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INTRODUCTION to PROBABILITY
Published by AMS
Solutions to the odd numbered exercises
SECTION 1.1
1. As n increases, the proportion of heads gets closer to 1/2, but the diﬀerence between the number
of heads and half the number of ﬂips tends to increase (although it will occasionally be 0). 3. (b) If one simulates a suﬃciently large number of rolls, one should be able to conclude that the
gamblers were correct. 5. The smallest n should be about 150. 7. The graph of winnings for betting on a color is much smoother (i.e. has smaller ﬂuctuations) than
the graph for betting on a number. 9. Each time you win, you either win an amount that you have already lost or one of the original
numbers 1,2,3,4, and hence your net winning is just the sum of these four numbers. This is not a
foolproof system, since you may reach a point where you have to bet more money than you have.
If you and the bank had unlimited resources it would be foolproof. 11. For two tosses, the probabilities that Peter wins 0 and 2 are 1/2 and 1/4, respectively. For four
tosses, the probabilities that Peter wins 0, 2, and 4 are 3/8, 1/4, and 1/16, respectively. 13. Your simulation should result in about 25 days in a year having more than 60 percent boys in the
large hospital and about 55 days in a year having more than 60 percent boys in the small hospital. 15. In about 25 percent of the games the player will have a streak of ﬁve. SECTION 1.2
1. P ({a, b, c}) = 1
P ({a, b}) = 5/6
P ({b, c}) = 1/2
P ({a, c}) = 2/3 3. (b), (d) 5. (a) 1/2
(b) 1/4
(c) 3/8
(d) 7/8 7. 11/12 9. P ({a}) = 1/2
P ({b}) = 1/3
P ({c}) = 1/6
P (φ) = 0 3/4, 1 11. 1 : 12, 1 : 3, 1 : 35 13. 11:4 15. Let the sample space be:
ω1 = {A, A} ω4 = {B, A} ω7 = {C, A}
1 ω2 = {A, B} ω5 = {B, B} ω8 = {C, B} ω3 = {A, C} ω6 = {B, C} ω9 = {C, C} where the ﬁrst grade is John’s and the second is Mary’s. You are given that
P (ω4 ) + P (ω5 ) + P (ω6 ) = .3,
P (ω2 ) + P (ω5 ) + P (ω8 ) = .4,
P (ω5 ) + P (ω6 ) + P (ω8 ) = .1.
Adding the ﬁrst two equations and subtracting the third, we obtain the desired probability as
P (ω2 ) + P (ω4 ) + P (ω5 ) = .6.
17. The sample space for a sequence of m experiments is the set of mtuples of S’s and F ’s, where S
represents a success and F a failure. The probability assigned to a sample point with k successes
and m − k failures is
1 k n − 1 m−k
.
n
n
(a) Let k = 0 in the above expression.
(b) If m = n log 2, then
lim 1 − n→∞ 1
n m 1− = lim n→∞ = lim ( 1 − n→∞ = e−1
= 1
n
1
n n n log 2 log 2 log 2 1
.
2 (c) Probably, since 6 log 2 ≈ 4.159 and 36 log 2 ≈ 24.953.
19. The leftside is the sum of the probabilities of all elements in one of the three sets. For the right
side, if an outcome is in all three sets its probability is added three times, then subtracted three
times, then added once, so in the ﬁnal sum it is counted just once. An element that is in exactly
two sets is added twice, then subtracted once, and so it is counted correctly. Finally, an element in
exactly one set is counted only once by the right side. 21. 7/212 23. We have ∞ ∞ r(1 − r)n = m(ωn ) =
n=0 25. n=0 r
=1.
1 − (1 − r) They call it a fallacy because if the subjects are thinking about probabilities they should realize
that
P (Linda is bank teller and in feminist movement) ≤ P (Linda is bank teller).
One explanation is that the subjects are not thinking about probability as a measure of likelihood.
For another explanation see Exercise 52 of Section 4.1. 27.
Px = P (male lives to age x) =
2 number of male survivors at age x
.
100, 000 Qx = P (female lives to age x) =
29. number of female survivors at age x
.
100, 000 (Solution by Richard Beigel)
(a) In order to emerge from the interchange going west, the car must go straight at the ﬁrst point
of decision, then make 4n + 1 right turns, and ﬁnally go straight a second time. The probability
P (r) of this occurring is
∞ (1 − r)2 r4n+1 = P (r) =
n=0 r(1 − r)2
1
1
,
=
−
1 − r4
1 + r2
1+r if 0 ≤ r < 1, but P (1) = 0. So P (1/2) = 2/15.
(b) Using standard methods from calculus, one can show that P (r) attains a maximum at the value r= √
1+ 5
−
2 √
1+ 5
≈ .346 .
2 At this value of r, P (r) ≈ .15.
31. (a) Assuming that each student gives any given tire as an answer with probability 1/4, then probability that they both give the same answer is 1/4.
(b) In this case, they will both answer ‘right front’ with probability (.58)2 , etc. Thus, the probability
that they both give the same answer is 39.8%. SECTION 2.1
The problems in this section are all computer programs. SECTION 2.2
1. (a) f (ω) = 1/8 on [2, 10] (b) P ([a, b]) =
3. (a) C= 1
log 5 b−a
8 . ≈ .621 (b) P ([a, b]) = (.621) log(b/a)
(c)
log 2
≈ .431
log 5
log(7/2)
≈ .778
P (x < 7) =
log 5
log(25/7)
≈ .791 .
P (x2 − 12x + 35 > 0) =
log 5
P (x > 5) = 5. (a) 1− (b) 1− (c) 1− 1
e1
1
e3
1
e1 ≈ .632
≈ .950
≈ .632
3 (d)
7. 1 (a) 1/3, (b) 1/2, (c) 1/2, (d) 1/3 13. 2 log 2 − 1. 15. Yes. SECTION 3.1
1. 24 3. 232 5. 9, 6. 7.
11. 5!
.
55
28
3n − 2 7
,
,
.
n3
27 1000 15.
17.
21.
23. (a) 263 × 103 (b) 13. 6
3 3
1 × 263 × 103 × (2n − 2)
.
3n 12 · 11 · . . . · (12 − n + 1)
, if n ≤ 12, and 1, if n > 12.
n12
They are the same. 1− 1 1
,
n n
(b) She will get the best candidate if the second best candidate is in the ﬁrst half and the best
candidate is in the secon half. The probability that this happens is greater than 1/4.
(a) SECTION 3.2
1. (a) 20
(b) .0064
(c) 21
(d) 1
(e) .0256
(f) 15
(g) 10 3.
5. 9
7 = 36 .998, .965, .729 7.
4 n j n−j
p q
j b(n, p, j)
=
b(n, p, j − 1) (n − j + 1)!(j − 1)! p
n!
j!(n − j)!
n!
q = n
pj−1 q n−j+1
j−1 . (n − j + 1) p
j
q = (n − j + 1) p
≥ 1 if and only if j ≤ p(n + 1), and so j = [p(n + 1)] gives b(n, p, j) its largest
j
q
value. If p(n + 1) is an integer there will be two possible values of j, namely j = p(n + 1) and
j = p(n + 1) − 1. But 9. n = 15, r = 7 11. Eight pieces of each kind of pie. 13. The number of subsets of 2n objects of size j is
2n
i
2n
i−1
Thus i = n makes 2n
i = 2n
.
j 1
2n − i + 1
≥1⇒i≤n+ .
i
2 maximum. 15. .3443, .441, .181, .027. 17. There are n ways of putting a diﬀerent objects into the 1st box, and then n−a ways of putting
a
b
b diﬀerent objects into the 2nd and then one way to put the remaining objects into the 3rd box.
Thus the total number of ways is
n−a
b n
a 19. 4
1 4
1 (a) (b) 4!
(c) 13
10
52
10
3
2 13
4 = n!
.
a!b!(n − a − b)! = 7.23 × 10−8 . 13 13
3
4
52
10
13 13
3
2
52
13 13
3 13
1 = .044. = .315. 21. 3(25 ) − 3 = 93 (We subtract 3 because the three pure colors are each counted twice.) 23. To make the boxes, you need n + 1 bars, 2 on the ends and n − 1 for the divisions. The n − 1 bars
and the r objects occupy n − 1 + r places. You can choose any n − 1 of these n − 1 + r places for the
bars and use the remaining r places for the objects. Thus the number of ways this can be done is
n−1+r
n−1
5 = n−1+r
.
r 25. (a) 6!
(b) 27. 10
/106 ≈ .1512
6
10
15
/
≈ .042
6
6 Ask John to make 42 trials and if he gets 27 or more correct accept his claim. Then the probability
of a type I error is
b(42, .5, k) = .044,
k≥27 and the probability of a type II error is
1− b(42, .75, k) = .042.
k≥27 n m
p (1 − p)n−m . Taking the derivative with respect to p and setting this equal to
m
0 we obtain m(1 − p) = p(n − m) and so p = m/n. 29. b(n, p, m) = 31. .999996. 33. By Stirling’s formula,
2n
n
4n
2n 35. 2 = √
2
2
(2n!) (2n!)
( 4πn(2n)2n e−2n )4
∼ √
=
n!4 (4n)!
( 2πn(nn )e−n )4 2π(4n)(4n)4n e−4n 2
.
πn Consider an urn with n red balls and n blue balls inside. The left side of the identity
n 2n
n =
j=0 n
j n 2 =
j=0 n
j n
n−j counts the number of ways to choose n balls out of the 2n balls in the urn. The right hand counts
the same thing but breaks the counting into the sum of the cases where there are exactly j red
balls and n − j blue balls.
38. Consider the Pascal triangle (mod 3) for example.
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14 1
1
1
1
1
1
1
1
1
1
1
1
1
1
1 1
2
0
1
2
0
1
2
0
1
2
0
1
2 1
0
0
1
0
0
1
0
0
1
0
0
1 1
1
1
2
2
2
0
0
0
1
1
1 1
2
0
2
1
0
0
0
0
1
2 1
0
0
2
0
0
0
0
0
1 1
1
1
0
0
0
0
0
0 1
2
0
0
0
0
0
0 1
0
0
0
0
0
0 1
1
1
1
1
1 1
2
0
1
2 1
01
011
1121
6 15
16
17
18 1
1
1
1 0
1
2
0 0
0
1
0 2
2
2
0 0
2
1
0 0
0
2
0 1
1
1
0 0
1
2
0 0
0
1
0 1
1
1
2 0
1
2
0 0
0
1
0 2
2
2
0 0
2
1
0 0
0
2
0 1
11
121
0001 Note ﬁrst that the entries in the third row are 0 for 0 < j < 3. Lucas notes that this will be true
for any p. To see this assume that 0 < j < p. Note that
p
j = p(p − 1) · · · p − j + 1
j(j − 1) · · · 1 is an integer. Since p is prime and 0 < j < p, p is not divisible by any of the terms of j!, and so
p
(p − 1)! must be divisible by j!. Thus for 0 < j < p we have
= 0 mod p. Let us call the
j
triangle of the ﬁrst three rows a basic triangle. The fact that the third row is
1001
produces two more basic triangles in the next three rows and an inverted triangle of 0’s between
these two basic triangles. This leads to the 6’th row
1002001
.
