Download 2007-12-07 1 of 3 1. 10 2. 1000-1130 3. 4. 1. Let Qn denote the

2007-12-07
九十六學年度國立中山大學應用數學系統計組推薦甄試試題
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注意事項：
1. 本試卷共六大題，每大題10分。
2. 考試時間： 10：00-11：30。
3. 請將姓名及報名編號寫在每頁頁尾指定處。
4. 請詳列計算過程書寫於題目下方空白處，並將答案書寫於右下方指定處。
1. Let Qn denote the probability that no run of 3 consecutive heads in n tosses of a fair
coin. Show that
1
1
1
Qn−1 + Qn−2 + Qn−3
2
4
8
= Q1 = Q2 = 1
Qn =
Q0
Find Q8 .
解答: Condition on when the first tail occurs. Let Ai denote that the first tail appears
in the ith toss. Then
Qn =
3
X
P {no run of 3 consecutive heads appears|Ai }P {Ai }
i=1
=
Q8 =
1
1
1
Qn−1 + Qn−2 + Qn−3
2
4
8
149
256
2. One thousand independent rolls of a fair die will be made. Compute an approximation
to the probability that number 6 will appear between 150 and 200 times inclusively.
If number 6 appears exactly 200 times, find the probability that number 5 will appear
less than 150 times.
解答:
(a) Let Xi = 1 if the outcome is 6 and 0 otherwise, i = 1, 2, . .P
. , 1000. Then Xi ’s are
iid Bernoulli distribution with probability 1/6. Let X = 1000
i=1 be B(1000, 1/6).
Note that E[X] = (1000)(1/6) = 500/3 and Var(X) = (1000)(1/6)(5/6)
=
p
1250/9. An application of Central Limit Theorem gives that (X−E[X])/ Var(X) ≈
N (0, 1). Thus


 149.5 − 1000
1000 
200.5 − 6
6
q
P {149.5 < X < 200.5} = P
<Z< q

1000 16 56
1000 61 56 
!
!
200.5 − 166.7
149.5 − 166.7
p
p
= Φ
−Φ
5000/36
5000/36
≈ Φ(2.87) + Φ(1.46) − 1 = .9258.
姓名:
編號:
2007-12-07
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九十六學年度國立中山大學應用數學系統計組推薦甄試試題
(b) Let Y denote the frequency of number 5 given that number 6 appears exactly
200 times. Then Y |X = 200 ∼ B(800, 1/5). Note that E[Y |X = 200] =
(800)(1/5) = 160 and Var(Y |X = 200) = (800)(1/5)(4/5) = 128. An application
p of Central Limit Theorem gives that ((Y |X = 200) − E[Y |X =
200])/ Var(Y |X = 200) ≈ N (0, 1). Thus



149.5 − 800(1/5) 
q
P {X < 149.5} = P Z <


800 1 4
55
= P {Z < −.93}
= 1 − Φ(.93) = .1762.
3. Consider the jointly distributed random variables X, Y with finite second moments
and positive variance. Find the values α̂ and β̂ for which E[Y − (αX + β)]2 is
minimized.
∂
解答: Set g(α, β) = E[Y − (αX + β)]2 , and solve for ∂α
g(α, β) = 0 and
to obtain
σ(Y )
β̂ = E(Y ) − α̂E(X), α̂ =
ρ(X, Y ).
σ(X)
And show that
∂2
g(α, β)
∂α2
> 0 and
∂2
∂2
g(α, β) ∂β
2 g(α, β)
∂α2
∂
g(α, β)
∂β
=0
2
∂
− ( ∂αβ
g(α, β))2 > 0.
4. Let X1 , X2 , . . . , Xn be i.i.d. U (0, 1) random variables. And let Yj denote the jth order
statistic of X1 , X2 , . . . , Xn . Find the probability density function of Y = Yn − Y1 .
解答: The joint density of Y1 and Yn is
n(n − 1)(yn − y1 )n−2 , 0 < y1 < yn < 1,
g1n (y1 , yn ) =
0,
otherwise.
Set Y = Yn − Y1 and
= Y1 , then Y1 = Z and Yn = Y + Z, and thus |J| = 1.
R 1−yZ n−2
fY (y) = n(n − 1) 0 y dz, 0 < y < 1; that is
n(n − 1)y n−2 (1 − y), 0 < y < 1
fY (y) =
0,
otherwise.
5. Suppose U1 , U2 , . . . are independent uniform (0, 1) random variables, and let N be
the first n ≥ 2 such that Un > Un−1 . For 0 ≤ u ≤ 1 :
(i) show that P (U1 ≤ u and N = 2) = u − u2 /2;
(ii) also show that P (U1 ≤ u and N = n) =
un−1
(n−1)!
−
un
,
n!
n ≥ 2;
(iii) find P (N = n), and show E(N ) = e.
解答:
RuR1
du2 du1 = 1/2 − (1 − u)2 /2 = u − u2 /2;
RuRu
Ru R1
(ii) P (U1 ≤ u and N = n) = 0 0 1 · · · 0 n−2 un−1 dun dun−1 · · · du1 =
un
, n ≥ 2;
n!
(i) P (U1 ≤ u and N = 2) =
姓名:
0
u1
編號:
un−1
(n−1)!
−
2007-12-07
九十六學年度國立中山大學應用數學系統計組推薦甄試試題
1
(iii) P (N = n) = (n−1)!
− n!1 , n ≥ 2;
P
P∞
P∞
1
n−1
)= ∞
E(N ) = n=2 n( n!1 − (n+1)!
n=2 n n! =
n=2
1
(n−2)!
3 of 3
= e.
6. Suppose X1 , . . . , Xn are independent exponential E(λ), λ > 0 random variables.
(i) Find the maximum likelihood estimate (MLE) of λ and its corresponding distribution;
(ii) Find an UMVUE estimate of λ based on the MLE found above, and see if it
achieves the Cramer-Rao lower bound.
解答: The density of Xi is
f (x) = λ e−λ x
for x > 0.
The log-likelihood function is
`(λ|x1 , . . . , xn ) = n log λ − λ
n
X
xi ,
i=1
then
n
1
n X
xi ⇒ λ̂ = .
` (λ|x1 , . . . , xn ) = −
λ i=1
X̄
0
P
Note that Note that E(λ) distribution is Gamma(1; λ). Then, ni=1 Xi follows
Gamma(n, λ) distribution and so X̄ ∼ Gamma(n; nλ), where the density is
fX̄ (x) =
Then
(nλ)n n−1 −nλx
x e
, x > 0.
Γ(n)
1
nλ
E
=
,
n−1
X̄
thus λ̂ is biased but T =
n−1
λ̂
n
is unbiased. Since can be computed as

!2 
"
2 #
n
X
n
d log `(λ|x1 , . . . , xn )
n
E
=E
−
Xi  = 2 .
dλ
λ i=1
λ
the Cramer-Rao lower bound is
λ2
.
n
E (T − λ)2 = E
Now consider
"
n−1
λ̂ − λ
n
2 #
=
λ2
,
n−2
which is greater than the CR lower bound and this is the UMVUE, the CR
lower bound can not be obtained.
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