STAT 610 Homework #5 solution
3.3. Let Xi be the event that a car passes in second i. Then
P (the pedastrian has to wait four seconds)
= P ( At least one car in the first three seconds ∩ a car in the fourth seconds ∩
no car in the last three seconds)
= P r((X1c ∩ X2c ∩ X3c )c ∩ X4 ∩ X5c ∩ X6c ∩ X7c )
= [1 − (1 − p)3 ]p(1 − p)3
3.8. (a) X ∼ Bin(1000, 1/2) since the 1000 customers choose theater randomly.
1000−x
1000
X 1000 1 x 1
< .01
1−
P (X > N ) =
2
2
x
x=N +1
which imples that
1000 1000
X 1000
1
< .01
2
x
x=N +1
By solving the last ineqality to get N , theh smallest integer that satisfies the last
inequality is 537.
(b) To use the normal approximation we take X ∼ N (500, 250), where µ = 1000( 21 ) = 500
and σ 2 = 1000( 12 )( 12 ) = 250. Then
X − 500
N − 500
√
P (X > N ) = P
> √
< .01
250
250
Thus,
P
N − 500
Z> √
250
< .01
where Z ∼ N (0, 1)
From the normal table P (Z > 2.33) ' .0099 ⇒
N√−500
250
= 2.33 ⇔ N ' 537.
3.11. (a)
lim
M/N →p,M →∞,N →∞
=
M
x
N −M
K −x
N
K
K!
M !(N − M )!(N − K)!
lim
x!(K − x)! M/N →p,M →∞,N →∞ N !(M − x)!(N − M − (K − x))!
In the limit, each of the factorial terms
be replaced by the approximation from
√ can
M +1/2 −M
Stirling’s formula, for example, M !/( 2πM
e
) → 1.
√
When this replacement is made, all the 2π and the exponential terms cancel. Thus,
M
N −M
x
K −x
lim
N
M/N →p,M →∞,N →∞
K
K
M M +1/2 (N − M )N −M +1/2 (N − K)N −K+1/2
=
lim
N
+1/2
x M/N →p,M →∞,N →∞ N
(M − x)M −x+1/2 (N − M − K + x)N −M −(K−x)+1/2
We can evaluate the limit by breaking the ratio into seven terms. In some limits we use
the fact that M → ∞, N → ∞ and M/N → p imply N − M = N (1 − M/N ) → ∞. The
first terms is
M
M
1
1
= lim
= −x = ex
lim
M
−x
M →∞ M − x
M →∞ 1 +
e
M
1
Similary we get more terms
lim
N −M →∞
N −M
N − M − (K − x)
and
lim
N →∞
N −K
N
N −M
N
= eK−x
= e−K
The product of above three terms is one. Three other temrs are
lim
M →∞
lim
N −M →∞
M
M −x
1/2
=1
N −M
N − M − (K − x)
and
lim
N →∞
N −K
N
1/2
=1
1/2
=1
The only term left is
(M − x)x (N − N − (K − x))k−x
(N − K)K
M/N →p,M →∞,N →∞
x K−x
N − M − (K − x)
M −x
=
lim
N −K
M/N →p,M →∞,N →∞ N − K
lim
= px (1 − p)k−x .
(b) In addition to condition of (a), we have K → ∞, p → 0, M K/N → pK → λ. By the
Poisson approximation to the Binomial, we heuristically get
M
N −M
K x
e−λ λx
x
K −x
→
p (1 − p)k−x →
N
x
x!
K
(c) Using Stirling’s formula as in (a), we can get
lim
N →∞,M →∞,K→∞,M/N →0,KM/N →λ
M
x
N −M
K −x
N
K
e−x K x ex M x ex (N − N )K−x eK−x
N K eK
N →∞,M →∞,K→∞,M/N →0,KM/N →λ x!
x K−x
1
KM
N −M
=
lim
x! N →∞,M →∞,K→∞,M/N →0,KM/N →λ
N
N
K
1
M K/N
= λx
lim
1−
x! N →∞,M →∞,K→∞,M/N →0,KM/N →λ
K
=
=
lim
e−λ λx
x!
2
3.12. Let X be the number of successes in n trals and Y be the number of faliures before the rth
successes.