This produces a basic triangle, a basic triangle multiplied by 2 (mod 3), and then another basic
triangle in the next three rows. Again these triangles are separated by inverted 0 triangles. We can
continue this way to construct the entire Pascal triangle as a bunch of multiples of basic triangles
separated by inverted 0 triangles. We need only know what the mutiples are. The multiples in row
np occur at positions 0, p, 2p, ..., np. Looking at the triangle we see that the multiple at position
(mp, jp) is the sum of the multiples at positions (j − 1)p and jp in the (m − 1)p’th row. Thus these
multiples satisfy the same recursion relation
n
j = n−1
n−1
+
j−1
j that determined the Pascal triangle. Therefore the multiple at position (mp, jp) in the triangle is
m
. Suppose we want to determine the value in the Pascal triangle mod p at the position (n, j).
j
Let n = sp + s0 and j = rp + r0 , where s0 and r0 are < p. Then the point (n, j) is at position
s
.
(s0 , r0 ) in a basic triangle multiplied by
r
Thus
s s0
n
=
.
r
j
r0
But now we can repeat this process with the pair (s, r) and continue until s < p. This gives us the
result:
k
n
si
=
(mod p) ,
j
rj
i=0
where s = s0 + s1 p1 + s2 p2 + · · · + sk pk ,
j = r0 + r1 p1 + r2 p2 + · · · + rk pk .
7 If rj > sj for some j then the result is 0 since, in this case, the pair (sj , rj ) lies in one of the inverted
0 triangles.
If we consider the row pk − 1 then for all k, sk = p − 1 and rk ≤ p − 1 so the product will be positive
resulting in no zeros in the rows pk − 1. In particular for p = 2 the rows pk − 1 will consist of all
1’s.
39.
1
2n!
2n(2n − 1) · · · 2 · 1
b(2n, , n) = 2−2n
=
2
n!n!
2n · 2(n − 1) · · · 2 · 2n · 2(n − 1) · · · 2
(2n − 1)(2n − 3) · · · 1
.
=
2n(2n − 2) · · · 2 SECTION 3.3
3. (a) 96.99%
(b) 55.16% SECTION 4.1
3. (a) 1/2
(b) 2/3
(c) 0
(d) 1/4 5. (a) (1) and (2)
7. (a) P (A ∩ B) = P (A ∩ C) = P (B ∩ C) = 1
,
4 1
,
4
1
1
P (A ∩ B ∩ C) = = P (A)P (B)P (C) = .
4
8
1
(b) P (A ∩ C) = P (A)P (C) = , so C and A are independent,
4
1
P (C ∩ B) = P (B)P (C) = , so C and B are independent,
4
1
1
P (C ∩ (A ∩ B)) = = P (C)P (A ∩ B) = ,
4
8
so C and A ∩ B are not independent. (b) (1) P (A)P (B) = P (A)P (C) = P (B)P (C) = 8 8. P (A ∩ B ∩ C) = P ({a}) =
P (A) = P (B) = P (C) = 1
,
8 1
.
2 Thus while P (A ∩ B ∩ C) = P (A)P (B)P (C) = 1
,
8 5
,
16
1
P (A)P (B) = P (A)P (C) = P (B)P (C) = .
4
Therefore no two of these events are independent. P (A ∩ B) = P (A ∩ C) = P (B ∩ C) = 9. (a) 1/3
(b) 1/2 13. 15. 1/2 (a) 48
11
52
−
13 4
2
≈ .307 .
48
13 48 3
11 1
≈ .328 .
(b)
51
12
˜
17(a) P (A ∩ B) = P (A) − P (A ∩ B) = P (A) − P (A)P (B)
= P (A)(1 − P (B))
˜
= P (A)P (B) .
˜ and B by A.
(b) Use (a), replacing A by B
19. .273. 21. No. 23. Put one white ball in one urn and all the rest in the other urn. This gives a probability of nearly
3/4, in particular greater than 1/2, for obtaining a white ball which is what you would have with
an equal number of balls in each urn. Thus the best choice must have more white balls in one urn
than the other. In the urn with more white balls, the best we can do is to have probability 1 of
getting a white ball if this urn is chosen. In the urn with less white balls than black, the best we
can do is to have one less white ball than black and then to have as many white balls as possible.
Our solution is thus best for the urn with more white balls than black and also for the urn with
more black balls than white. Therefore our solution is the best we can do. 25. We must have
p n − 1 k−1 n−k
n k n−k
p
=p
q
.
p q
k−1
j This will be true if and only if np = k. Thus p must equal k/n.
27.
(a) P (Pickwick has no umbrella, given that it rains)= 2
.
9 (b) P (It does not rain, given that he brings his umbrella)=
9 5
.
12 29. 2
.
3
The events ‘Accepted by Dartmouth’ and ‘Accepted by Harvard’ are not independent.
P (Accepted by Dartmouth  Accepted by Harvard) = 31. The probability of a 60 year old male living to 80 is .41, and for a female it is .62. 33. You have to make a lot of calcuations, all of which are like this:
˜
P (A1 ∩ A2 ∩ A3 ) = P (A2 )P (A3 ) − P (A1 )P (A2 )P (A3 )
= P (A2 )P (A3 )(1 − P (A1 ))
˜
= P (A1 )P (A2 )P (A3 ). 35. The random variables X1 and X2 have the same distributions, and in each case the range values are
the integers between 1 and 10. The probability for each value is 1/10. They are independent. If the
ﬁrst number is not replaced, the two distributions are the same as before but the two random vari37.
P (max(X, Y ) = a) = P (X = a, Y ≤ a) + P (X ≤ a, Y = a) − P (X = a, Y = a).
ables are not independent.
P (min(X, Y ) = a) = P (X = a, Y > a) + P (X > a, Y = a) + P (X = a, Y = a).
Thus P (max(X, Y ) = a) + P (min(X, Y ) = a) = P (X = a) + P (Y = a)
and so u = t + s − r. 39. (a) 1/9
(b) 1/4
(c) No
(d) pZ = 43. −2
1
6 −1
1
6 0 1 2 4 1
6 1
6 1
6 1
6 .710. 45.
(a) The probability that the ﬁrst player wins under either service convention is equal to the probability that if a coin has probability p of coming up heads, and the coin is tossed 2N + 1 times, then
it comes up heads more often than tails. This probability is clearly greater than .5 if and only if
p > .5.
(b) If the ﬁrst team is serving on a given play, it will win the next point if and only if one of the
following sequences of plays occurs (where ‘W’ means that the team that is serving wins the play,
and ‘L’ means that the team that is serving loses the play):
W, LLW, LLLLW, . . . .
The probability that this happens is equal to
p + q2 p + q4 p + . . . ,
which equals p
1
.
=
2
1−q
1+q Now, consider the game where a ‘new play’ is deﬁned to be a sequence of plays that ends with a
point being scored. Then the service convention is that at the beginning of a new play, the team
that won the last new play serves. This is the same convention as the second convention in the
preceding problem.
¿From part a), we know that the ﬁrst team to serve under the second service convention will win
the game more than half the time if and only if p > .5. In the present case, we use the new value
of p, which is 1/(1 + q). This is easily seen to be greater than .5 as long as q < 1. Thus, as long as
p > 0, the ﬁrst team to serve will win the game more than half the time.
10 47. (a) P (Y1 = r, Y2 = s) = P (Φ1 (X1 ) = r, Φ2 (X2 ) = s)
P (X1 = a, X2 = b) . =
Φ1 (a)=r
2 (b)=s (b) If X1 , X2 are independent, then
P (Y1 = r, Y2 = s) = P (X1 = a, X2 = b)
Φ1 (a)=r
Φ2 (b)=s P (X1 = a)P (X2 = b) =
Φ1 (a)=r
Φ2 (b)=s P (X1 = a) =
Φ1 (a)=r P (X2 = b)
Φ2 (b)=s = P (Φ1 (X1 ) = r)P (Φ2 (X2 ) = s)
= P (Y1 = r)P (Y2 = s) ,
so Y1 and Y2 are independent.
49. P (both coins turn up using (a)) = 1 p2 + 1 p2 .
2 1
2 2
P (both coins turn up heads using (b)) = p1 p2 .
Since (p1 − p2 )2 = p2 − 2p1 p2 + p2 > 0, we see that p1 p2 < 1 p2 + 1 p2 , and so (a) is better.
1
2
2 1
2 2 51. ˜
˜
P (A) = P (AC)P (C) + P (AC)P (C)
˜
˜
≥ P (BC)P (C) + P (BC)P (C) = P (B) . 53. We assume that John and Mary sign up for two courses. Their cards are dropped, one of the cards
gets stepped on, and only one course can be read on this card. Call card I the card that was not
stepped on and on which the registrar can read government 35 and mathematics 23; call card II the
card that was stepped on and on which he can just read mathematics 23. There are four possibilities
for these two cards. They are:
Card I
Mary(gov,math)
Mary(gov,math)
John(gov,math)
John(gov,math) Card II
John(gov, math)
John(other,math)
Mary(gov,math)
Mary(other,math) Prob.
.0015
.0025
.0015
.0012 Cond. Prob.
.224
.373
.224
.179 In the third column we have written the probability that each case will occur. For example,
for the ﬁrst one we compute the probability that the students will take the appropriate courses:
.5 × .1 × .3 × .2 = .0030 and then we multiply by 1/2, the probability that it was John’s card that
was stepped on. Now to get the conditional probabilities we must renormalize these probabilities
so that they add up to one. In this way we obtain the results in the last column. From this we
see that the probability that card I is Mary’s is .597 and that card I is John’s is .403, so it is more
likely that that the card on which the registrar sees Mathematics 23 and Government 35 is Mary’s.
55.
P (R1 ) = 4
52
5 P (R2 ∩ R1 ) = 11 = 1.54 × 10−6 .
4·3
52 47
5
5 . Thus
P (R2  R1 ) = 3
47
5 = 1.96 × 10−6 . Since P (R2 R1 ) > P (R1 ), a royal ﬂush is attractive.
4 4
3 2
.
P (player 2 has a full house) =
52
5
P (player 1 has a ﬂush and player 2 has a full house) =
13 · 12 4·8·7 4
3 4
4
3
3
+4·8·5
·
+4·5·8·
2
3
2
3
52 47
5
5 4
3
+4·5·4
2
3 3
2 . Taking the ratio of these last two quantities gives:
P(player 1 has a royal ﬂush  player 2 has a full house) = 1.479 × 10−6 .
