IEOR 6711: Stochastic Models I
Fall 2012, Professor Whitt
Solutions to Homework Assignment 3 due on Tuesday, September 25.
Problem 2.1 The conditions (i) and (ii) of denition 2.1.2 are apparent from the denition 2.1.1.
Hence it is sucient to show that denition 2.1.1 implies the last two conditions of denition
2.1.2.
P(N(h) = 1) = h + o(h) :
lim
h0
P(N(h) = 1) h
h
= lim
h0
e
h
h h
h
= lim
h0
_
e
h
1
_
= 0 .
P(N(h) 2) = o(h) :
lim
h0
P(N(h) 2)
h
= lim
h0
1 e
h
e
h
h
h
= lim
h0
1 e
h
h
lim
h0
e
h
=
1 (1 h + o(h))
h
(1 o(h)) = +
o(h)
h
+ o(h) 0.
Or using e
ax
= 1 + ax + o(x), o(x) o(x) = o(x), and f(x) o(x) = o(x) for any f(x)
satisfying lim
x0
f(x) is nite,
P(N(h) = 1) = e
h
h = (1 h + o(h))h = h + o(h) .
P(N(h) 2) = 1e
h
e
h
h = 1(1+h)(1h+o(h)) =
2
h
2
+o(h) = o(h) .
Problem 2.2 For s < t,
P(N(s) = k|N(t) = n) =
P(N(s) = k, N(t) = n)
P(N(t) = n)
=
P(N(s) = k, N(t) N(s) = n k)
P(N(t) = n)
=
P(N(s) = k)P(N(t) N(s) = n k)
P(N(t) = n)
=
P(N(s) = k)P(N(t s) = n k)
P(N(t) = n)
=
_
e
s
(s)
k
k!
__
e
(ts)
((t s))
(nk)
(n k)!
__
e
t
(t)
n
n!
_
1
=
n!
k!(n k)!
s
k
(t s)
nk
t
n
=
_
n
k
_
_
s
t
_
k
_
1
s
t
_
nk
.
1
Problem 2.4 Let {X(t) : t 0} be a stochastic process having stationary independent incre-
ments and X(0) = 0. (People call it Levy process.) Two typical Levy processes are Poisson
and Brownian motion processes. They are representatives of purely discrete and purely
continuous continuous time stochastic processes, respectively. Furthermore, it is not easy
to nd any non-trivial Levy process except them. Now lets try to express E[X(t)X(t +s)]
by moments of X using only the properties of Levy process.
E[X(t)X(t + s)] = E[X(t)(X(t + s) X(t) + X(t))]
= E[X(t)(X(t + s) X(t)) + X(t)
2
]
= E[X(t)(X(t + s) X(t))] +E[X(t)
2
]
= E[X(t)]E[(X(t + s) X(t))] +E[X(t)
2
] by independent increment
= E[X(t)]E[(X(s))] +E[X(t)
2
] by stationary increment
Now return to our original process, Poisson process. By substituting E[N(t)] = t, E[N(t)
2
] =
t + (t)
2
,
E[N(t)N(t + s)] =
2
st + t +
2
t
2
.
A digression : if X(t) Normal(0, t), what is the result? This is the Brownian motion case.
Problem 2.5 {N
1
(t) + N
2
(t), t 0} is a Poisson process with rate
1
+
2
.
Axioms (i) and (ii) of denition e.1.2 easily follow. Letting N(t) = N
1
(t) + N
2
(t),
P(N(h) = 1) = P(N
1
(h) = 1, N
2
(h) = 0) +P(N
1
(h) = 0, N
2
(h) = 1)
=
1
h(1
2
h) +
2
h(1
1
h) + o(h)
= (
1
+
2
)h + o(h)
and
P(N(h) = 2) = P(N
1
(h) = 1, N
2
(h) = 1)
= (
1
h + o(h))(
2
h + o(h))
=
1
2
h
2
+ o(h) = o(h) .
The probability that the rst event of the combined process comes from {N
1
(t), t 0}
is
1
/(
1
+
2
), independently of the time of the event. Let X
i
and Y
i
are the i-th inter
arrival times of N
1
and N
2
, respectively. Then
P(rst from N
1
|rst at t) = P(X
1
< Y
1
| min{X
1
, Y
1
} = t)
=
1
1
+
2
where the last equality comes from our old homework 1.1.34.
2
Problems 2.62.9 Answers in back of the book.
Problem 2.10 (a) First note that the time until next bus arrival follows exponential distribution
with rate by the memoryless property of exponential distribution. Let X be the time
until next bus arrival. Then T, the random variable representing the time spent to
reach home is
T =
_
X + R if X s ,
s + W if X > s
= (X + R)1
{Xs}
+ (s + W)1
{X>s}
.
Hence
E[T] = E[(X + R)1
{Xs}
] +E[(s + W)1
{X>s}
]
= E[X1
{Xs}
] + RE[1
{Xs}
] + (s + W)E[1
{X>s}
]
=
_
s
0
xe
x
dx + RP(X s) + (s + W)P(X > s)
=
_
s
0
xe
x
dx + R(1 e
s
) + (s + W)e
s
= xe
x
s
0
e
x
s
0
+ R + (s + W R)e
s
=
1
(1 e
s
) se
s
+ R + (s + W R)e
s
=
1
+ R +
_
W R
1
_
e
s
.
(b) Considering
d
ds
E[T] = (1 (W R))e
s
_
_
> 0 if W <
1
+ R ,
= 0 if W =
1
+ R ,
< 0 if W >
1
+ R ,
we get
argmin
0s<
E[T] =
_
_
0 if W <
1
+ R ,
any number [0, ) if W =
1
+ R ,
if W >
1
+ R ,
(c) Since the time until the bus arrives is exponential, it follows by the memoryless property
that if it is optimal to wait any time then one should always continue to wait for the
bus.
Problem 2.11 Conditioning on the time of the next car yields
E[wait] =
_
0
E[wait|car at x]e
x
dx .
3
Now,
E[wait|car at x] =
_
x +E[wait] if x < T ,
0 if x T
and so
E[wait] =
_
T
0
xe
x
dx +E[wait](1 e
T
)
or
E[wait] =
1
e
T
T
1
.
Problem 2.13 First note that T is (unconditionally) exponential with rate p, N is geometric
with parameter p, and the distribution of T given that N = n is gamma with parameter n
and , we obtain
P(N = n|t < T t) =
P(t < T t|N = n)
P(t < T t)
e
t
(t)
n1
(n 1)!
p(1 p)
n1
p e
pt
=
e
t(1p)
[t(1 p)]
n1
(n 1)!
.
Hence, given that T = t, N has the distribution of X + 1, where X is a Poisson random
variable with mean t(1 p). A simpler argument is to note that the occurrences of failure
causing shocks and non-failure causing shocks are independent Poisson processes. Hence, the
number of non-failure causing shocks by time t is Poisson with mean (1 p)t, independent
of the event that the rst failure shock occurred at that time.
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