Download Stat 421 Solutions for Homework Set 1 Page 15 Exercise 1

Stat 421 Solutions for Homework Set 1
Page 15 Exercise 1: Suppose that A ⊂ B. Show that B c ⊂ Ac .
Proof. Assume that x ∈ B c . We need to show that x ∈ Ac . Let us show this indirectly. By way of
contradiction we assume, that x ∈ A. Then x ∈ B because A ⊂ B. ⇒⇐ (contradiction) because we assumed
that x ∈ B c . Hence x ∈
/ A implies x ∈ Ac .
Page 15 Exercise 9: Let∪S be a given sample ∩
space and let A1 , A2 , ... be and infinite sequence of events.
∞
∞
For n = 1, 2, ... , let Bn = i=n Ai and let Cn = i=n Ai .
(a): Show that B1 ⊃ B2 ⊃ · · · and that C1 ⊂ C2 ⊂ · · ·.
Solution: For each n, Bn = Bn+1 ∪ An , hence Bn ⊃ bn+1 ∀n. For each n, Cn+1 ∩ An = Cn , so Cn ⊂ Cn+1 .
(b): Show that an outcome in S belongs to the event
of the events A1 , A2 , ....
∩∞
n=1
Bn if and only if it belongs to an infinite number
Proof. ’⇒’Suppose that x ∈ ∩Bn . To prove that x belongs to infinitely many of A1 , A2 , . . . ,, suppose that x
does not belong to infinitely many of A1 , A2 , . . . ,. Then there exists some m such that x ∈
/ Aj ∀j ≥ m ⇒ x ∈
/
∞
∩∞
A
=
B
⇒
x
∈
/
∪
A
=
B
⇒
x
∈
/
∩B
.
Since
this
conclusion
contradicts
the
initial
supposition,
m
m
n
j=m j
j=m j
it must be true that x does belong to infinitely many of A1 , A2 , . . . ,.
’⇐’ Now we assume that x is in infinitely many Ai . Then, for every n, there is a value of j > n such that
∞
x ∈ Aj , hence x ∈ ∪∞
i=n Ai = Bn for every n and x ∪n=1 Bn .
∪
(c): Show that an outcome in S belongs to the event n=1 ∞Cn if and only if it belongs to all the events
A1 , A2 , ... except possibly a finite number of those events.
∞
Proof. Suppose that x ∩∞
n=1 Cn . That is, there exists n such that x ∈ Cn = ∩n=1 Ai , so x ∈ Ai , for all
i ≥ n. So, there are at most finitely many i (a subset of 1, ..., n − 1) such that x ∈
/ Ai . Finally, suppose that
x ∈ Ai for all but finitely may i. Let k be the last i such that x ∈
/ Ai . then x ∈ Ai for all i ≥ k + 1, hence
∞
x ∈ ∩∞
i=k+1 Ai = Ck+1 . hence x ∩n=1 Cn .
Page 21 Exercise 3: Consider two events A and B such that P r(A) = 1/3 and P r(A) = 1/2. Determine
the value of P r(B ∩ Ac ) for each of the following conditions: (a) A and B are disjoint; (b) A ⊂ B; (c)
P r(A ∪ B) = 1/8.
(a): If A and B are disjoint then B ⊂ Ac and B ∩ Ac = B, so P r(B ∩ Ac ) = P r(B) = 1/2.
(b): If A ⊂ B, then B = A sup(B ∩ Ac ) with A and B ∩ Ac disjoint. So P r(B) = P r(A) + P r(B ∩ Ac ). That
is, 1/2 = 1/3 + P r(B ∩ Ac ), so P r(B ∩ Ac ) = 1/6.
(c): Recall from Theorem 1.4.11, that B = (B ∩ A) + (B ∩ Ac ). Also, B ∩ A and B ∩ Ac are disjoint so,
P r(B) = P r(B ∩ A) + P r(B ∩ Ac ). That is, 1/2 = 1/8 + P r(B ∩ Ac ), so P r(B ∩ Ac ) = 3/8.
Page 21 Exercise 7: We have Pr(A) = .4 and Pr(B) = .7. Determine the maximum and minimum
possible values for Pr(AB) and the conditions under which these values are attained.
Note first that Pr(AB) = Pr(A) + Pr(B) − Pr(A ∪ B) = 1.1 − Pr(A ∪ B), so the min and max values are
determined by Pr(A ∪ B). The smallest possible value of Pr(A ∪ B) occurs when A ⊂ B, and the value is
Pr(A ∪ B) = Pr(B) = .7, and then Pr(AB) = 1.1 − .7 = .4 is the maximum. The largest possible value of
Pr(A ∪ B) occurs when A ∪ B = S, and the value is Pr(A ∪ B) = 1, and then Pr(AB) = 1.1 − 1 = .1 is the
minimum.
Page 21 Exercise 9: Prove that for every two events A and B, the probability that exactly one of the
two events will occur is given by the expression P r(A) + P r(B) − 2P r(A ∪ B).
1
Proof. We need: P r( ”everything in A that is not in B”) +(P r( ”everything in B that is not in A”), which
is:
P r(A ∩ B c ) + P r(Ac ∩ B)
= [P r(A) − P r(A ∩ B)] + [P r(B) − P r(A ∩ B)]
= P r(A) + P r(B) − 2P r(A ∩ B)
Page 21 Exercise 11: A point (x, y) is to be selected from the square S containing all points (x, y) such
that 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. Suppose that the probability that the selected point will belong to each
specified subset of S is equal to the area of that subset. Find the probability of each of the following subsets:
1
1
1
1
3
(a) the subset of points such that (x − )2 + (y − )2 ≥ ; (b) the subset of points such that < x + y < ;
2
2
4
2
2
(c) the subset of points such that y ≤ 1 − x2 ; (d) the subset of points such that x = y.
Solution (a): The set of points for which (x− 1/2)2 +(y − 1/2)2 < 1/4 is the interior of a circle
√that is√contained
1
1
in the unit square. The previous equation gives us the center at ( /2, /2) and radius r = r2 = 1/4 = 1/2.
the area of this circle is π/4, so the area of the remaining region is 1 − π/4 (what we needed!)
Solution (b): We need the area of the region between the two lines y = 1/2 − x and y = 3/2 − x. the
remaining area is the union of the two right triangles with base and height both equal to 1/2. Each triangle
has area 1/8, so the region between the two lines has area 1 − 2/8 = 3/4.
Solution (c): Here we can use calculus. we want the area under the curve y = 1 − x2 between x = 0 and
x = 1. This equals
∫ 1
x3 1
2
(1 − x2 )dx = x −
=
3
3
x=0
0
Solution (d): Since the area of a line is 0, the probability of a line segment is 0.
2