Since this probability is less than the probability that player 1 has a royal ﬂush (1.54 × 10−6 ), a
full house repels a royal ﬂush.
57. 59. P (BA) ≤ P (B) and P (BA) ≥ P (A)
⇔ P (B ∩ A) ≤ P (A)P (B) and P (B ∩ A) ≥ P (A)P (B)
⇔ P (A ∩ B) = P (A)P (B) .
Since A attracts B, P (BA) > P (A) and
P (B ∩ A) > P (A)P (B) ,
and so
P (A) − P (B ∩ A) < P (A) − P (A)P (B) .
Therefore,
˜
˜
P (B ∩ A) < P (A)P (B) ,
˜
˜
P (BA) < P (B) ,
˜
and A repels B. 61. Assume that A attracts B1 , but A does not repel any of the Bj ’s. Then
P (A ∩ B1 ) > P (A)P (B1 ),
and
P (A ∩ Bj ) ≥ P (A)P (Bj ),
Then 1 ≤ j ≤ n. P (A) = P (A ∩ Ω)
= P (A ∩ (B1 ∪ . . . ∪ Bn ))
= P (A ∩ B1 ) + · · · + P (A ∩ Bn )
> P (A)P (B1 ) + · · · + P (A)P (Bn )
= P (A) P (B1 ) + · · · + P (Bn )
= P (A) ,
12 which is a contradiction. SECTION 4.2
1. (a) 2/3
(b) 1/3
(c) 1/2
(d) 1/2 3. (a) .01
(b) e−.01 T where T is the time after 20 hours.
(c) e−.2 ≈ .819
(d) 1 − e−.01 ≈ .010 5. (a) 1
(b) 1
(c) 1/2
(d) π/8
(e) 1/2 7. P (X > 2
1
,Y > ) =
3
3 But P (X >
11. 1 1 dydx =
1
3 2
3 2
.
9 2 1
2
1
)P (Y > ) = · , so X and Y are independent.
3
3
3 3 If you have drawn n times (total number of balls in the urn is now n + 2) and gotten j black balls,
(total number of black balls is now j + 1), then the probability of getting a black ball next time is
(j + 1)/(n + 2). Thus at each time the conditional probability for the next outcome is the same in
the two models. This means that the models are determined by the same probability distribution,
so either model can be used in making predictions. Now in the coin model, it is clear that the
proportion of heads will tend to the unknown bias p in the long run. Since the value of p was
assumed to be unformly distributed, this limiting value has a random value between 0 and 1. Since
this is true in the coin model, it is also true in the Polya Urn model for the proportion of black
balls.(See Exercise 20 of Section 4.1.) SECTION 4.3
1. 2/3 3. (a) Consider a tree where the ﬁrst branching corresponds to the number of aces held by the player,
and the second branching corresponds to whether the player answers ‘ace of hearts’ or anything
else, when asked to name an ace in his hand. Then there are four branches, corresponding to the
numbers 1, 2, 3, and 4, and each of these except the ﬁrst splits into two branches. Thus, there are
seven paths in this tree, four of which correspond to the answer ‘ace of hearts.’ The conditional
13 probability that he has a second ace, given that he has answered ‘ace of hearts,’ is therefore
1 3
48
+
12
2 1 1 3
48
+
11
3 2 1 3
48
+
10
4 3 48
9 52
13 51
12 52
13 ≈ .6962 . (b) This answer is the same as the second answer in Exercise 2, namely .5612.
5. Let x = 2k . It is easy to check that if k ≥ 1, then
px/2
3
= .
px/2 + px
4
If x = 1, then px/2
=0.
px/2 + px Thus, you should switch if and only if your envelope contains 1. SECTION 5.1
1. (a), (c), (d) 3. Assume that X is uniformly distributed, and let the countable set of values be {ω1 , ω2 , . . .}. Let p
be the probability assigned to each outcome by the distribution function f of X. If p > 0, then
∞ ∞ f (ωi ) =
i=1 p,
i=1 and this last sum does not converge. If p = 0, then
∞ f (ωi ) = 0 .
i=1 So, in both cases, we arrive at a contradiction, since for a distribution function, we must have
∞ f (ωi ) = 1 .
i=1 5. (b) Ask the Registrar to sort by using the sixth, seventh, and ninth digits in the Social Security
numbers.
(c) Shuﬄe the cards 20 times and then take the top 100 cards. (Can you think of a method of
shuﬄing 3000 cards? 7. (a) pj (n) = 1 5
6 6 n−1 for j = 0, 1, 2, . . . . 125
5
.
(b) P (T > 3) = ( )3 =
6
216
5
125
.
(c) P (T > 6  T > 3) = ( )3 =
6
216
14 9. (a) 1000
(b) 100
10 N −100
90
N
100 (c) N = 999 or N = 1000
13. .7408, .2222, .0370 17. 649741 19. The probability of at least one call in a given day with n hands of bridge can be estimated by
−12
1
1 − e−n·(6.3×10 ) . To have an average of one per year we would want this to be equal to 365 . This
would require that n be about 400,000,000 and that the players play on the average 8,700 hands a
day. Very unlikely! It’s much more likely that someone is playing a practical joke. 21. (a) b(32, j, 1/80) = 1 j 79 32−j
32
j
80
80
(b) Use λ = 32/80 = 2/5. The approximate probability that a given student is called on j times
is e−2/5 (2/5)j /j! . Thus, the approximate probability that a given student is called on more than
twice is
(2/5)1
(2/5)2
(2/5)0
+
+
≈ .0079 .
1 − e−2/5
0!
1!
2! 23.
P (outcome is j + 1)/P(outcome is j) = mj+1 e−m
(j + 1)! m
mj e−m
=
.
j!
j+1 Thus when j + 1 ≤ m, the probability is increasing, and when j + 1 ≥ m it is decreasing. Therefore,
j = m is a maximum value. If m is an integer, then the ratio will be one for j = m − 1, and so
both j = m − 1 and j = m will be maximum values. (cf. Exercise 7 of Chapter 3, Section 2)
25. Without paying the meter Prosser pays
2· 52 e−5
5n e−5
5e−5
+ (5 · 2)
+ · · · (5 · n)
+ · · · = 25 − 15e−5 = $24.90.
1!
2!
n! He is better oﬀ putting a dime in the meter each time for a total cost of $10.
26.
number
0
1
2
3
4
5
27. observed
229
211
93
35
7
1 expected
227
211
99
31
9
1 m = 100 × (.001) = .1. Thus P (at least one accident) = 1 − e−.1 = .0952. Here m = 500 × (1/500) = 1, and so P (at least one fake) = 1 − e−1 = .632. If the king tests two
2
= 1, and so the answer is again .632.
coins from each of 250 boxes, then m =250 ×
500
31. The expected number of deaths per corps per year is 29. 1· 32
11
2
91
+2·
+3·
+4·
= .70.
280
280
280
280 The expected number of corps with x deaths would then be 280 ·
the following comparison:
15 (.70)x e−(.70)
. From this we obtain
x! Number of deaths Corps with x deaths 0
1
2
3
≥4 Expected number of Corps 144
91
32
11
2 139.0
97.3
34.1
7.9
1.6 The ﬁt is quite good.
33. Poisson with mean 3. 35. (a) In order to have d defective items in s items, you must choose d items out of D defective ones
and the rest from S − D good ones. The total number of sample points is the number of ways
to choose s out of S. (b) Since
min(D,s)
P (X = j) = 1,
j=0 we get
min(D,s)
j=0 D
j s−D
s−j = S
s . 37. The maximum likelihood principle gives an estimate of 1250 moose. 43. If the traits were independent, then the probability that we would obtain a data set that diﬀers
from the expected data set by as much as the actual data set diﬀers is approximately .00151. Thus,
we should reject the hypothesis that the two traits are independent. SECTION 5.2
1. 2.
5. 7.
9. (a) f (x) = 1 on [2, 3]; F (x) = x − 2 on [2, 3].
1
(b) f (x) = x−2/3 on [0, 1]; F (x) = x1/3 on [0, 1].
3
1
1
(a) F (x) = 2 − x , f (x) = x2 on [ 1 , 1].
2
(b) F (x) = ex − 1, f (x) = ex on
(a) F (x) = 2x,
f (x) = 2 on
√
1
f (x) = √x on
(b) F (x) = 2 x, [0, log 2].
[0, 1 ].
2
[0, 1 ].
4 Using Corollary 5.2, we see that the expression
(a) F (y) = y2
2 , 0 ≤ y ≤ 1;
(2−y)2
,
2 1− (b) F (y) = 2y − y ,
2 13.
(a) F (r) = F (t) = f (y) = 2 − 2y, √
r, (b) F (s) = 1 −
(c) 1 ≤ y ≤ 2, √ 1 − 4s , t
,
1+t f (y) = √
rnd will simulate the given random variable.
y,
2−y 0 ≤ y ≤ 1;
1 ≤ y ≤ 2. 0 ≤ y ≤ 1. 1
f (r) = √ ,
on
2 r
2
,on
f (s) = √
1 − 4s
1
f (t) =
2 , on
(1 + t)
16 [0,1].
[0, 1 ].
4
[0, ∞]. 15. F (d) = 1 − (1 − 2d)2 , 17. (a) f (x) = π
2 sin(πx), 0, f (d) = 4(1 − 2d) on [0, 1 ].
2 0 ≤ x ≤ 1;
otherwise. sin2 ( π ) = .146.
8 (b) 1
w−b
a fX ( a ), a = 0: fW (w) = 0 if w = 0. 19. a = 0 : fW (w) = 21. P (Y ≤ y) = P (F (X) ≤ y) = P (X ≤ F −1 (y)) = F (F −1 (y)) = y 23. The mean of the uniform density is (a + b)/2. The mean of the normal density is µ. The mean of
the exponential density is 1/λ. 25. (a) .9773, (b) .159, 27. A: 15.9%, B: 34.13%, 29. e−2 , e−2 . 31. 1
2. 35. P (size increases) = P (Xj < Yj ) = λ/(λ + µ). (c) .0228,
C: (d) 34.13%, on [0, 1]. .6827.
D: 13.59%, F: 2.28%. P (size decreases) = 1 − P (size increases) = µ/(λ + µ).
37. FY (y) = √ log2 (y)
1
e− 2 , for y > 0.
2πy SECTION 6.1
1. 1/9 3. 5 10.1” 5. 1/19 7. Since X and Y each take on only two values, we may choose a, b, c, d so that
U= Y +c
X +a
,V =
b
d take only values 0 and 1. If E(XY ) = E(X)E(Y ) then E(U V ) = E(U )E(V ). If U and V are
independent, so are X and Y . Thus it is suﬃcient to prove independence for U and V taking on
values 0 and 1 with E(U V ) = E(U )E(V ).Now
E(U V ) = P (U = 1, V = 1) = E(U )E(V ) = P (U = 1)P (V = 1),
and P (U = 1, V = 0) = P (U = 1) − P (U = 1, V = 1)
= P (U = 1)(1 − P (V = 1)) = P (U = 1)P (V = 0). Similarly,
P (U = 0, V = 1) = P (U = 0)P (V = 1)
P (U = 0, V = 0) = P (U = 0)P (V = 0).
