̂/

√2.

(e) EV(θ, η). µ′1 = η−γθ = X and µ′2 = σ2 +µ2 = π2θ2/6+(η−γθ)2 = M ′2,

θ̂ = (√

6/π)σ̂ and η̂ = X + γ(√

6/π)σ̂.(f) PAR(θ, κ). µ′1 = θ/(κ − 1) = X and µ′2 = σ2 + µ2 = θ2κ/[(κ − 2)(κ −

1)2]+θ2/(κ−1)2 = M ′2, so κ̂/(κ̂−2) = σ̂2/X2, and κ̂ = (2σ̂2/X

2)/(σ̂2/X

2−1)

and θ̂ = X(κ̂− 1).

3. (a) L = θn(x1 · · ·xn)θ−1, so lnL = n ln θ+(θ−1)(lnx1 + · · ·+lnxn), and

(∂/∂θ) lnL = n/θ+(lnx1+· · ·+lnxn) = 0 implies θ̂ = −n/(lnX1+· · ·+lnXn).This is justified by noticing that L→ 0 as θ → 0 and as θ →∞, since 0 < xi < 1for i = 1, . . . , n.

(b) L = (θ + 1)n(x1 · · ·xn)−θ−2, so lnL = n ln(θ + 1) − (θ + 2)(lnx1 +· · · + lnxn), and (∂/∂θ) lnL = n/(θ + 1) − (lnx1 + · · · + lnxn) = 0 implies

θ̂ = n/(lnX1 + · · · + lnXn) − 1. This is justified by noticing that L → 0 asθ → −1 and as θ →∞. Here we need to extend the parameter space to (−1,∞),for otherwise our MLE might not belong to the parameter space. (If we haddone that in Problem 1, the mean would not necessarily exist.)

(c) L = θ2nx1 · · ·xne−θ(x1+···+xn), so lnL = 2n ln θ + lnx1 + · · · + lnxn −θ(x1 + · · ·+xn), and (∂/∂θ) lnL = 2n/θ− (x1 + · · ·+xn) = 0 implies θ̂ = 2/X.This is justified by noticing that L→ 0 as θ → 0 and as θ →∞.

4. (a) BIN(1, p). L = px1+···+xn(1 − p)n−(x1+···+xn), so lnL = (x1 + · · · +xn) ln p+ (n− (x1 + · · ·+ xn)) ln(1− p), and (∂/∂p) lnL = (x1 + · · ·+ xn)/p−(n− (x1 + · · ·+ xn))/(1− p) = 0 implies p̂ = X.

(b) GEO(p). L = (1− p)x1+···+xn−npn, so lnL = (x1 + · · ·+ xn − n) ln(1−p) + n ln p, and (∂/∂p) lnL = −(x1 + · · · + xn − n)/(1 − p) + n/p = 0 impliesp̂ = 1/X.

(c) NB(3, p). L =(x1−1

2

)· · ·(xn−1

2

)p3n(1−p)x1+···+xn−3n, so lnL = ln

(x1−1

2

)+

· · · + ln(xn−1

2

)+ (3n) ln p + (x1 + · · · + xn − 3n) ln(1 − p), and (∂/∂p) lnL =

3n/p− (x1 + · · ·+ xn − 3n)/(1− p) = 0 implies p̂ = 3/X.

(d) N(0, θ). L = (2πθ)−n/2e−(x21+···+x

2n)/(2θ), so lnL = −(n/2)(ln(2π) +

ln θ)−(x21+· · ·+x2n)/(2θ), and (∂/∂θ) lnL = −n/(2θ)+(x21+· · ·+x2n)/(2θ2) = 0

implies θ̂ = (X21 + · · ·+X2

n)/n.

1

(e) GAM(θ, 2). L = θ−2nx1 · · ·xne−(x1+···+xn)/θ, so lnL = −2n ln θ+lnx1+· · ·+lnxn− (x1 + · · ·+xn)/θ, and (∂/∂θ) lnL = −2n/θ+(x1 + · · ·+xn)/θ2 = 0

implies θ̂ = X/2.(f) DE(θ, 0). L = (2θ)−ne−(|x1|+···+|xn|)/θ, so lnL = −n(ln 2 + ln θ)− (|x1|+

· · · + |xn|)/θ, and (∂/∂θ) lnL = −n/θ + (|x1| + · · · + |xn|)/θ2 = 0 implies

θ̂ = (|X1|+ · · ·+ |Xn|)/n.

