Math 5010-1, Spring 2005
Assignment 4
Problems
#1, p. 171. Write O, W, B respectively for “orange,” “white,” and “black.” The possible values
of X are: 0, 1, ±2, 4. The respective probabilities are as follows:
8
2
14 ,
2
P {X = −2} = P {2 W’s} =
P {X = −1} = P {1 W and 1 O} =
8
1
2
14
2
1 ,
2
2
14 ,
2
P {X = 0} = P {2 O’s} =
P {X = 1} = P {1 B and 1 W} =
P {X = 2} = P {1 B and 1 O} =
8
1
4
1
14 ,
2 4 2
1 1
14 ,
2
4
2
14 .
2
P {X = 4} = P {2 B’s} =
#7, p. 171.
(a)
(b)
(c)
(d)
1, 2, 3, 4, 5, 6.
1, 2, 3, 4, 5, 6.
2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
0, ±1, ±2, ±3, ±4, ±5.
#14, p. 172. Let Nk denote the number that Player k draws. It follows that P {X = 0} = P {N1 <
N2 }. By relabeling the Players you find that this is the same as P {N2 < N1 }. But
P {N1 < N2 }+P {N2 < N1 } = 1 because one of the two events must occur. Therefore,
P {X = 0} = 12 .
Similiarly, P {X = 1} = P {N2 < N1 < N3 }. Relabel the players to find that this is
the same as the probability of any given win order for three players. Because there are 3!
1
such relabelings, we have shown that P {X = 1} = 3!
= 16 . If you have successfully this
far then you probably have noticed that in general P {X = i} = i!1 for i = 0, . . . , 5.
#17, p. 172. The pmf p satisfies: p(x) = 0 unless F jumps at x. In that case, p(x) = F (x)−F (x−).
1
(a) The pmf is:
p(x) =
1


4




 16
if x = 1,


1


12




0
if x = 3,
if x = 2,
otherwise.
1
For instance, p(3) = F (3) − F (3−) = 1 − 11
12 = 12 . Note also that p(0) = F (0) −
F (0−) = 0 because F is continuous at zero.
(b) P { 21 < X < 32 } is the probability that X = 1. This is p(1) = 14 .
#18, p. 173. The possible values of X −2 are: 0, ±1, ±2. Moreover, X is a binomial with parameters
n = 4 and p = 12 . Therefore,
p(x) =



























4
0
2−4 =
1
16
if x = −2,
4
1
2−4 =
1
4
if x = −1,
4
2
2−4 =
3
8
if x = 0,
4
3
2−4 =
1
4
if x = 1,
4
4
2−4 =
1
16
if x = 2.
For instance, p(−2) = P {X − 2 = −2} = P {X = 0} =
4
0
( 21 )0 (1 − 12 )4−0 =
1
16 .
Theoretical Exercises
#1, p. 180. As in Example 1e (p. 125), we compute P {T > n} first. Let Ai be as in that example,
and note that
N
[
P {T > n} = P
!
Ai
i=1
=
N
X
P (Ai ) −
i=1
+
XX
P (Ai ∩ Aj ) + · · ·
1≤i<j≤N
X
···
X
(−1)k+1 P (Ai1 ∩ · · · ∩ Aik ) + · · · +
1≤i1 <···<ik ≤N
+ (−1)N +1 P (A1 ∩ · · · ∩ AN ).
2
So far, this is the same as before. But now note that P (Ai ) = (1 − Pi )n , P (Ai ∩ Aj ) =
(1 − Pi − Pj )n , etc. Plug to find that
N
[
P {T > n} = P
!
Ai
i=1
=
N
X
i=1
+
XX
n
(1 − Pi ) −
n
(1 − Pi − Pj ) + · · ·
1≤i<j≤N
X
···
X
n
(−1)k+1 (1 − Pi1 − · · · − Pik ) + · · · +
1≤i1 <···<ik ≤N
+ (−1)N +1 0
=
N
−1
X
X
···
X
n
(−1)k+1 (1 − Pi1 − · · · − Pik ) .
k=1 1≤i1 <···<ik ≤N
Therefore,
P {T = n} = P {T > n − 1} − P {T > n}
=
N
−1
X
X
···
X
n
(−1)k+1 (1 − Pi1 − · · · − Pik )
k=1 1≤i1 <···<ik ≤N
−
N
−1
X
X
···
X
n−1
(−1)k+1 (1 − Pi1 − · · · − Pik )
.
k=1 1≤i1 <···<ik ≤N
But (1 − α)n−1 − (1 − α)n = (1 − α)n−1 α. Apply this with α := Pi1 + · · · + Pik to find
that
!n−1 k
k
N
−1 X
X
X
X
X
k+1
Pi`
Piν .
···
(−1)
1−
P {T = n} =
k=1 1≤i1 <···<ik ≤N
`=1
ν=1
#2, p. 180. Note that if b ≤ c then {X ≤ b} ⊆ {X ≤ c}. That is, as b decreases to −∞, the events
{X ≤ b} decrease to ∩−∞<b<∞ {X ≤ b} = ∩∞
n=−∞ {X ≤ n} (why?). The point is that
the latter intersection is taken over the integers, which is a countable set. Therefore,
by the continuity properties of P , limb→−∞ P {X ≤ b} = P (∩∞
n=−∞ {X ≤ n}). But
this is the probability that X = −∞ (why?). The latter probability is zero because
X is a bona fide random variable.
3