Thus U and V are independent, and hence X and Y are also.
9. The second bet is a fair bet so has expected winning 0. Thus your expected winning for the two
bets is the same as the original bet which was 7/498 = .0141414... On the other hand, you bet
1 dollar with probability 1/3 and 2 dollars with probability 2/3. Thus the expected amount you
17 bet is 1 2 dollars and your expected winning per dollar bet is .0141414/1.666667 = .0085 which
3
makes this option a better bet in terms of the amount won per dollar bet. However, the amount
of time to make the second bet is negligible, so in terms of the expected winning per time to make
one play the answer would still be .0141414.
11. The roller has expected winning .0141; the pass bettor has expected winning .0136. 13. 45 15. E(X) = 1 , so this is a favorable game.
5
k−1 times 17. pk = p( S · · · S F ) = pk−1 (1 − p) = pk−1 q, k = 1, 2, 3, . . . .
∞ ∞ pk = q pk = q
k=1 k=0 1
=1.
1−p ∞ kpk−1 = E(X) = q
k=1 q
1
= . (See Example 6.4.)
2
(1 − p)
q 19.
E(X) = 4
4
4
4 (3 − 3) +
+ 23. 3
2
4
3
3
2
4
2 (3 − 2) +
(0 − 2) + 3
3
4
3
3
0
4
1 (0 − 3) +
(3 − 0) + 3
1
4
2
3
1
4
1 (3 − 1)
.
(0 − 1) = 0 10 25.
(b) Let S be the number of stars and C the number of circles left in the deck. Guess star if S > C
and guess circle if S < C. If S = C toss a coin.
(d) Consider the recursion relation:
h(S, C) = max(S, C)
S
C
+
h(S − 1, C) +
h(S, C − 1)
S+C
S+C
S+C and h(0, 0) = h(−1, 0) = h(0, −1) = 0. In this equation the ﬁrst term represents your expected
winning on the current guess and the next two terms represent your expected total winning
on the remaining guesses. The value of h(10, 10) is 12.34.
27. (a) 4
4 (b) 4 +
x=1 29. 4
x 4
x
8
x = 5.79 . If you have no tencards and the dealer has an ace, then in the remaining 49 cards there are 16 ten
cards. Thus the expected payoﬀ of your insurance bet is:
2· 16
33
1
−1·
=−
.
49
49
49 If you are playing two hands and do not have any tencards then there are 16 tencards in the
remaining 47 cards and your expected payoﬀ on an insurance bet is:
2· 16
31
1
−1·
=
.
47
47
47 Thus in the ﬁrst case the insurance bet is unfavorable and in the second it is favorable.
18 31. (a) 1 − (1 − p)k .
N
· (k + 1)(1 − (1 − p)k ) + (1 − p)k .
k
(c) If p is small, then (1 − p)k ∼ 1 − kp, so the expected number in (b) is
√
1
∼ N [kp + k ], which will be minimized when k = 1/ p.
(b) 33. We begin by noting that
P (X ≥ j + 1) = P ((t1 + t2 + · · · + tj ) ≤ n) .
Now consider the j numbers a1 , a2 , · · · , aj deﬁned by
a1 = t1
a2 = t1 + t2
a3 = t1 + t2 + t3
.
.
.
.
.
.
.
.
.
aj = t1 + t2 + · · · + tj .
The sequence a1 , a2 , · · · , aj is a monotone increasing sequence with distinct values and with successive diﬀerences between 1 and n. There is a onetoone correspondence between the set of all such
sequences and the set of possible sequences t1 , t2 , · · · , tj . Each such possible sequence occurs with
probability 1/nj . In fact, there are n possible values for t1 and hence for a1 . For each of these there
are n possible values for a2 corresponding to the n possible values of t2 . Continuing in this way we
see that there are nj possible values for the sequence a1 , a2 , · · · , aj . On the other hand, in order to
have t1 + t2 + · · · + tj ≤ n the values of a1 , a2 , · · · , aj must be distinct numbers lying between 1 to
n and arranged in order. The number of ways that we can do this is n . Thus we have
j
n 1
.
j nj P (t1 + t2 + · · · + tj ≤ n) = P (X ≥ j + 1) = E(X) = P (X = 1) + P (X = 2) + P (X = 3) · · ·
+ P (X = 2) + P (X = 3) · · · .
+ P (X = 3) · · · .
If we sum this by rows we see that
n−1 P (X ≥ j + 1) . E(X) =
j=0 Thus, n E(X) =
j=1 n
j 1
n j = 1+ 1
n n . The limit of this last expression as n → ∞ is e = 2.718... .
There is an interesting connection between this problem and the exponential density discussed in
Section 2.2 (Example 2.17). Assume that the experiment starts at time 1 and the time between
occurrences is equally likely to be any value between 1 and n. You start observing at time n. Let
T be the length of time that you wait. This is the amount by which t1 + t2 + · · · + tj is greater than
n. Now imagine a sequence of plays of a game in which you pay n/2 dollars for each play and for
the j’th play you receive the reward tj . You play until the ﬁrst time your total reward is greater
than n. Then X is the number of times you play and your total reward is n + T . This is a perfectly
19 fair game and your expected net winning should be 0. But the expected total reward is n + E(T ).
Your expected payment for play is n E(X). Thus by fairness, we have
2
n + E(T ) = (n/2)E(X) .
Therefore, n
E(X) − n .
2
We have seen that for large n, E(X) ∼ e. Thus for large n,
E(T ) = e
E(waiting time) = E(T ) ∼ n( − 1) = .718n .
2
Since the average time between occurrences is n/2 we have another example of the paradox where
we have to wait on the average longer than 1/2 the average time time between occurrences.
35. One can make a conditionally convergent series like the alternating harmonic series sum to anything
one pleases by properly rearranging the series. For example, for the order given we have
∞ (−1)n+1 2n 1
·
n 2n (−1)n+1 E= 1
= log 2 .
n n=0
∞ =
n=0 But we can rearrange the terms to add up to a negative value by choosing negative terms until they
add up to more than the ﬁrst positive term, then choosing this positive term, then more negative
terms until they add up to more than the second positive term, then choosing this positive term,
etc.
37. (a) Under option (a), if red turns up, you win 1 franc, if black turns up, you lose 1 franc, and if 0
turns up, you lose 1/2 franc. Thus, the expected winnings are
1 18
−1
18
+ (−1)
+
37
37
2 1
≈ −.0135 .
37 (b) Under option (b), if red turns up, you win 1 franc, if black turns up, you lose 1 franc, and if 0
comes up, followed by black or 0, you lose 1 franc. Thus, the expected winnings are
1 18
1
18
+ (−1)
+ (−1)
37
37
37 19
≈ −.0139 .
37 (c)
39. (Solution by Peter Montgomery) The probability that book 1 is in the right place is the probability
that the last phone call referenced book 1, namely p1 . The probability that book 2 is in the right
place, given that book 1 is in the right place, is
p2 + p2 p1 + p2 p2 + . . . =
1 p2
.
(1 − p1 ) Continuing, we ﬁnd that
P = p1 p3
pn
p2
···
.
(1 − p1 ) (1 − p1 − p2 )
(1 − p1 − p2 − . . . − pn−1 Now let q be a real number between 0 and 1, let
p1 = 1 − q ,
20 p2 = q − q 2 ,
and so on, and ﬁnally let
pn = q n−1 .
Then
P = (1 − q)n−1 ,
so P can be made arbitrarily close to 1. SECTION 6.2
1.
3.
5. 7.
9.
11. 2
2
, σ = D(X) =
.
3
3
−1
−1
, E(Y ) =
, V (X) = 33.21, V (Y ) = .99 .
E(X) =
19
19
(a) E(F ) = 62, V (F ) = 1.2 .
(b) E(T ) = 0, V (T ) = 1.2 .
10
50
, V (C) =
.
(c) E(C) =
3
27
√
3
3
.
V (X) = , D(X) =
4
2
√
2 5
3
.
V (X) = , D(X) =
4
3
E(X) = (1 + 2 + · · · + n)/n = (n + 1)/2. E(X) = 0, V (X) = V (X) = (12 + 22 + · · · + n2 )/n − (E(X))2
= (n + 1)(2n + 1)/6 − (n + 1)2 /4 = (n + 1)(n − 1)/12.
13. Let X1 , . . . , Xn be identically distributed random variables such that
P (Xi = 1) = P (Xi = −1) = 1
.
2 n Then E(Xi ) = 0, and V (Xi ) = 1. Thus Wn = j=1 Xi . Therefore
n
n
E(Wn ) = i=1 E(Xi ) = 0, and V (Wn ) = i=1 V (Xi ) = n.
15. (a) PXi =
(b) PXi Xj = 0 1 n−1
n 1
n . Therefore, E(Xi )2 = 1/n for i = j. 0
1− 1
n(n−1) Therefore, E(Xi Xj ) = 1 for i = j . 1
n(n−1) 1
.
n(n − 1) (c)
E(Sn )2 = E(Xi )2 +
i =n· E(Xi Xj )
i j=i 1
1
+ n(n − 1) ·
=2.
n
n(n − 1) (d)
V (Sn ) = E(Sn )2 − E(Sn )2
= 2 − (n · (1/n))2 = 1 .
21 16. (a) For p = .5: 10
N 30
50 1
.656
.638
.678 k
2
.979
.957
.967 3
.998
.999
.997 1
.772
.749
.629 k
2
.967
.964
.951 3
.994
.997
.997 For p = .2: 10
N 30
50 (b) Use Exercise 12 and the fact that E(Sn ) = np and V (Sn ) = npq. The two examples in (a)
suggests that the probability that the outcome is within k standard deviations is approximately
the same for diﬀerent values of p. We shall see in Chapter 9 that the Central Limit Theorem
explains why this is true.
19. Let X1 , X2 be independent random variables with
−1 pX1 = pX2 =
Then −2 pX1 +X2 = 1 0 2 1
2 1
4 1
2 1
4 . 1
2 . Then
¯
¯
σX1 = σX2 = 1, σX1 +X2 = 1 .
¯
Therefore
V (X1 + X2 ) = 1 = V (X1 ) + V (X2 ) = 2 ,
and
σX1 +X2 = 1 = σX1 + σX2 = 2 .
¯
¯
¯
21. f (x) = − 2(X(ω) − x)p(ω)
ω = −2 X(ω)p(ω) + 2x
ω p(ω)
ω = −2µ + 2x .
Thus x = µ is a critical point. Since f (x) ≡ 2, we see that x = µ is the minimum point.