(g) WEI(θ, 1/2). L = 2−nθ−n/2(x1 · · ·xn)−1/2e−(x1/21 +···+x1/2

n )/√θ, so lnL =

−n ln 2 − (n/2) ln θ − (lnx1 + · · · + lnxn)/2 − (x1/21 + · · · + x

1/2n )/

√θ, and

(∂/∂θ) lnL = −n/(2θ) + (x1/21 + · · · + x

1/2n )/(2θ3/2) = 0 implies θ̂ = ((X

1/21 +

· · ·+X1/2n )/n)2.

(h) PAR(1, κ). L = κn[(1+x1) · · · (1+x)]−κ−1, so lnL = n lnκ−(κ+1)(ln(1+x1)+ · · ·+ln(1+xn)), and (∂/∂κ) lnL = n/κ−(ln(1+x1)+ · · ·+ln(1+xn)) = 0implies κ̂ = n/(ln(1 +X1) + · · ·+ ln(1 +Xn)).

5. f(x; θ) = 2θ2/x3, x ≥ θ, θ > 0. L(θ) = 2nθ2n/(x1 · · ·xn)3 if x1 ≥θ, . . . , xn ≥ θ, that is, if x1:n ≥ θ. L(θ) is maximized at θ̂ = X1:n.

6. (a) L(θ1, θ2) = 1/(θ2 − θ1)n if x1, . . . , xn all belong to [θ1, θ2], or equiva-lently if θ1 ≤ x1:n and xn:n ≤ θ2. Since L is maximized when θ1 is maximizedand θ2 is minimized, we must have θ̂1 = X1:n and θ̂2 = Xn:n.

(b) L(θ, η) = θnηnθ(x1 · · ·xn)−θ−1 if x1, . . . , xn are all in [η,∞), that is, ifx1:n ≥ η. For each θ, L(θ, η) is maximized as a function of η at η̂ = X1:n.

Next, to find θ̂ we maximize lnL(θ, η̂) = n ln θ + nθ ln η̂ − (θ + 1)∑ni=1 lnxi.

Differentiate with respect to θ to get

∂

∂θL =

n

θ+ n ln η̂ −

n∑i=1

lnxi = 0,

hence

θ̂ =1

1n

∑ni=1 ln(Xi/X1:n)

.

2

Assignment 7, 5080-2, F 2015. Chapter 9, Exercises 13, 17, 21, 23, 26.

13.

L = (2πσ21)−n/2e−(2σ

21)

−1∑n

i=1(xi−µ)2(2πσ2

2)−m/2e

−(2σ22)

−1∑m

j=1(yj−µ)2

Take the partials of lnL with respect to µ, σ21 , and σ

22 , and set them equal to

0. We get

1

σ21

n∑i=1

(xi − µ) +1

σ22

m∑j=1

(yj − µ) = 0,

− n

2σ21

+1

2σ41

n∑i=1

(xi − µ)2 = 0,

and

− m

2σ22

+1

2σ42

m∑j=1

(yj − µ)2 = 0.

However, it is not clear how to solve this nonlinear system simultaneously forµ, σ2

1 , and σ22 . Therefore, we leave this one incomplete.

17. (a) E[X] = [(θ − 1) + (θ + 1)]/2 = θ.(b) Fn:n(t) = (t− (θ − 1))n/2n, so fn:n(t) = n(t− θ + 1)n−1/2n and

E[Xn:n] =

∫ θ+1

θ−1t[n(t− θ + 1)n−1/2n] dt

=

∫ θ+1

θ−1[n(t− θ + 1)n/2n] dt+ (θ − 1)

∫ θ+1

θ−1[n(t− θ + 1)n−1/2n] dt

= θ +2n

n+ 1− 1.

Similarly,

E[X1:n] = θ − 2n

n+ 1+ 1,

so E[(X1:n +Xn:n)/2] = θ.

