TEST BANK FOR Introduction to Stochastic Processes with R By Robert P. Dobrow

,Instructor’s Solutions Manual

for


Introduction to Stochastic Processes with R


by Robert P. Dobrow




March 6, 2016




1

,Chapter 1
1.1 For the following scenarios identify a stochastic process {Xt , t ∈ I}, describing (i) Xt
in context, (ii) state space, and (iii) index set. State whether the state space and index
set are discrete or continuous.

b) Xt is the student’s status at the end of year t. State space (discrete): S =
{Drop Out, Frosh, Sophomore, Junior, Senior, Graduate}. Index set (discrete): I =
{0, 1, 2, . . .}.
c) Xt is the magnitude for an earthquake which occurs at time t. State space (contin-
uous): (0, 10). Index set (continuous): [0, ∞).
d) Xt is the circumference of the tree at location t. State space (continuous): (0, ∞).
Index set (continuous): [0, 2] × [0, 2]; or the x − y coordinates of a location in the
arboretum.
e) Xt is the arrival time of student t. State space (continuous): [0, 60]. Index set
(discrete): {1, 2, . . . , 30}.
f) Xt is the order of the deck of cards after t shuffles. State space (discrete): Set of
all orderings of the deck (52! elements). Index set (discrete): {0, 1, 2, . . .}.

1.2 A regional insurance company insures homeowners against flood damage. Half of their
policyholders are in Florida, 30% in Louisiana, and 20% in Texas.

a) Let A be the event that a claim is filed for flood damage. Then

P (A) = P (A|F )P (F ) + P (A|L)P (L) + P (A|T )P (T )
= (0.03)(0.50) + (0.015)(0.30) + (0.02)(0.20) = 0.0235.


b) P (T |A) = P (A|T )P (T )/P (A) = (0.02)(0.20)/0.0235 = 0.17.

1.3 Let B1 , . . . , Bk be a partition of the sample space. For events A and C, prove the law
of total probability for conditional probability.


k k   
X X P (A ∩ Bi ∩ C) P (Bi ∩ C)
P (A|Bi ∩ C)P (Bi |C) =
i=1 i=1
P (Bi ∩ C) P (C)
k k
X P (A ∩ Bi ∩ C) 1 X
= = P (A ∩ Bi ∩ C)
i=1
P (C) P (C) i=1
P (A ∩ C)
= = P (A|C).
P (C)


1.4 See Exercise 1.2. Among policyholders who live within 5 miles of the coast, 75% live
in Florida, 20% live in Louisiana, and 5% live in Texas. Suppose a policyholder lives
within 5 miles of the coast. Use the law of total probability for conditional probability
to find the chance they will file a claim for flood damage next year.


2

, Let A be the event that a claim is filed. Let C be the event that a claimant lives
within five miles of the coast. Then

P (A|C) = P (A|F, C)P (F |C) + P (A|L, C)P (L|C) + P (A|T, C)P (T |C)
= (0.10)(0.75) + (0.06)(0.20) + (0.06)(0.05) = 0.09.



1.5 Two fair, six-sided dice are rolled. Let X1 , X2 be the outcomes of the first and second
die, respectively.

a) Uniform on {1, 2, 3, 4, 5, 6}.
b) Uniform on {2, 3, 4, 5, 6}.
1.6 Bob has n coins in his pocket. One is two-headed, the rest are fair. A coin is picked
at random, flipped, and shows heads.

Let A be the event that the coin is two-headed.
P (H|A)P (A) (1)(1/n) 2
P (A|H) = c c
= = .
P (H|A)P (A) + P (H|A )P (A ) (1)(1/n) + (1/2)((n − 1)/n) n+1


1.7 A rat is trapped in a maze with three doors and some hidden cheese.

Let X denote the time until the rat finds the cheese. Let 1, 2, and 3 denote each door,
respectively. Then

E(X) = E(X|1)P (1) + E(X|2)P (2) + E(X|3)P (3)
1 1 1 2
= (2 + E(X)) + (3 + E(X)) + (1) = 2 + E(X) .
3 3 3 3
Thus, E(X) = 6 minutes.
1.8 A bag contains 1 red, 3 green, and 5 yellow balls. A sample of four balls is picked. Let
G be the number of green balls in the sample. Let Y be the number of yellow balls in
the sample.

a)
P (G = 1|Y = 2) = P (G = 2|Y = 2) = 1/2.

b) The conditional distribution of G is binomial with n = 2 and p = 3/9 = 1/3.
 
2
P (G = k|Y = 2) = (1/3)k (2/3)2−k , for k = 0, 1, 2.
k


1.9 Suppose X is uniformly distributed on {1, 2, 3, 4}. If X = x, then Y is uniformly
distributed on {1, . . . , x}.

a) P (Y = 2|X = 2) = 1/2.


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