23. If X and Y are independent, then
Cov(X, Y ) = E(X − E(X)) · E(Y − E(Y )) = 0 .
Let U have distribution
pU = π/2 π 3π/2
1/4 1/4 1/4 0
1/4 . Then let X = cos(U ) and Y = sin(U ). X and Y have distributions
pX = 1
1/4
22 0
1/4 −1
1/4 0
1/4 , 0
1/4 pY = 1
1/4 0
1/4 −1
1/4 . Thus E(X) = E(Y ) = 0 and E(XY ) = 0, so Cov(X, Y ) = 0. However, since
sin2 (x) + cos2 (x) = 1, X and Y are dependent.
25. (a) The expected value of X is
5000 iP (X = i) . µ = E(X) =
i=1 The probability that a white ball is drawn is
n P (white ball is drawn) = P (X = i)
i=1 Thus
P (white ball is drawn) = i
.
5000 µ
.
5000 (b) To have P (white,white) = P (white)2 we must have
n 5000 (
i=1 i 2
i
) P (X = i) = (
P (X = i))2 .
5000
5000
i=1 But this would mean that E(X 2 ) = E(X)2 , or V (X) = 0. Thus we will have independence only if
X takes on a speciﬁc value with probability 1.
(c) From (b) we see that
P (white,white) =
Thus
V (X) =
27. 1
E(X 2 ) .
50002 (σ 2 + µ2 )
.
50002 The number of boxes needed to get the j’th picture has a geometric distribution with
p= (2n − k + 1)
.
2n Thus
V (Xj ) = 2n(k − 1)
.
(2n − k + 1)2 Therefore, for a team of 26 players the variance for the number of boxes needed to get the ﬁrst half
of the pictures would be
13
26(k − 1)
= 7.01 ,
(26 − k + 1)2
k=1 and to get the second half would be
26 k=14 26(k − 1)
= 979.23 .
(26 − k + 1)2 Note that the variance for the second half is much larger than that for the ﬁrst half.
23 SECTION 6.3
µ = 0,
µ = 0,
µ = 0,
µ = 0, σ 2 = 1/3
σ 2 = 1/2
σ 2 = 3/5
σ 2 = 3/5 1. (a)
(b)
(c)
(d) 3. µ = 40, σ 2 = 800 5. (d) a = −3/2, b = 0, c = 1
3
45
, b = 0, c =
(e) a =
48
16 7. f (a) = E(X − a)2 = (x − a)2 f (x)dx . Thus
f (a) = − 2(x − a)f (x)dx = −2 xf (x)dx + 2a f (x)dx = −2µ(X) + 2a .
Since f (a) = 2, f (a) achieves its minimum when a = µ(X).
9. (a) 3µ, 3σ 2
(b) E(A) = µ, V (A) = σ2
3 σ2
+ µ2
3
In the case that X is uniformly distributed on [0, 100], one ﬁnds that
1
b2 + (100 − b)2 ,
E(X − b) =
200
which is minimized when b = 50.
When fX (x) = 2x/10,000, one ﬁnds that
(c) E(S 2 ) = 3σ 2 + 9µ2 , E(A2 ) = 11. 200
b3
E(X − b) =
−b+
,
3
15000
√
which is minimized when b = 50 2.
13. Integrating by parts, we have
∞ E(X) = xdF (x)
0 = −x(1 − F (x))
∞ = ∞
0 ∞ + (1 − F (x))dx 0 (1 − F (x))dx . 0 To justify this argment we have to show that a(1 − F (a)) approaches 0 as a tends to inﬁnity. To
see this, we note that
∞ ∞ a xf (x)dx =
0 xf (x)dx + xf (x)dx
a 0
a ≥ a xf (x)dx +
0 af (x)dx
0 a xf (x)dx + a(1 − F (a)) . =
0 24 Letting a tend to inﬁnity, we have that
E(X) ≥ E(X) + lim a(1 − F (a)) .
a→∞ Since both terms are nonnegative, the only way this can happen is for the inequality to be an
equality and the limit to be 0.
To illustrate this with the exponential density, we have
∞ ∞ (1 − F (x))dx = 0 e−λx dx = 0 1
= E(X) .
λ 15. E(Y ) = 9.5, E(Z) = 10, E(X − Y ) = 1/2, E(X − Z) = 1/2 .
Z is better, since it has the same expected value as X and the variance is only slightly larger. 17. (a)
Cov(X, Y ) = E(XY ) − µ(X)E(Y ) − E(X)µ(Y ) + µ(X)µ(Y )
= E(XY ) − µ(X)µ(Y ) = E(XY ) − E(X)E(Y ) .
(b) If X and Y are independent, then E(XY ) = E(X)E(Y ), and so Cov(X, Y ) = 0.
(c)
V (X + Y ) = E(X + Y )2 − (E(X + Y ))2
= E(X 2 ) + 2E(XY ) + E(Y 2 )
− E(X)2 − 2E(X)E(Y ) − E(Y )2
= V (X) + V (Y ) + 2Cov(X, Y ) . 19. (a) 0
1
(b) √
2
1
(c) − √
2
(d) 0 21. We have
fXY (x,y)
=
fY (y) 2π −2ρxy+y
· exp −(x 2(1−ρ2 )
√
2
2π · exp(− y2 )
2 √1 1 = 2 ) 1−ρ2 2π(1 − ρ2 ) · exp −(x − ρy)2
2(1 − ρ2 ) which is a normal density with mean ρy and variance 1 − ρ2 . Thus,
∞ E(XY = y) = x
−∞
∞ = ρy 1
2π(1 −
1 2π(1 − −∞ =
27. ρ2 ) · exp ρ2 ) −(x − ρy)2
dx
2(1 − ρ2 ) · exp(−(x − ρy)2 ) ρy < y, if 0 < ρ < 1;
y,
if ρ = 1. Let Z represent the payment. Then
P (Z = kX = x) = P (Y1 ≤ x, Y2 ≤ x, . . . , Yk ≤ x, Yk+1 > x)
= xk (1 − x) .
25 Therefore,
1 xk (1 − x) dx P (Z = k) =
0 1
1
xk+1 −
xk+2
k+1
k+2
1
1
−
=
k+1 k+2
1
=
.
(k + 1)(k + 2) 1
= Thus, ∞ 1
(k + 1)(k + 2) k E(Z) =
k=0 0 , which diverges. Thus, you should be willing to pay any amount to play this game. SECTION 7.1
1. (a) .625
(b) .5 3.
5. 0 1 2 3 4 1
64 3
32 17
64 3
8 1
4 3 4 5 6 1
12 4
12 4
12 3
12 1 2 3 4 1
12 4
12 4
12 3
12 (a)
(b) 7. (a) P (Y3 ≤ j) = P (X1 ≤ j, X2 ≤ j, X3 ≤ j) = P (X1 ≤ j)3 .
Thus
1
2
3
4
5
pY3 =
7
19
37
61
1
216 216 216 216 6
91
216 216 . This distribution is not bellshaped.
(b) In general,
P (Yn ≤ j) = P (X1 ≤ j)3 =
Therefore,
P (Yn = j) = j
n n − j
n j−1
n n . n . This distribution is not bellshaped for large n.
9. Let p1 , . . . , p6 be the probabilities for one die and q1 , . . . , q6 be the probabilities for the other die.
Assume ﬁrst that all probabilities are positive. Then p1 q1 > p1 q6 , since there is only one way to
get a 2 and several ways to get a 7. Thus q1 > q6 . In the same way q6 q6 > q1 p6 and so q6 > q1 .
This is a contradiction. If any of the sides has probability 0, then we can renumber them so that
it is side 1. But then the probability of a 2 is 0 and so all sums would have to have probability 0,
which is impossible.
26 Here’s a fancy way to prove it. Deﬁne the polynomials
5 p(k+1) xk p(x) =
k=0 and 5 q(k+1) xk . q(x) =
k=0 Then we must have 10 p(x)q(x) =
k=0 (1 − x11 )
xk
=
.
11
(1 − x) The left side is the product of two ﬁfth degree polynomials. A ﬁfth degree polynomial must have
a real root which will not be 0 if p1 > 0. Consider the right side as a polynomial. For x to be a
nonzero root of this polynomial it would have to be a real eleventh root of unity other than 1, and
there are no such roots. Hence again we have a contradiction. SECTION 7.2
3. (a)
fZ (x) = x3 /24,
x − x3 /24 − 4/3, if 0 ≤ x ≤ 2;
if 2 ≤ x ≤ 4. (b)
fZ (x) = (x3 − 18x2 + 108x − 216)/24, if 6 ≤ x ≤ 8;
(−x3 + 18x2 − 84x + 40)/24, if 8 ≤ x ≤ 10. (c)
fZ (x) =
5. x2 /8,
if 0 ≤ x ≤ 2;
1/2 − (x − 2)2 /8, if 2 ≤ x ≤ 4. (a)
λµ λx
µ+λ e ,
λµ −µx
,
µ+λ e fZ (x) = x < 0;
x ≥ 0. (b)
fZ (x) =
7. 1 − e−λx ,
0 < x < 1;
(eλ − 1)e−λx , x ≥ 1. We ﬁrst ﬁnd the density for X 2 when X has a general normal density
fX (x) = 2
2
1
√ e−(x−µ) /2σ dx .
σ 2π Then (see Theorem 1 of Chapter 5, Section 5.2 and the discussion following) we have
2
fX (x) = √
√
1
1
√
√ exp(−x/2σ 2 − µ2 /2σ 2 ) exp( xµ/σ 2 ) + exp(− xµ/σ 2 ) .
σ 2π 2 x
27 Replacing the last two exponentials by their series representation, we have
2 fX (x) = e −µ/2σ 2 ∞ r µ
2σ 2 r=0 where 1
G(1/2σ 2 , r + 1/2, x) ,
r! ap −ax p−1
e
x
Γ(p) G(a, p, x) = is the gamma density. We now consider the original problem with X1 and X2 two random variables
with normal density with parameters µ1 , σ1 and µ2 , σ2 . This is too much generality for us, and we
shall assume that the variances are equal, and then for simplicity we shall assume they are 1. Let
µ2 + µ2 .
1
2 c=
We introduce the new random variables
Z1 = 1
(µ1 X1 + µ2 X2 ) ,
c 1
(µ2 X1 − µ1 X2 ) .
c
Then Z1 is normal with mean c and variance 1 and Z2 is normal with mean 0 and variance 1. Thus,
Z2 = 2
fZ1 = e−c 2 ∞ c2
2 /2
r=0 r 1
G(1/2, r + 1/2, x) ,
r! and
2
fZ2 = G(1/2, 1/2, x) . Convoluting these two densities and using the fact that the convolution of a gamma density G(a, p, x)
and G(a, q, x) is a gamma density G(a, p + q, x) we ﬁnally obtain
2
2
2
2
fZ1 +Z2 = fX1 +X2 = e−c 2 ∞
/2
r=0 c2
2 r 1
G 1/2, r + 1, x .
r! (This derivation is adapted from that of C.R. Rao in his book Advanced Statistical Methods in
Biometric Research, Wiley, l952.)