21. (a) f(x; p) = px(1− p)1−x, hence (∂/∂p) ln f(x; p) = x/p− (1− x)/(1−p) = (x − p)/[p(1 − p)], hence I1(p) = Var(X)/[p(1 − p)]2 = 1/[p(1 − p)], andCLRB = p(1− p)/n.

(b) CRLB for p(1− p) is (1− 2p)2p(1− p)/n.(c) UMVUE of p is p̂ = X since Var(X) = p(1− p)/n = CRLB.

23. (a)

L(θ) = (2πθ)−n/2e−(1/θ)∑n

i=1x2i ,

∂

∂θlnL(θ) = − n

2θ+

1

2θ2

n∑i=1

x2i = 0,

1

so θ̂ = (1/n)∑ni=1 x

2i . It is clearly unbiased.

(b)

I1(θ) = E

[(X2

1 − θ

2θ2

)2]=

Var(X21 )

4θ4=E[X4

1 ]− (E[X21 ])

2

4θ4=

3θ2 − θ2

4θ4=

1

2θ2,

so CRLB = 2θ2/n. But Var(θ̂) = (1/n)Var(X21 ) = 2θ2/n = CRLB, so this is a

UMVUE of θ.

26. (a) We’ve seen that MLE = θ̂ = Xn:n.(b) Since µ′1 = θ/2, MME = θ̃ = 2X.

(c) E[θ̂] = [n/(n+ 1)]θ, so no.(d) E[θ̃] = θ, so yes.(e)

MSE(θ̂) = Var(θ̂) + (nθ/(n+ 1)− θ)2

=

[n

n+ 2−(

n

n+ 1

)2

+1

(n+ 1)2

]θ2 =

2θ2

(n+ 1)(n+ 2)

and

MSE(θ̃) = Var(2X) = 4Var(X) =4

nVar(X1) =

4

n

θ2

12=θ2

3n.

The MSE of θ̂ is smaller unless n = 1.

2

Assignment 8, 5080-2, F 2015. Chapter 9, Exercises 28, 34, 40, 44.

28. (a) Var(θ̂1) = Var(X) = Var(X1)/n = θ2/n and Var(θ̂2) = (n/(n +1))2Var(X) = nθ2/(n+ 1)2.

(b) MSE(θ̂1) = Var(θ̂1)+b(θ̂1)2 = Var(θ̂1) = θ2/n and MSE(θ̂2) = Var(θ̂2)+

b(θ̂2)2 = nθ2/(n+ 1)2 + (n/(n+ 1)− 1)2θ2 = (n+ 1)θ2/(n+ 1)2 = θ2/(n+ 1).

(c) Var(θ̂1) = θ2/2 > Var(θ̂2) = 2θ2/9.

(d) MSE(θ̂1) = θ2/2 > MSE(θ̂2) = θ2/3.(e) Assume squared error loss. Rθ̂1(θ) = RX(θ) = E[(X − θ)2] = Var(X) =

θ2/n, so the Bayes risk with an EXP(2) prior is∫∞0Rθ̂1(θ)(1/2)e−θ/2 dθ =

E[θ2]/n = 2(22)/n = 8/n.

34. (a) L(p) = pn(1 − p)∑Xi so lnL(p) = n ln p + (

∑Xi) ln(1 − p) and

(∂/∂p) lnL(p) = n/p − (∑Xi)/(1 − p) = 0. Conclude that (1 − p̂)/p̂ = X or

p̂ = (1 +X)−1.

(b) θ̂ = (1− p̂)/p̂ = X by invariance.(c) (∂/∂p) ln f = [(1−p)/p−X1]/(1−p), hence I1(p) = Var(X1)/(1−p)2 =

[(1 − p)/p2]/(1 − p)2 = 1/[p2(1 − p)]. If τ(p) = (1 − p)/p, then τ ′(p) = −1/p2

and the CRLB of θ is τ ′(p)2/(nI1(p)) = (1− p)/(np2) = θ/(np) = θ(θ + 1)/n.(d) The MLE is X and Var(X) = Var(X1)/n = (1−p)/(np2), which matches

the CRLB. Since X is unbiased for θ, it is a UMVUE.(e) MSE(X) = E[(X − θ)2] = Var(X) = (1− p)/(np2)→ 0. Yes.(f) X is asymptotically normal with asymptotic mean θ and asymptotic

variance θ(θ + 1)/n.(g) For X we get E[(X − θ)2]/[θ(θ + 1)] = Var(X)/[θ(θ + 1)] = 1/n. For

θ̃ = nX/(n+1) we get E[(nX/(n+1)−θ)2]/[θ(θ+1)] which after much algebrareduces to [n+ θ/(θ + 1)]/(n+ 1)2, which is less than 1/(n+ 1).

