Solutions to Mathematical Statistics (II)
Midterm Exam.
Oct. 22, 2018
1.(50) Let X1,···, X25 be r.s. from N(µ, 1). Consider a test of size α= 0.05
for testing H0:µ= 0 vs H1:µ > 0. Compute power of the test when
µ= 0.329 (Hint : z.05 = 1.645).
(Sol) Under H0,Z=¯
X−µ
σ/√n=¯
X
1/√25 ∼N(0,1), and we reject if Z > z.05 =
1.645. Now, the power at µ= 0.329 is P(¯
X
0.2>1.645) = P(¯
X−0.329
0.2>1.645 −
0.329
0.2) = P(Z > 1.645 −1.645) = 0.5.
2.(180) Let X∼Exp(λ), i.e., the pdf of Xis f(x) = 1
λe−x/λI(x > 0).
(1)(40) Find the pdf of T= 1 −e−x/λ.
(Sol) The inverse function is x=−λlog(1 −t), and Jacobian is λ/(1 −t).
Hence the pdf of Tis g(t) = f(−λlog(1 −t))|J|=I(0 <t<1).
(2)(50) Let X1,···, Xnbe r.s. from Exp(λ), and Y1<·· · < Ynbe the order
statistics of X0
is. Let W1=nY1, W2= (n−1)(Y2−Y1), W3= (n−2)(Y3−
Y2),·· · , Wn=Yn−Yn−1. Show that W1, W2,· · · , Wnare iid Exp(λ).
(Sol) The inverse functions are y1=1
nw1, y2=1
nw1+1
n−1w2,·· · , yn=
1
nw1+1
n−1w2+·· · +wn, and |J|= 1/n!. Hence, the jpdf of W1,···, Wnis
g(w1,·· · , wn) = f(1
nw1,1
nw1+1
n−1w2,·· ·)/n! = n!λ−ne−Pwi/λ /n! = Q1
λe−wi/λ.
(3)(50) Show that 2 Pn
i=1 Wi/λ ∼χ2(r), and find r.
(Sol) Note that Wi= Γ(1, λ), so that V≡PWi∼Γ(n, λ). Let Z= 2V/λ,
i.e., v=λz/2, and |J|=λ/2. Hence, the pdf of Zis g(z) = f(λz/2)|J|=
z
2n
2
−1e−z/2
Γ( 2n
2)22n/2∼χ2(2n).
(4)(40) Using the result of (3), find 100(1 −α)% C.I. for λ.
(Sol) Since 2v/λ ∼χ2(2n), we have 1 −α=P(χ2
1−α
2≤2v
λ≤χ2
α
2). Hence,
100(1 −α)% C.I. for λis (2v/χ2
α
2,2v/χ2
1−α
2).
3.(90) Let X1,··· , X10 be r.s. from P oisson(θ).
(1)(50) Consider a test H0:θ= 0.2 vs H1:θ > 0.2. Find a randomized test
of size α= 0.05. (When Y∼P oisson(2), P (Y≤4) = 0.947, P (Y≤5) =
0.983).
(Sol) Note that Y=P10
i=1 Xi∼P oisson(2), and P(Y≥5) = 0.053, P (Y≥
−1−