9. P (X10 > 22) = .341 by numerical integration. This also could be estimated by simulation. 11. 10 hours 13. Y1 = −log(X1 ) has an exponential density fY1 (x) = e−x . Thus Sn has the gamma density
fSn (x) =
Therefore
fZn (x) = 19. 1
1
log
(n − 1)!
x n−1 The support of X + Y is [a + c, b + d]. 21. xn−1 e−x
.
(n − 1)! (a)
fA (x) = √
28 2
1
e−x /(2n) .
2πn . (b) fA (x) = nn xn e−nx /(n − 1)! . SECTION 8.1
1.
3. 1/9
We shall see that Sn − n/2 tends to inﬁnity as n tends to inﬁnity. While the diﬀerence will be small
compared to n/2, it will not tend to 0. On the other hand the diﬀerence Sn /n − 1/2 does tend to
0. 5. k = 10 7.
1
1
−
− p + p2
4
4
1
1
1
= − ( − p)2 ≤ .
4
2
4 p(1 − p) = Thus, max p(1 − p) =
0≤p≤1 1
. From Exercise 6 we have that
4
P 9.  Sn
− p ≥
n ≤ 1
p(1 − p)
≤
.
n 2
4n 2 P (Sn ≥ 11) = P (Sn − E(Sn ) ≥ 11 − E(Sn ))
= P (Sn − E(Sn ) ≥ 10)
V (Sn )
≤
= .01.
102 11. No, we cannot predict the proportion of heads that should turn up in the long run, since this will
depend upon which of the two coins we pick. If you have observed a large number of trials then, by
the Law of Large Numbers, the proportion of heads should be near the probability for the coin that
you chose. Thus, in the long run, you will be able to tell which coin you have from the proportion
of heads in your observations. To be 95 percent sure, if the proportion of heads is less than .625,
predict p = 1/2; if it is greater than .625, predict p = 3/4. Then you will get the correct coin if the
proportion of heads does not deviate from the probability of heads by more than .125. By Exercise
7, the probability of a deviation of this much is less than or equal to 1/(4n(.125)2 ). This will be
less than or equal to .05 if n > 320. Thus with 321 tosses we can be 95 percent sure which coin we
have. 15. Take as Ω the set of all sequences of 0’s and 1’s, with 1’s indicating heads and 0’s indicating
tails. We cannot determine a probability distribution by simply assigning equal weights to all
inﬁnite sequences, since these weights would have to be 0. Instead, we assign probabilities to ﬁnite
sequences in the usual way, and then probabilities of events that depend on inﬁnite sequences can
be obtained as limits of these ﬁnite sequences. (See Exercise 28 of Chapter 1, Section 1.2.) 17. For x ∈ [0, 1], let us toss a biased coin that comes up heads with probability x. Then
E f (Sn )
→ f (x).
n
29 But
E f (Sn )
=
n n f
k=0 k
n n k
x (1 − x)n−k .
k The right side is a polynomial, and the left side tends to f(x). Hence
n f
k=0 k
n n k
x (1 − x)n−k → f (x).
k This shows that we can obtain obtain any continuous function f (x) on [0,1] as a limit of polynomial
functions. SECTION 8.2
1. (a) 1
(b) 1
(c) 100/243
(d) 1/12 3.
f (x) = 1 − x/10,
0 if 0 ≤ x ≤ 10;
otherwise.
100
g(x) = 2 .
3x 5. (a) 1, 1/4, 1/9
(b) 1 vs. .3173, .25 vs. .0455, .11 vs. .0027 7. (b) 1, 1, 100/243, 1/12 9. (a) 0
(b) 7/12
(c) 11/12 11. (a) 0
(b) 7/12 13. (a) 2/3
(b) 2/3
(c) 2/3 17. E(X) = ∞
−∞ xp(x)dx. Since X is nonnegative, we have
xp(x)dx ≥ aP (X ≥ a) . E(X) ≥
x≥a SECTION 9.1(The answers to the problems in this chapter do not use the ‘1/2 correction mentioned
in Section 9.1.
30 1. (a) .158655
(b) .6318
(c) .0035
(d) .9032 3. (a) P (June passes) ≈ .985
(b) P (April passes) ≈ .056 5. Since his batting average was .267, he must have had 80 hits. The probability that one would
obtain 80 or fewer successes in 300 Bernoulli trials, with individual probability of success .3, is
approximately .115. Thus, the low average is probably not due to bad luck (but a statistician
would not reject the hypothesis that the player has a probability of success equal to .3). 7. .322 9. (a) 0
(b) 1 (Law of Large Numbers)
(c) .977 (Central Limit Theorem)
(d) 1 (Law of Large Numbers)
∗
S1900 ≥ √ 115 − 95
1900 · .05 · .95 ∗
= P (S1900 ≥ 2.105) = .0176. 13. P (S1900 ≥ 115) = P 17. √
2 pq
1
√
We want √
= .01. Replacing pq by its upper bound 1 , we have √ = .01. Thus we would
2
n
n
need n = 10,000. Recall that by Chebyshev’s inequality we would need 50,000. SECTION 9.2
1. (a) .4762
(b) .0477 3. (a) .5
(b) .9987 5. (a) P (S210 < 700) ≈ .0757.
(b) P (S189 ≥ 700) ≈ .0528
(c) P (S179 < 700, S210 ≥ 700) = P (S179 < 700) − P (S179 < 700, S210 < 700)
= P (S179 < 700) − P (S210 < 700)
≈ .9993 − .0757 = .9236 . 7. (a) Expected value = 200, variance = 2
(b) .9973 11. n
Sn − nµ
√
.
≥√
2
nσ
nσ 2
By the Central√
Limit Theorem, this probability is approximated by the area under the normal
n
and inﬁnity, and this area approaches 0 as n tends to inﬁnity.
curve between
σ
Her expected loss is 60 dollars. The probability that she lost no money is about .0013. 13. p = .0056 9. P Sn
−µ ≥
n =P Sn − nµ ≥ n =P 31 SECTION 9.3
1. 1
1
(E(X) − µ) = (µ − µ) = 0 ,
σ
σ
X −µ 2
1 2
2
∗
= 2σ = 1 .
σ (X ) = E
σ
σ
E(X ∗ ) = 3. 11. Sn − nµ
. Since each Yj has mean 0 and variance 1, E(Tn ) = 0 and
Tn = Y1 + Y2 + · · · + Yn =
σ
Tn
Sn − nµ
∗
∗
√
= Sn .
V (Tn ) = n. Thus Tn = √ =
n
σ n
(a) .5
(b) .148
(c) .018 13. .0013 SECTION 10.1
1. In each case, to get g(t) just replace z by et in h(z).
1
(a) h(z) = (1 + z)
2
6 zj (b) h(z) =
j=1 (c) h(z) = z 3
k (d) h(z) = 3. 1
zn
zj
k + 1 j=1 (e) h(z) = z n (pz + q)k
2
(f) h(z) =
3−z
1
1 1
(a) h(z) = + z + z 2 .
4 2
4
1 1 t 1 2t
(b) g(t) = h(et ) = + e + e .
4 2
4
(c) g(t) = 1 1
+
4 2
∞ =1+
k=1 ∞
k=0 k−2 2
1
+
2k!
k! Thus µ0 = 1, and µk =
(d) p0 = 1
,
4 p1 = 1
,
2 ∞ 1
tk
+
k!
4 1
2 p2 = k=0 2k k
t
k!
∞ tk = 1 +
k=1
k−2 +2 µk k
t .
k! for k ≥ 1. 1
.
4 5. (a) µ1 (p) = µ1 (p ) = 3, µ2 (p) = µ2 (p ) = 11
µ3 (p) = 43, µ3 (p ) = 47
µ4 (p) = 171, µ4 (p ) = 219 7. (a) g−X (t) = g(−t)
32 (b) gX+1 (t) = et g(t)
(c) g3X (t) = g(3t)
(d) gaX+b = ebt g(at)
6 9. 6 aj z j , hX (z) = (a) j=1 j=1
6 6 aj z j hZ (z) = (b) j=1 (c) 6 6 bj z j−1 = j=1
6 j=1
6 aj z j−1 or
j=1 . j=1 aj z j−1 Either bj z j hZ (z) = (z 2 + · · · + z 12 )/11 . Assume that Then bj z j . hY (z) = z 11 − 1
1 + z + · · · z 10
=
.
11
11(z − 1) bj z j−1 is a polynomial of degree 5 (i.e., either a6 = 0 or b6 = 0). Suppose
j=1 6 aj z j−1 is a polynomial of degree 5. Then it must have a real root, which is a real root of that
j=1 (z 11 − 1)/(z − 1). However (z 11 − 1)/(z − 1) has no real roots. This is because the only real root
of z 11 − 1 is 1, which cannot be a real root of (z 11 − 1)/(z − 1). Thus, we have a contradiction.
This means that you cannot load two dice in such a way that the probabilities for any sum from 2
to 12 are the same. (cf. Exercise 11 of Section 7.1).
11. Let pn = probability that the gambler is ruined at play n.
Then
if n is even,
pn = 0,
p1 = q,
pn = p(p1 pn−2 + p3 pn−4 + · · · + pn−2 p1 ),
Thus if n > 1 is odd. 2 h(z) = qz + pz h(x)
so 1− h(z) = 1 − 4pqz 2
.
2pz By Exercise 10 we have
q/p,
1, h(1) = , if q ≤ p,
if q ≥ p, 1/(q − p), if q > p,
∞,
if q = p. h (1) = This says that when q > p, the gambler must be ruined, and the expected number of plays before
ruin is 1/(q − p). When p > q, the gambler has a probability q/p of being ruined. When p = q, the
gambler must be ruined eventually, but the expected number of plays until ruin is not ﬁnite.
13. (a) From the hint:
hk (z) = hU 1 (z) · · · hU k (z) = h(z) k . (b)
k hk (1) = h(1)
33 = (q/p)k
1 if q ≤ p,
if q ≥ p. h (1) = k/(q − p) if q > p,
∞
if q = p. Thus the gambler must be ruined if q ≥ p. The expected number of plays in this case is k/(q − p)
if q > p and ∞ if q = p. When q < p he is ruined with probability (q/p)k . SECTION 10.2
1. (a) d = 1
(b) d = 1
(c) d = 1
(d) d = 1
(e) d = 1/2
(f) d ≈ .203 3. (a) 0
(b) 276.26 5. Let Z be the number of oﬀspring of a single parent. Then the number of oﬀspring after two
generations is
SN = X1 + · · · + XN ,
where N = Z and Xi are independent with generating function f . Thus by Exercise 4, the
generating function after two generations is h(z) = f (f (z)). 7. Let N be the time she needs to be served. Then the number of customers arriving during this time
is X1 + · · · + XN , where Xi are identically distributed independent of N . P (X0 = 0) = p, P (Xi =
1) = q. Thus by Exercise 4, h(z) = g(f (z)). 9. If there are k oﬀspring in the ﬁrst generation, then the expected total number of oﬀspring will be
kN , where N is the expected total numer for a single oﬀspring. Thus we can compute the expected
total number by counting the ﬁrst oﬀspring and then the expected number after the ﬁrst generation.