40. (a) E[(X − c)2] = var(X) + (E[X]− c)2 makes the result obvious.(b) E[|X − c|] =

∫|x− c|f(x) dx =

∫ c−∞(c− x)f(x) dx+

∫∞c

(x− c)f(x) dx.Now differentiate with respect to c and set the derivative to 0 and solve for c.It is basically a calculus problem.

44. (a) Posterior is proportional to e−βθθne−θ∑xi = θn+1−1e−θ(β+

∑xi),

which is the stated distribution.(b) Mean of posterior is (n+ 1)/(β +

∑xi).

(c) The −1 moment of GAM(θ, κ) is E[X−1] =∫∞0xκ−2e−x/θ dx/(θκΓ(κ)).

This is θκ−1Γ(κ− 1)/(θκΓ(κ)) = 1/[(κ− 1)θ]. Apply this rule to the posteriorand we get (β +

∑xi)/n.

(d) Now GAM(θ, κ)/(θ/2) = GAM(2, 2κ/2) = χ2(2κ), so GAM(θ, κ) =(θ/2)χ2(2κ). Here the Bayes estimator is the median of the posterior, which is(1/2)(β +

∑xi)−1 times the median of the χ2(2n+ 2) distribution.

(e) Here the Bayes estimator is the median of the reciprocal of the posteriorrandom variable, which is the reciprocal of the median. So the answer is thereciprocal of the answer in (d).

1

Assignment 9. 5080-2, F 2015. Chapter 10, Exercises 5, 10, 12, 13, 14.

5. (a) S is GAM(θ, 2n), so the conditional density is

n∏i=1

θ−2xie−xi/θ

/θ−2nΓ(2n)−1s2n−1e−s/θ,

where s = x1 + · · · + xn, which does not depend on θ.(b)

∏ni=1 θ

−2xie−xi/θ = θ−2nx1x2 · · ·xne−s/θ, so S = X1 + · · · + Xn is sufficient by

the factorization theorem.

10. (a)

f(x1, . . . , xn;µ) = (2πσ2)−n/2e−(2σ2)−1

∑n

i=1(xi−µ)2

= (2πσ2)−n/2e−(2σ2)−1

∑n

i=1x2i e(σ

2)−1µ∑n

i=1xie−(2σ

2)−1nµ2

.

Since σ2 is a constant,∑ni=1Xi is sufficient.

(b) f(x1, . . . , xn;µ) = (2πσ2)−n/2e−(2σ2)−1

∑n

i=1(xi−µ)2 . By the factorization theorem,∑n

i=1(Xi − µ)2 is sufficient, and it is a statistic since µ is known.

12.

f(x1, . . . , xn; θ, η) = (1/θ)ne−∑

(xi−η)/θn∏1

1{xi>η}

= (1/θ)ne−(∑

xi−nη)/θ1{x1:n>η},

so X and X1:n are jointly sufficient.

13.

f(x1, . . . , xn; θ1, θ2) =

(Γ(θ1 + θ2)

Γ(θ1)Γ(θ2)

)n(x1 · · ·xn)θ1−1[(1 − x1) · · · (1 − xn)]θ2−1,

so∏n

1 Xi and∏n

1 (1 −Xi) are jointly sufficient.

14.

f(x1, . . . , xn; θ) = (1/θ)nn∏1

1{θ<xi<2θ}

= (1/θ)n1{θ<x1:n and xn:n<2θ},

so X1:n and Xn:n are jointly sufficient, but there is no single sufficient statistic for θ.

1

Assignment 10. Math 5080-2, F 2015. Chapter 10, Exercises 17–22.

