Stat 252.01 Winter 2006
Assignment #3 Solutions
(6.64) If Y1 , Y2 , . . . , Yn are all independent and identically distributed beta(2, 2) random variables, then each has density function
fY (y) =
Γ(4)
y(1 − y) = 6y(1 − y), 0 < y < 1.
Γ(2)Γ(2)
(a) If Y(n) = max{Y1 , . . . , Yn }, then
P (Y(n) ≤ t) = P (Y1 ≤ t, . . . , Yn ≤ t) = P (Y1 ≤ t) · · · P (Yn ≤ t) = [P (Y1 ≤ t)]n
since the Yi are independent (the second equality) and identically distributed (the third
equality). Now, for any 0 < t < 1,
Z t
Z t
Z t
Z t
P (Y1 ≤ t) =
fY (y) dy =
6y(1 − y) dt =
6y dy −
6y 2 dy = 3t2 − 2t3 = t2 (3 − 2t)
0
0
0
0
so that the distribution function of Y(n) is
F (t) = [P (Y(n) ≤ t)]n = t2n (3 − 2t)n , 0 < t < 1.
(Of course, F (t) = 0 for t ≤ 0, and F (t) = 1 for t ≥ 1.
(b) The density function of Y(n) is therefore
d
d
F (t) = t2n (3−2t)n = 2nt2n−1 (3−2t)n −2nt2n (3−2t)n−1 = 6nt2n−1 (3−2t)n−1 (1−t)
dt
dt
for 0 < t < 1 and 0 otherwise.
f (t) =
(c) For n = 2, the expected value E(Y(2) ) is
Z 1
Z 1
Z 1
36 60 24
22
4
E(Y(2) ) =
t f (t) dt =
12t (3−2t)(1−t) dt =
(36t4 −60t5 +24t6 ) dt = − + = .
5 6 7
35
0
0
0
(6.65) If Y1 , Y2 , . . . , Yn are all independent and identically distributed exponential(β) random
variables, then each has density function
fY (y) =
1 −y/β
e
, 0 < y < ∞.
β
(a) If Y(1) = min{Y1 , . . . , Yn }, then
P (Y(1) > t) = P (Y1 > t, . . . , Yn > t) = P (Y1 > t) · · · P (Yn > t) = [P (Y1 > t)]n
since the Yi are independent (the second equality) and identically distributed (the third
equality). Now, for any 0 < t < ∞,
Z ∞
Z
1 ∞ −y/β
P (Y1 > t) =
fY (y) dy =
e
dy = e−t/β
β
t
t
so that
P (Y(1) > t) = e−tn/β .
Thus, the distribution function of Y(n) is
F (t) = P (Y(1) ≤ t) = 1 − P (Y(1) > t) = 1 − e−tn/β = 1 − e−t/(β/n)
which is the distribution function of an exponential random variable with mean β/n.
(b) If n = 5, β = 2, then the distribution function of Y(1) is
F (t) = 1 − e−5t/2 , 0 < t < ∞
so that the corresponding density function is
5
f (t) = e−5t/2 , 0 < t < ∞.
2
Hence,
P (Y(1)
5
≤ 3.6) =
2
Z
3.6
−5t/2
e
3.6
−9
=1−e .
−5t/2 dt = −e
0
0
2. (a) Since Y1 ∼ U(0, θ), we have
(
1/θ,
fY (y) =
0,
0 ≤ y ≤ θ,
otherwise.
(b) If f (t) denotes the density of θ̂ = min(Y1 , . . . , Yn ), then f (t) = F 0 (t), where
F (t) = P (θ̂ ≤ t) = 1 − P (θ̂ > t) = 1 − P (Y1 > t, . . . , Yn > t) = 1 − [P (Y1 > t)]n .
Note that in the last step we have used the fact that Yi are iid. Next, for 0 ≤ t ≤ θ, we compute
Z ∞
Z θ
1
θ−t
P (Y1 > t) =
fY (y) dy =
dy =
.
θ
t
t θ
Thus, we conclude
d
f (t) =
dt
θ−t
1−
θ
n = nθ−n (θ − t)n−1 , 0 ≤ t ≤ θ.
(c) By definition,
Z
∞
E(θ̂) =
Z
tf (t) dt =
−∞
θ
nθ−n t(θ − t)n−1 dt.
0
This last integral is solved with a simple substitution. Let u = θ − t so that du = −dt. Thus,
Z θ
Z 0
Z θ
nθ−n t(θ − t)n−1 dt = −nθ−n
(θ − u)un−1 du = nθ−n
θun−1 − un du
θ
0
0
θn+1
θn
−n
= nθ
θ·
−
n
n+1
θ
=
.
n+1
(d) From (c), we clearly see that θ̂ is NOT an unbiased estimator of θ. However,
θ̃ = (n + 1) min(Y1 , . . . , Yn )
IS an unbiased estimator of θ. (You should check that 2Y is also an unbiased estimator of θ.
Why?)