This gives the formula
N =1+
kpk = 1 + mN .
k ¿From this it follows that N is ﬁnite if and only if m < 1, in which case
N = 1/(1 − m). SECTION 10.3
1
1.
(a) g(t) = (e2t − 1)
2t
e2t (2t − 1) + 1
(b) g(t) =
2t2
2t
e − 2t − 1
(c) g(t) =
2t2
2t
e (ty − 1) + 2et − t − 1
(d) g(t) =
t2
2t
2
e (4t − 4t + 2) − 2
(e) (3/8)
t3
34 3. 5. 2
2−t
4 − 3t
(b) g(t) =
2(1 − t)(2 − t)
4
λ
(c) g(t) =
(d) g(t) =
,
2
(2 − t)
λ+t
(a) g(t) = t<λ. 1 2iτ
(e − 1)
2iτ
e2iτ (2iτ − 1) + 1
k(τ ) =
−2τ 2
2iτ
e − 2iτ − 1
k(τ ) =
−2τ 2
2iτ
e (iτ − 1) + 2eiτ − iτ − 1
k(τ ) =
−τ 2
2iτ
e (−4τ 2 − 4iτ + 2
k(τ ) = (3/8)
−iτ 3 (a) k(τ ) =
(b)
(c)
(d)
(e) 1 − e−t
t
e2t − et
(b) et g(t) =
t
e3t − 1
(c) g(et) =
3t
eb (eat − 1)
(d) eb g(at) =
at 7. (a) g(−t) = 9. (a) g(t) = et 2 +t 2 (b) g(t) (c) g(t) (d) g(t/n) n
n 2 (e) et /2 SECTION 11.1
1. w(1) = (.5, .25, .25)
w(2) = (.4375, .1875, .375)
w(3) = (.40625, .203125, .390625) 3. Pn = P for all n. 5. 1 7. (a) Pn = P
(b) Pn = 9. p2 + q 2 , q 2 , P,
I,
0
1 if n is odd,
if n is even.
0 1
p q
q p
35 11. .375 19. (a) 5/6. (b) The ‘transition matrix’ is
H
P=
T H
5/6
1/2 T
1/6
1/2 . (c) 9/10. (d) No. If it were a Markov chain, then the answer to (c) would be the same as the answer to (a). SECTION 11.2
1. a = 0 or b = 0 3. Examples 11.10 and 11.11 5. The transition matrix in canonical form is
GG, Gg
GG, Gg
1/2
GG, gg 0 Gg, Gg 1/4 P=
Gg, gg 0 GG, GG 0
gg, gg
0 GG, gg
0
0
1/8
0
0
0 Gg, Gg
1/4
1
1/4
1/4
0
0 Gg, gg
0
0
1/4
1/2
0
0 GG, GG
1/4
0
1/16
0
1
0 GG, Gg
GG, Gg
1/2
GG, gg 0 Q = Gg, Gg 1/4 Gg, gg 0
. GG, gg
0
0
1/8
0 Gg, Gg
1/4
1
1/4
1/4 Gg, gg 0
0 1/4 , 1/2 gg, gg 0
0 1/16 .
1/4 0 1 Thus and
GG, Gg
GG, Gg
8/3
GG, gg 4/3 =
Gg, Gg 4/3
Gg, gg
2/3 N = (I − Q)−1 GG, gg
1/6
4/3
1/3
1/6 Gg, Gg
4/3
8/3
8/3
4/3 From this we obtain GG, Gg 29/6
GG, gg 20/3 ,
t = Nc =
Gg, Gg 17/3 Gg, gg
29/6
and
GG, GG
GG, Gg
3/4
GG, gg 1/2 B = NR =
Gg, Gg 1/2
Gg, gg
1/4 36 gg, gg 1/4
1/2 .
1/2 3/4 Gg, gg
2/3
4/3
4/3
8/3 . 7. 9.
12. 2.5 3 1.5
N= 2 4 2 1.5 3 2.5 7
Nc = 8 7 5/8 3/8
B = 1/2 1/2 3/8 5/8
2.08
ABC
ABC 5/18
AC 0 BC 0 0
P= A 0
B 0
C
none
0 1.385 .659 .692
N= 0
1.714
0 0
0
2.25 2.736
Nc = 1.714 2.25 AC
BC
5/18 4/18
5/12
0
0
10/18
0
0
0
0
0
0
0
0 A
B
C
0
0
4/18
5/2
0
1/12
0
5/18 2/18
1
0
0
0
1
0
0
0
1
0
0
0 none 0
1/12 1/18 0 0 0 1 A
B
C
none ABC .275 .192 .440 .093
B = AC .714
0
.143 .143 BC
0
.625 .25 .125
13. Using timid play, Smith’s fortune is a Markov chain with transition matrix
1 2 3 4
1 0 .4 0 0
2 .6 0 .4 0 3 0 .6 0 .4 4 0 0 .6 0 P = 5 0 0 0 .6 6 0 0 0 0 7 0 0 0 0 0 0 0 0 0
8 0 0 0 0 5 6 7 0
0 0 0 .6
0 0 0 0
0 0 0 0
.4 0 0 0
0 .4 0 0
.6 0 .4 0
0 .6 0 .4
0 0 0 1
0 0 0 0 For this matrix we have
0
1 .98
2 .95 3 .9 B = 4 .84 5 .73 6 .58
7 .35 37 8 .02
.05 .1 .16 . .27 .42 .65 8 0
0 0 0 0 . 0 0 0
1 For bold strategy, Smith’s fortune is governed instead by the transition matrix
1 2
1 0 .4
20 0 P = 40 0 00 0
8 0 0 4
0
.4
0
0
0 0 8 .6 0
.6 0 .6 .4 , 1 0
0 1 with
0
1 .936
B = 2 .84
4
.6 8 .064
.16 .
.4 From this we see that the bold strategy gives him a probability .064 of getting out of jail while the
timid strategy gives him a smaller probability .02. Be bold!
15. (a)
3
4
5 3
0
2/3
0
4 1/3
0
2/3 P = 5 0
2/3
0 1 0
0
0
2
0
0
0 1
1/3
0
0
1
0 2 0
0 1/3 . 0 1 (b)
3
3 5/3
N = 4 1
5 2/3 4
2
3
2 5 4/3
2 ,
7/3 3 5
t = 4 6 ,
5 5
1 3 5/9
B = 4 1/3
5 2/9 2 4/9
2/3 .
7/9 (c) Thus when the score is deuce (state 4), the expected number of points to be played is 6, and
the probability that B wins (ends in state 2) is 2/3.
17. For the colorblindness example, we have g, GG
G, Gg B=
g, Gg G, gg G, GG
2/3
2/3
1/3
1/3 g, gg 1/3
1/3 ,
2/3 2/3 and for Example 9 of Section 11.1, we have
GG, GG
GG, Gg
3/4
GG, gg 1/2 B=
Gg, Gg 1/2
Gg, gg
1/4 38 gg, gg 1/4
1/2 .
1/2 3/4 In each case the probability of ending up in a state with all G’s is proportional to the number of
G’s in the starting state. The transition matrix for Example 9 is
GG, GG GG, GG
1
GG, Gg 1/4 GG, gg 0 P=
Gg, Gg 1/16 Gg, gg 0
gg, gg
0 GG, Gg
0
1/2
0
1/4
0
0 GG, gg
0
0
0
1/8
0
0 Gg, Gg
0
1/4
1
1/4
1/4
0 Gg, gg
0
0
0
1/4
1/2
0 gg, gg 0
0 0 .
1/16 1/4 1 Imagine a game in which your fortune is the number of G’s in the state that you are in. This is a
fair game. For example, when you are in state Gg,gg your fortune is 1. On the next step it becomes
2 with probability 1/4, 1 with probability 1/2, and 0 with probability 1/4. Thus, your expected
fortune after the next step is equal to 1, which is equal to your current fortune. You can check that
the same is true no matter what state you are in. Thus if you start in state Gg,gg, your expected
ﬁnal fortune will be 1. But this means that your ﬁnal fortune must also have expected value 1.
Since your ﬁnal fortune is either 4 if you end in GG, GG or 0 if you end in gg, gg, we see that the
probability of your ending in GG, GG must be 1/4.
19. (a)
1
2 1
0
2/3
2 2/3
0
P= 0 0
0
3
0
0 . 0
3 1/3
0
0
1/3 .
1
0 0
1 (b)
1
2 B= 2
6/5
,
9/5 1
2 N= 1
9/5
6/5
0
3/5
2/5 3
2/5
,
3/5 t= 1
2 3
.
3 (c) The game will last on the average 3 moves.
(d) If Mary deals, the probability that John wins the game is 3/5.
21. The problem should assume that a fraction
qi = 1 − qij > 0
j of the pollution goes into the atmosphere and escapes.
(a) We note that u gives the amount of pollution in each city from today’s emission, uQ the
amount that comes from yesterday’s emission, uQ2 from two days ago, etc. Thus
wn = u + uQ + · · · uQn−1 .
(b) Form a Markov chain with Qmatrix Q and with one absorbing state to which the process
moves with probability qi when in state i. Then
I + Q + Q2 + · · · + Qn−1 → N ,
39 so
w(n) → w = uN .
(c) If we are given w as a goal, then we can achieve this by solving w = Nu for u, obtaining
u = w(I − Q) .
27. Use the solution to Exercise 24 with w = f. 29. For the chain with pattern HTH we have already veriﬁed that the conjecture is correct starting in
HT. Assume that we start in H. Then the ﬁrst player will win 8 with probability 1/4, so his expected
winning is 2. Thus E(T H) = 10 − 2 = 8, which is correct according to the results given in the
solution to Exercise 28. The conjecture can be veriﬁed similarly for the chain HHH by comparing
the results given by the conjecture with those given by the solution to Exercise 28. 31. You can easily check that the proportion of G’s in the state provides a harmonic function. Then
by Exercise 27 the proportion at the starting state is equal to the expected value of the proportion
in the ﬁnal aborbing state. But the proportion of 1s in the absorbing state GG, GG is 1. In the
other absorbing state gg, gg it is 0. Thus the expected ﬁnal proportion is just the probability of
ending up in state GG, GG. Therefore, the probability of ending up in GG, GG is the proportion
of G genes in the starting state.(See Exercise 17.) 33. In each case Exercise 27 shows that
f (i) = biN f (N ) + (1 − biN )f (0) .