17. (a) X1:n has density ne−n(y−η), y > η, so

E[g(X1:n)] =

∫ ∞η

g(y)ne−n(y−η)dy = enηn

∫ ∞η

g(y)e−nydy.

If this is 0, then the integral is 0, and so is its derivative, hence g(η)e−nη = 0 a.e., henceg = 0 a.e.

(b)

E[X1:n] =

∫ ∞η

yne−n(y−η)dy = n

∫ ∞0

(z + η)e−nzdz =

∫ ∞0

(y/n+ η)e−ydy = 1/n+ η,

which suffices.(c) 1− p = 1−F (xp) = e−(xp−η), so xp = η− ln(1− p). Hence X1:n− 1/n− ln(1− p)

is an unbiased estimator of xp and a function of X1:n.

18. (a) f(x; θ) = (2πθ)−1/2e−x2/(2θ), so X2 is complete and sufficient by the REC the-

orem. Alternatively, we know X2/θ is χ2(1) or GAM(2, 1/2), hence X2 is GAM(2θ, 1/2).Let u be a function on the positive reals such that E[u(X2)] = 0 for all θ > 0, or∫∞0u(y)y−1/2e−y/(2θ)dy = 0 for all θ > 0. This implies that u(y)y−1/2 has zero Laplace

transform, hence is the zero function a.e.(b) X is N(0, θ) and E[X] = 0 for all θ > 0, hence completeness fails, because we

have a nontrivial unbiased estimator of 0.

19. The N(µ, µ2) distribution has density f(x;µ) = (2πµ2)−1/2 exp{−(2µ2)−1(x −µ)2} = (2πµ2)−1/2 exp{−(2µ2)−1x2 + µ−1x − 1/2}. The sum in the exponential has twoterms but µ is a scalar, not two-dimensional.

20. (a) f(x; p) = px(1 − p)1−x = (1 − p) exp{x ln(p/(1 − p))}, so∑n

1 Xi is completeand sufficient.

(b) f(x;µ) = µxe−µ/x! = (e−µ/x!)ex lnµ, so∑n

1 Xi is complete and sufficient.(c) f(x; p) =

(x−1r−1)prqx−r =

(x−1r−1)(p/q)rex ln q, so

∑n1 Xi is complete and sufficient.

(d) f(x : µ, σ2) = (2πσ2)−1/2 exp{−(x − µ)2/(2σ2)} = (2πσ2)−1/2 exp{−x2/(2σ2) +µx/σ2 − µ2/(2σ2)}, so (

∑n1 Xi,

∑n1 X

2i ) is complete and sufficient.

(e) f(x; θ) = (1/θ)e−x/θ, so∑n

1 Xi is complete and sufficient.(f) f(x; θ, κ) = {θκΓ(κ)}−1xκ−1e−x/θ = {θκΓ(κ)}−1x−1 exp{κ lnx − x/θ}, so

(∑n

1 Xi,∑n

1 lnXi) is complete and sufficient.(g) f(x; θ1, θ2) = C(θ1, θ2)xθ1−1(1 − x)θ2−1 = C(θ1, θ2)x−1(1 − x)−1 exp{θ1 lnx +

θ2 ln(1− x), so (∑n

1 lnXi,∑n

1 ln(1−Xi)) is complete and sufficient.

(h) f(x; θ) = (β/θβ)xβ−1e−xβ/θβ , so

∑n1 X

βi is complete and sufficient.

21. (a) By 20(a), S =∑n

1 Xi is complete and sufficient for p, so any function of thisstatistic that is unbiased for p(1− p) or p2 will work. Now E[S] = np, Var(S) = np(1− p),and E[S2] = np(1− p) + (np)2.

(a) (S2−S)/[n(n−1)] works because its expectation is (−np2 +n2p2)/[n(n−1)] = p2.(b) S/n minus the answer to (a) works, and

S

n− S2 − Sn(n− 1)

=nS − S2

n(n− 1).

22. By 33(g) of Chapter 9, [(n−1)/n]∑

Xi is unbiased for e−µ. It is also a function ofa complete sufficient statistic. Sufficiency of

∑ni=1Xi follows by the factorization theorem.