FX (r − x)
=
P (X ≤ r − 1)
=
P (rth success on (n + 1) th or later trial)
=
P (at leastn + 1 − r faliures before the rth successes)
=
P (Y ≥ n − r + 1)
=
1 − P (Y ≤ n − r)
=
1 − FY (n − r)
3.13. For any X with support 0, 1, · · · , we have the mean and variance of the 0-truncated XT are
given by
EXT =
∞
X
xP (XT = x) =
x=1
=
∞
X
P (X = x)
x
P (X > 0)
x=1
∞
∞
X
X
1
1
EX
xP (X = x) =
xP (X = x) =
P (x > 0) x=1
P (X > 0) x=0
P (X > 0)
In a similar way we get EXT2 =
EX 2
P (X>0) .
V arXt =
Thus,
EX 2
−
P (X > 0)
EX
P (X > 0)
(a) For Poisson(λ), P (X > 0) = 1 − P (X = 0) = 1 −
P (XT = x)
EXT
=
e−λ λ0
0!
2
= 1 − e−λ , therefore
e−λ λx
, x = 1, 2, · · ·
x!(1 − e−λ
= λ/(1 − e−λ )
(λ2 + λ)/(1 − e−λ ) − (λ/(1 − e−λ ))2
r
0
r
(b) For negative binomial(r, p), P (X > 0) = 1 − P (X = 0) = 1 − r−1
0 p (1 − p) = 1 − p ,
Then
r+x−1 r
p (1 − p)x
x
, x = 1, 2, · · ·
P (XT = x) =
1 − pr
r(1 − p)
EXT =
p(1 − pr )
r(1 − p)
r(1 − p) + r2 (1 − p)2
V arXT =
−
p2 (1 − pr )
p(1 − pr )2
V arXT
3.14. (a)
P∞
x=1
−(1−p)x
x log p
=
1
log p
P∞
x=1
=
−(1−p)x
x
= 1, since the sum is the Tylor series for log p.
(b)
"∞
#
"∞
#
−1 X
−1 X
−1 1
−1 1 − p
x
x
EX =
(1 − p) =
(1 − p) − 1 =
−1 =
log p x=1
log p x=0
log p p
log p
p
Since the geometric series converes uniformly,
∞
EX 2
=
=
∞
−1 X
(1 − p) X d
x(1 − p)x =
(1 − p)x
log p x=1
log p x=1 dp
∞
(1 − p) d X
(1 − p) 1 − p
−(1 − p)
(1 − p)x =
= 2
log p dp x=1
log p
p
p log p
3
Thus
−(1 − p)
1−p
1
+
p2 log p
log p
V arX =
Alternatively, the mgf can be calculated,
MX (t) =
x
∞ −1 X (1 − p)et
log(1 + pet − et )
=
log p x=1
x
log p
and can be differentiated to obtain the moments.
3.15. The moment generating function for the negative binomial is
r r
1 r(1 − p)(et − 1)
p
=
1
+
M (t) =
1 − (1 − p)et
r 1 − (1 − p)et
the term
r(1 − p)(et − 1)
λ(et − 1)
→
= λ(et − 1) as r → ∞, p → 1andr(1 − p) → λ
t
1 − (1 − p)e
1
Thus by Lemma 2.3.14, the negative binomial moment generating function converges to
t
eλ(e −1) , the Poisson mement generating function.
3.16. (a) Using integration by parts with, u = tα and dv = e−t dt, we obtain
∞ Z ∞
Z ∞
Γ(α + 1) =
t(α+1)−1 e−t dt = tα (−et ) −
αtα−1 (−et )dt = 0 + αΓ(α) = αΓ(α)
0
0
(b) Making the change of variable z =
Z ∞
Z ∞√
t−1/2 e−t dt =
Γ(1/2) =
0
0
√
√
0
2t, i.e., t = z 2 /2, we obtain
√
2 −z2 /2
e
zdz = 2
z
Z
∞
e−z
2
/2
dz =
0
√ 1 Z ∞ −z2 /2
2
e
dz
2 −∞
√
2√
2π = π
=
2
The fourth eqaulity is obtained since e−z
2
/2
is even function.
r
p
, and,
3.18. If Y ∼ negaive binomial(r, p), its moment generating function is MY (t) = 1−(1−p)e
t
r
p
from Theorem 2.3.15, MpY (t) = 1−(1−p)ept . Now use L’Hôpital’s rule to calculate
lim
p→0
p
1 − (1 − p)ept
= lim
p→1
1
1
,
=
(p − 1)tept + ept
1−t
so the moment generating function converges to (1 − t)−r , the moment generating function
of a gamma(r, 1).
4