Math 180B, Winter 2015
Homework 9 Solutions
Ex. 5.2.1.
(a) P[X = 0] = (.94)2 0 = 0.2901; P[X = 1] = 20(.06)(.94)19 = 0.3703;
P[X = 2] = 190(.06)2 (.94)18 = 0.2246.
(b) P[X = 0] = (.97)40 = 0.2957; P[X = 1] = 40(.03)(.97)39 = 0.3658; P[X = 2] =
780(.03)2 (.97)38 = 0.2206.
(c) P[X = 0] = e−1.2 = 0.3012; P[X = 1] = e−1.2 (1.2) = 0.3614; P[X = 2] =
e−1.2 (1.2)2 /2 = 0.2169.
Pr. 5.2.4. The exact distribution of the number of points that fall in [0, 1) is binomial
with parameters N and p = 1/N . Because N p = 1, the limit distribution, as N → ∞, is
the Poisson distribution with λ = 1.
Pr. 5.2.5.
The exact distribution of the number of points that fall within one unit
of the origin is binomial with parameters N and p = (π · 12 )/(πr2 ) = r−2 . Because
N p = N r−2 = λπ, the limit distribution, as N → ∞, is the Poisson distribution with
parameter λπ.
Pr. 5.2.8. We have
P[Sn = 0] = .3024,
P[X(1) = 0] = 0.3679
P[Sn = 1] = .4404,
P[X(1) = 1] = 0.3679
P[Sn = 2] = .2144,
P[X(1) = 2] = 0.1839
P[Sn = 3] = .0404,
P[X(1) = 3] = 0.0613
P[Sn = 4] = .0024,
P[X(1) = 4] = 0.0153.
These are in accord with the error bound (5.7) (page 233) because
|P[Sn = 0] − P[X(1) = 0]| = 0.0655
|P[Sn = 1] − P[X(1) = 1]| = 0.0725
|P[Sn = 2] − P[X(1) = 2]| = 0.0305
|P[Sn = 3] − P[X(1) = 3]| = 0.0209
|P[Sn = 4] − P[X(1) = 4]| = 0.0129.
1
P4
i=1
p2i = .3. Indeed,
Ex. 5.3.8.
P[X(1) = 5, X(2) − X(1) = 7]
P[X(2) = 12]
−5 5
e 5 /5! · e−5 57 /7!
=
e−10 1012 /12!
12
=
(1/2)12 = 0.1934.
5
P[X(1) = 5|X(2) = 12] =
(j)
be the arrival time of the k th “mark” in Poisson process number j
(j)
has the exponential distribution with parameter λj , this probability
Pr. 5.3.9. Let Wk
(j = 1, 2).
(a) Because W1
is
(1)
P[W1
<
(2)
W1 ]
∞
Z
∞
Z
λ1 λ2 e−λ1 s−λ2 t dt ds
=
Z0 ∞
=
s
λ1 e−(λ1 +λ2 )s ds
0
=
(j)
(b) Because W2
λ1
.
λ1 + λ2
has the gamma distribution with parameters (2, λj ), this probability
is
(1)
P[W2
<
(2)
W2 ]
Z
∞
Z
=
0
∞
λ21 λ22 ste−λ1 s−λ2 t dt ds
s
The inner integral can be integrated by parts:
Z ∞
∞ s
1
1
t
−λ2 t −λ2 t
+ 2 e
+ 2 e−λ2 s .
te
dt = −
=
λ
λ
λ
λ2
s
2
2
s
2
Therefore,
(1)
P[W2
<
(2)
W2 ]
Z
∞
λ21 (λ2 s2 + s)e−(λ1 +λ2 )s ds
0
Z ∞
Z
2
2 −(λ1 +λ2 )s
2
= λ1 λ2
s e
ds + λ1
=
=
0
2
2λ1 λ2
∞
se−(λ1 +λ2 )s ds
0
λ21
+
(λ1 + λ2 )3
(λ1 + λ2 )2
2λ2 λ2 + λ31 + λ21 λ2
= 1
(λ1 + λ2 )3
λ31 + 3λ21 λ2
=
.
(λ1 + λ2 )3
Second Solution. Let X = {X(t) : t ≥ 0} be a Poisson process with rate λ1 + λ2 . Let
p = λ1 /(λ1 + λ2 ) and q = 1 − p = λ2 /(λ1 + λ2 ). Using a device discussed in class, color
2
the points of X at random, either Red (with probability p) or Green (with probability q),
the different points being colored independently of each other. Then the Red points form
a Poisson process X1 of rate p · (λ1 + λ2 ) = λ1 , the Green points form a Poisson process
X2 of rate q · (λ1 + λ2 ) = λ2 , and X1 and X2 are independent.
(a) Thought of in terms of the coloring construction just discussed, the required probability is just the probability that the first point of X is colored Red; this probability is
p = λ1 /(λ1 + λ2 ).
(b) In terms of the coloring construction just discussed, we now seek the probability
that as the points are colored, two are colored Red before two are colored Green. This
can happen in one of three ways: the points are colored RR or RGR or GRR. These three
events have respective probabilities p2 , pqp = p2 q, and qp2 . The required probability is
therefore
λ21
2λ21 λ2
p + 2p q =
+
(λ1 + λ2 )2
(λ1 + λ2 )3
λ2 (λ1 + λ2 ) + 2λ21 λ2
= 1
(λ1 + λ2 )3
λ3 + 3λ21 λ2
= 1
.
(λ1 + λ2 )3
2
2
3