Completeness follows by the REC property, or we can verify it directly. Notice that∑ni=1Xi is Poisson(nµ). So we must show that if u is a function on the nonnegative

integers such that E[u(∑Xi)] = 0 for every µ, then u is the zero function. We’re given

that∑∞j=0 u(j)e−nµ(nµ)j/j! = 0 for all µ > 0. But if a power series is 0 throughout its

region of convergence, then all its coefficients are 0, hence u = 0 identically.

Assignment 11. Math 5080-2, F 2015. Chapter 11, Exercises 1, 5, 7, 13.

1. (a) 19.3± 1.645 · 3/√

16 or (18.1, 20.5).(b) lower: 19.3− 1.282 · 3/

√16 = 18.34. upper: 19.3 + 1.282 · 3/

√16 = 20.26.

(c) For x±z1−α/2σ/√n we have λ = 2z1−α/2σ/

√n, hence n = 4z21−α/2σ

2/λ2. For the

second question, this formula gives 4 · 1.6452 · 9/22 = 24.35 so take n = 25.(d) 19.3± 1.753

√10.24/

√16 or (17.9, 20.7).

(e) Use (11.3.6): (15(10.24)/32.8, 15(10.24)/4.6) = (4.68, 33.39).

5. (a) X1 has density e−(x−η), x > η; so X1:n has density ne−n(x−η), x > η; hence Qhas density ne−nx, x > 0. Thus, its distribution is EXP(1/n).

(b) P (x(1−γ)/2 < X1:n − η < x(1+γ)/2) = γ, so the confidence interval is (X1:n −x(1+γ)/2, X1:n − x(1−γ)/2). Now EXP(1/n) has cdf Fn(x) = 1 − e−nx, and Fn(xα) = αimplies e−nxα = 1−α, hence xα = −(1/n) ln(1−α). Thus, the confidence interval becomes(X1:n + (1/n) ln((1− γ)/2), X1:n + (1/n) ln((1 + γ)/2)).

(c) We assume the data Xi are from EXP(850, η). We can apply the results of (b) tothe observations Xi/850, which are EXP(1, η/850). We get a confidence interval for η/850,and multiplying by 850, we have a confidence interval for η:

850

(162

850+

1

19ln

(1− 0.9

2

),

162

850+

1

19ln

(1 + 0.9

2

)= (27.98, 159.7).

The answer in the book seems to be wrong.

7. (a) If X is WEI(θ, 2), then X2 is EXP(θ2) by the transformation of densitiestheorem. Therefore

∑X2i is GAM(θ2, n), so Q = (2/θ2)

∑X2i is χ2(2n).

(b) Get a confidence interval for θ2, then take square roots:(√2∑X2i

χ2(1+γ)/2(2n)

,

√2∑X2i

χ2(1−γ)/2(2n)

).

(c) A lower confidence limit for θ is√2∑X2i

χ2γ(2n)

, (1)

so since e−(t/θ)2 is an increasing function of θ, it suffices to substitute (1) into e−(t/θ)2 forθ.

(d) We solve P (X > tp) = 1 − p for tp. From e−(tp/θ)2

= 1 − p we get (tp/θ)2 =

− ln(1 − p) = | ln(1 − p)|, so tp = θ√| ln(1− p)|. It suffices to substitute (1) with γ

replaced by 1− γ in for θ in the formula for tp.

13. (a) f(x; θ) = θ−κΓ(κ)−1e(κ−1) ln x−x/θ, so∑Xi is sufficient for θ with κ known.

Now∑Xi is GAM(θ, nκ), hence (2/θ)

∑Xi is GAM(2, nκ) or χ2(2nκ) if 2nκ is an integer.

Therefore,

P

(χ2α/2(2nκ) < (2/θ)

∑Xi < χ2

1−α/2(2nκ)

)= 1− α,

or

P

(2∑

Xi/χ21−α/2(2nκ) < θ < 2

∑Xi/χ

2α/2(2nκ)

)= 1− α.

(b) If X1 is GAM(1, κ), then 2X1 is GAM(2, κ) or χ2(2κ) if 2κ is an integer. So

P (χ20.05(2κ) < 2X1 < χ2

0.95(2κ)) = 0.90.