Thus
biN = f (i) − f (0)
.
f (N ) − f (0) Substituting the values of f in the two cases gives the desired results. SECTION 11.3
1. (a), (f) 3. (a) a = 0 or b = 0
(b) a = b = 1
(c) (0 < a < 1 and 0 < b < 1) or (a = 1 and 0 < b < 1) or (0 < a < 1 and b = 1). 5. (a) (2/3, 1/3)
(b) (1/2, 1/2)
(c) (2/7, 3/7, 2/7) 7. The ﬁxed vector is (1, 0) and the entries of this vector are not strictly positive, as required for the
ﬁxed vector of an ergodic chain. 9. Let p11
P = p21
p31 p12
p22
p32 p13
p23 ,
p33 with column sums equal to 1. Then
3 3 pj1 , 1/3 (1/3, 1/3, 1/3)P = (1/3
j=1 = (1/3, 1/3, 1/3) .
40 3 pj2 , 1/3
j=1 pj3 )
j=1 The same argument shows that if P is an n × n transition matrix with columns that add to 1 then
w = (1/n, · · · , 1/n)
is a ﬁxed probability vector. For an ergodic chain this means the the average number of times in
each state is 1/n.
11. In Example 11.11 of Section 11.1, the state (GG, GG) is absorbing, and the same reasoning as in
the immediately preceding answer applies to show that this chain is not ergodic. 13. The ﬁxed vector is w = (a/(b + a), b/(b + a)). Thus in the long run a proportion b/(b + a) of the
people will be told that the President will run. The fact that this is independent of the starting
state means it is independent of the decision that the President actually makes. (See Exercise 2 of
Section 11.1) 15. It is clearly possible to go between any two states, so the chain is ergodic. From 0 it is possible to
go to states 0, 2, and 4 only in an even number of steps, so the chain is not regular. For the general
Erhrenfest Urn model the ﬁxed vector must statisfy the following equations:
1
w1 = w0 ,
n
wj+1 j+1
n−j+1
+ wj−1
= wj ,
n
n
1
wn−1 = wn .
n if 0 < j < n, It is easy to check that the binomial coeﬃcients satisfy these conditions.
17. Consider the Markov chain whose state is the value of Sn mod(7), that is, the remainder when Sn
is divided by 7. Then the transition matrix for this chain is
0 0
0
1 1/6 2 1/6 P = 3 1/6 4 1/6 5 1/6
6 1/6 1
1/6
0
1/6
1/6
1/6
1/6
1/6 2
1/6
1/6
0
1/6
1/6
1/6
1/6 3
1/6
1/6
1/6
0
1/6
1/6
1/6 4
1/6
1/6
1/6
1/6
0
1/6
1/6 5
1/6
1/6
1/6
1/6
1/6
0
1/6 6 1/6
1/6 1/6 1/6 . 1/6 1/6 0 Since the column sums of this matrix are 1, the ﬁxed vector is
w = (1/7, 1/7, 1/7, 1/7, 1/7, 1/7, 1/7) .
19.
(a) For the general chain it is possible to go from any state i to any other state j in r2 −2r+2 steps.
We show how this can be done starting in state 1. To return to 1, circle (1, 2, .., r − 1, 1) r − 2
times (r2 − 3r + 2 steps) and (1, ..., r, 1) once (r steps). For k = 1, ..., r − 1 to reach state k + 1,
circle (1, 2, . . . , r, 1) r −k times (r2 −rk steps) then (1, 2, . . . , r −1, 1) k −2 times (rk −2r −k +2
steps) and then move to k + 1 in k steps.You have taken r 2 − 2r + 2 steps in all. The argument
is the same for any other starting state with everything translated the appropriate amount.
(b) 0
P = ∗
∗ ∗
∗ 0
0 ∗ , P2 = ∗
0
0 0
41 0
∗
∗ ∗
∗
0 , P3 = ∗
0
∗ ∗
∗
0 0
∗,
∗ ∗
P4 = ∗
∗ ∗
∗
∗ , P5 = ∗
0
∗ ∗
∗
∗ ∗
∗
∗ ∗
∗ .
∗ 25. To each Markov chain we can associate a directed graph, whose vertices are the states i of the chain,
and whose edges are determined by the transition matrix: the graph has an edge from i to j if and
only if pij > 0. Then to say that P is ergodic means that from any state i you can ﬁnd a path
following the arrows until you reach any state j. If you cut out all the loops in this path you will
then have a path that never interesects itself, but still leads from i to j. This path will have at
most r − 1 edges, since each edge leads to a diﬀerent state and none leads to i. Following this path
requires at most r − 1 steps. 27. If P is ergodic it is possible to go between any two states. The same will be true for the chain
with transition matrix 1 (I+P). But for this chain it is possible to remain in any state; therefore,
2
by Exercise 26, this chain is regular. 29.
(b) Since P has rational transition probabilities, when you solve for the ﬁxed vector you will
get a vector a with rational components. We can multiply through by a suﬃciently large
integer to obtain a ﬁxed vector u with integer components such that each component of u
is an integer multiple of the corresponding component of a. Let a(n) be the vector resulting
from the nth iteration. Let b(n) = a(n) P. Then a(n+1) is obtained by adding chips to b(n+1) .
We want to prove that a(n+1) ≥ a(n) . This is true for n = 0 by construction. Assume that
it is true for n. Then multiplying the inequality by P gives that b(n+1) ≥ b(n) . Consider
(n+1)
(n+1)
. This is obtained by adding chips to bj
until we get a multiple of
the component aj
(n) aj . Since bj
(n+1) aj (n+1) ≤ bj , any multiple of aj that could be obtained in this manner to deﬁne
(n) could also have been obtained to deﬁne aj by adding more chips if necessary. Since
(n) we take the smallest possible multiple aj , we must have aj ≤ an+1 . Thus the results after
j
each iteration are monotone increasing. On the other hand, they are always less than or equal
to u. Since there are only a ﬁnite number of integers between components of a and u, the
iteration will have to stop after a ﬁnite number of steps.
31. If the maximum of a set of numbers is an average of other elements of the set, then each of the
elements with positive weight in this average must also be maximum. By assumption, Px = x.
This implies Pn x = x for all n. Assume that xi = M , where M is the maximum value for the xk ’s,
and let j be any other state. Then there is an n such that pn > 0. The ith row of the equation
ij
Pn x = x presents xi as an average of values of xk with positive weight,one of which is xj . Thus
xj = M , and x is constant. SECTION 11.4
1.
3. 1/3
1/3
For regular chains, only the constant vectors are ﬁxed column vectors. SECTION 11.5
1.
Z= 11/9
−1/9
42 −2/9
10/9 . and
0
4 M= 2
0 . 3. 2 5. The ﬁxed vector is w = (1/6,1/6,1/6,1/6,1/6,1/6), so the mean recurrence time is 6 for each state. 7. (a)
1 1
0
2 0 3 1/4 4 0 5 0
6
0 2
0
0
1/4
0
0
0 3
4
1
0
1
0
0
1/4
1/2
0
1/2
0
0
1/2 5
6 0
0
0
0 1/4
0 0
1/2 0
1/2 1/2
0 (b) The rat alternates between the sets {1, 2, 4, 5} and {3, 6}.
(c) w = (1/12, 1/12, 4/12, 2/12, 2/12, 2/12).
(d) m1,5 = 7
9. (a) if n is odd, P is regular. If n is even, P is ergodic but not regular.
(b) w = (1/n, · · · , 1/n).
(c) From the program Ergodic we obtain
0
0 0
14 M = 26 36
4 4 1
4
0
4
6
6 2
6
4
0
4
6 3
6
6
4
0
4 4 4
6 6 . 4
0 This is consistent with the conjecture that mij = d(n − d), where d is the clockwise distance from
i to j.
11. Yes, the reverse transition matrix is the same matrix. 13. Assume that w is a ﬁxed vector for P. Then
wi p∗ =
ij
i i wi wj pji
=
wi wj pji = wj ,
i so w is a ﬁxed vector for P*. Thus if w* is the unique ﬁxed vector for P* we must have w = w*.
15. If pij = pji then P has column sums 1. We have seen (Exercise 9 of Section 11.3) that in this case
the ﬁxed vector is a constant vector. Thus for any two states si and sj , wi = wj and pij = pji .
Thus wi pij = wj pji , and the chain is reversible. 17. We know that wZ = w. We also know that mki = (zii − zki )/wi and wi = 1/ri . Putting these in
the relation
wk mki + wi ri , mi =
¯
k 43 we see that
wk mi =
¯
k zii − zki
+1
wi 1
zii
wk −
wk zki + 1
=
wi
wi
k
k
zii
zii
−1+1=
.
=
wi
wi
18. Form a Markov chain whose states are the possible outcomes of a roll. After 100 rolls we may
assume that the chain is in equilibrium. We want to ﬁnd the mean time in equilibrium to obtain
snake eyes for the ﬁrst time. For this chain mki is the same as ri , since the starting state does not
eﬀect the time to reach another state for the ﬁrst time. The ﬁxed vector has all entries equal to
1/36, so ri = 36. Using this fact, we obtain
mi =
¯ wk mki + wi ri = 35 + 1 = 36.
k We see that the expected time to obtain snake eyes is 36, so the second argument is correct.
19. Recall that zjj − zij
.
wj mij =
j Multiplying through by wj summing on j and, using the fact that Z has row sums 1, we obtain
zjj − mij =
j zjj − 1 = K, zij =
j j which is independent of i.
21. The transition matrix is
GO
GO 1/6
A 1/6 P=
B 1/3
C
1/3 A
B
C 1/3 1/3 1/6
1/6 1/3 1/3 .
1/6 1/6 1/3 1/3 1/6 1/6 Since the column sums are 1, the ﬁxed vector is
w = (1/4, 1/4, 1/4, 1/4) .
From this we see that wf = 0. From the result of Exercise 20 we see that your expected winning
starting in GO is the ﬁrst component of the vector Zf where 15 −30 f = .
−5
20
Using the program ergodic we ﬁnd that the long run expected winning starting in GO is 10.4.
23. (n) Assume that the chain is started in state si . Let Xj
step and 0 otherwise. Then
(n) Sj (0) = Xj (1) + Xj and (n) equal 1 if the chain is in state si on the nth
(2) + Xj n
E(Xj ) = Pij . 44 (n) + . . . Xj . Thus n
(n) (n) E(Sj ) = pij .
h=0 If now follows then from Exercise 16 that
(n) E(Sj )
= wj .
n→∞
n
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This note was uploaded on 04/09/2012 for the course ECON 101 taught by Professor Gottlieb during the Spring '08 term at Rutgers.
 Spring '08
 Gottlieb

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