Now if x1 = 10, we must solve the equations χ20.05(2κ) = 20 and χ2

0.95(2κ) = 20 byinterpolation. Get 2κ from the table on page 604 and divide by 2 to get κ. The book gives(5.62, 15.94).

Assignment 12. Math 5080-2, F 2015. Chapter 11, Exercises 8, 10, 11, 15,19.

8. (a) P (Xn:n < θ < 2Xn:n) = P (θ < 2Xn:n) = P (Xn:n > θ/2) = 1 −P (Xn:n ≤ θ/2) = 1 − P (X1 ≤ θ/2, . . . Xn ≤ θ/2) = 1 − P (X1 ≤ θ/2)n =1 − (1/2)n.

(b) 1 − α = P (Xn:n < θ < cXn:n) = P (θ < cXn:n) = P (Xn:n > θ/c) =1 − P (Xn:n ≤ θ/c) = 1 − P (X1 ≤ θ/c, . . .Xn ≤ θ/c) = 1 − P (X1 ≤ θ/c)n =1 − (1/c)n, so α = (1/c)n or c = α−1/n.

10. (a) Use p̂− z0.95√p̂(1 − p̂)/50 = 0.8 − 1.645

√0.8(1 − 0.8)/50 = 0.707.

(b) e−5/θ > 0.707 is equivalent to θ > −5/ ln(0.707) = 14.4. Answer iscoincidentally the same as in Exercise 3.

11. Use p̂±z0.95√p̂(1 − p̂)/40 = 1/8±1.645

√(1/8)(7/8)/40 = 0.125±0.086,

which is [0.039, 0.211].

15. (a) To use this method, we need to know the distribution function ofS1 = X1:n. By (6.5.5), S1 has CDF G(y) = 1−(1−F (y))n, where F is the CDF

of WEI(θ, β). Now F (x) = 1−e−(x/θ)β , x > 0. Hence G(y) = 1−e−n(x/θ)β . Nowwe need to find h1 and h2 such that G(h1(θ)) = α/2 and G(h2(θ)) = 1 − α/2.Solving, we get

h1(θ) = θ

[− ln(1 − α/2)

n

]1/β, h2(θ) = θ

[− ln(α/2)

n

]1/β.

If S1 lies between these two limits with probability 1 − α, then θ belongs to(S1

/[− ln(α/2)

n

]1/β, S1

/[− ln(1 − α/2)

n

]1/β)with probability 1 − α.

(b) Y = Xβ has density fY (y) = fX(y1/β)(1/β)y1/β−1 = (1/θβ)e−y/θβ

,y > 0, which is EXP(θβ). Therefore, S2 is GAM(θβ , n). Since we don’t knowan explicit form for the CDF of the gamma distribution, it is better to use thepivotal quantity method here. 2S2/θ

β is GAM(2, n) or χ2(2n). Thus,

P (χ2α/2(2n) < 2S2/θ

β < χ21−α/2(2n)) = 1 − α,

soP ([2S2/χ

21−α/2(2n)]1/β < θ < [2S2/χ

2α/2(2n)]1/β) = 1 − α.

19. We know that∑n1

i=1(Xi−µ1)2/σ21 and

∑n2

j=1(Yj −µ2)2/σ22 are indepen-

dent χ2(n1) and χ2(n2), and their numerators are sufficient statistics. Theirratio (first over second) is a pivotal quantity for σ2

2/σ21 . And if we adjust it

1

by dividing by n1/n2, the pivotal quantity then has an F (n1, n2) distribution.Thus,

P

(Fα/2(n1, n2) <

(1/n1)∑n1

i=1(Xi − µ1)2

(1/n2)∑n2

j=1(Yj − µ2)2σ22

σ21

< F1−α/2(n1, n2)

)= 1 − α,

implying that our confidence interval is(Fα/2(n1, n2)

/((1/n1)

∑n1

i=1(Xi − µ1)2

(1/n2)∑n2

j=1(Yj − µ2)2

), F1−α/2(n1, n2)

/((1/n1)

∑n1

i=1(Xi − µ1)2

(1/n2)∑n2

j=1(Yj − µ2)2

)).

2