Assignment 1-Probability: basic definition
Due: 17.3.2016. Need to do 3,5,6,7,8. 1,2, 4 are optional.
1. Prove from basic definitin that P(Ac ) = 1−P(A) (use only the three properties defining
a probability measure)
Solution:
1 = P (Ω) = P (A ∪ Ac ) = P (A) + P (Ac )
First equality form property 1 of the probability measure the second from property 2.
2. Prove by induction on n that
P (∪ni=1 Ai )
=
n
X
P(Ai )−
i=1
X
P(Ai1 ∩Ai2 )+
i1 <i2
X
P(Ai1 ∩Ai2 ∩Ai3 )+· · ·+(−1)n+1 P(A1 ∩· · ·∩An ).
i1 <i2 <i3
Solution: For n = 2:
A1 ∪ A2 = (A1 ∩ Ac2 ) ∪ (Ac1 ∩ A2 )
Thus
P (A1 ∪ A2 ) = P (A1 ∩ Ac2 ) = P (Ac1 ∩ A2 )
(0.1)
Thus
A1 = A1 ∩ Ω = A1 ∩ (A2 ∪ Ac2 ) = (A1 ∩ A2 ) ∪ (A1 ∩ Ac2 )
thus,
P (A1 ) = P (A1 ∩ A2 ) + P (A1 ∩ Ac2 )
similarily
P (A2 ) = P (A1 ∩ A2 ) + P (Ac1 ∩ A2 )
Thus
P (A1 ∩ Ac2 ) = P (A1 ) − P (A1 ∩ A2 )
P (Ac1 ∩ A2 ) = P (A2 ) − P (A1 ∩ A2 )
Substitute these equalities in (0.1) to get the reult for n = 2 Assume the equality holds
for n. Let Bn = ∪ni=1 Ai Then
P (Bn ∪ An+1 ) = P (Bn ) + P (An+1 ) − P (Bn ∩ An+1 )
(0.2)
Bn ∪ An+1 = (∪ni=1 (Ai ) ∩ An+1 )
The induction hypohesis yields
P (Bn ∩ An+1 ) =
n
X
P(Ai ∩ An+1 ) −
i=1
+
X
X
P(Ai1 ∩ Ai2 ∩ An+1 ])
(0.3)
i1 <i2
P(Ai1 ∩ Ai2 ∩ Ai3 ∩ An+1 ]) + · · · + (−1)n+1 P(A1 ∩ · · · ∩ An ∩ An+1 )
i1 <i2 <i3
Subtitute (0.3) in 0.2) yeilds the result for n + 1
1
3. You toss a fair dice 10 times. What is the probability that ”6” appears at least once?
Solution: Ai the event ”6” in the ith toss. P (∪6i=1 )Ai = 1 − P (∩Aci ) = 1 − (5/6)10 (we
use independence)
4. Let Fi , i = 1, 2, · · · be a sequence of σ-algebra defined on the same sample space Ω .
Prove that ∩∞
i=1 Fi is a σ-algebra.
Solution:
(a) Since Φ ∈ Fi for all i then Φ ∈ ∩∞
i=1 Fi
c
(b) Assume that A ∈ ∩∞
i=1 Fi ⇒ for all i A ∈ Fi ⇒ A ∈ Fi (since Fi σ-algebra).Thus,
c
∞
A ∈ ∩i=1 Fi
(c) Let Aj , j = 1, 2, · · · be in ∩∞
i=1 Fi . Thus for each i and each j, Aj ∈ Fi , thus for
∞
∞
each i, ∪j=1 Aj ∈ Fi , thus ∪j=1 Aj ∈ ∩∞
i=1 Fi
Is ∪∞
i=1 a σ algebra? prove or give a counter example.
Solution : not nessarily : Ω = {1, 2, 3, 4} F1 = {Φ, Ω, {1}, {2, 3, 4}}
F2 = {Φ, Ω, {2}, {1, 3, 4}} and Fi = {Φ, Ω}, i ≥ 3. Then ∪∞
i=1 Fi = {Φ, Ω, {1}, {2}, {1, 3, 4}, {2, 3, 4}}
is not a σ algebra. For example does not include {1, 2}.
5. Let A and B be events. P(A) = 3/4 and P(B) = 1/3 show that 1/12 ≤ P(A∩B) ≤ 1/3
and give example that both extremes are possible. Find corresponding bounds on
P(A ∪ B).
Solution: P (A ∩ B) ≤ P (A) = 1/3 1 ≥ P (A ∪ B) = P (A) + P (B) − P (A ∩ B) =
13/12 − P (A ∩ B) thus P (A ∩ B) ≥ 1/12. Consider Ω = [0, 1] with Lebesgue measure
as probability. Let A = [0, 3/4], B = [, 1/3] then P (A ∩ B) = P (B) = 1/3
If A = [0, 3/4], B = [2/3, 1] P (A ∩ B) = P (2/3, 3/4) = 1/12 To get similar bounds
on P (A ∪ B) notice that : P (A ∪ B) ≥ max(P (A), P (B)) = 3/4 P (A ∪ B) = P (A) +
P (B) − P (A ∩ B) ≤ min(1, 13/12 − 1/12) = 1 Thus 3/4 ≤ P (A ∪ B) ≤ 1
6. Assume that you throw a dice once.
(i) Describe the sample space.
Ω = {1, 2, 3, 4, 5, 6}
(ii) Find the σ algebra generated by the sets {6}, {1, 3, 5}
F = {Φ, Ω, {6}, {1, 3, 5}, {1, 2, 3, 4, 5}, {2, 4, 6}, {1, 3, 5, 6}, {2, 4}}
7. Let X be a Normal distributed random variable with mean µ and variance σ 2 . Find
E[etX ] . This is the moment generating function of X.
8. For each positive integer n define fn to be the normal density with mean 0 and variance
n., i.e.
x2
1
fn (x) = √
e− 2n
2πn
What is f (x) = limn→∞ fn (x)
2
R∞
What is limn→∞ −∞ fn (x)dx?
Solution: The limit is 0,
Is
Z
∞
∞
f (x)dx ?
fn (x)dx =
lim
n→∞
Z
−∞
−∞
The limit is 1 (inegrating density function ) Explain why this do violated the
monotone convergence Theorem?
√
Solution: Because
it
is
not
monotone
in
n
for
every
x.
For
example
2πf2 =
√
√
0.2601300474 < 2πf5 = 0.2997762378 > 2πf10 = 0.2589053970
9. Let P be the Lebesgue measure on [0, 1], (i.e. the probability space is ([0, 1], B, L)).
Define:
0
if 0 ≤ w < 1/2
Z(w) =
2
if 1/2 ≤ w ≤ 1
For A ∈ B[0, 1] , define
Z
Z(w)dP(w)
P̃(A) =
A
(i) Show that P̃ is a probability measure.
Solution: 1. can do it from scratch as in class. 2. Use the theroem given in class:
Zis non-negative. E(Z) = 1. Thus P̃ (A) = E[Z1A ] is a probability measure.
(ii) Show that if P(A) = 0, then P̃(A) = 0. We say that P̃is absolutely
continuous
R
with respect to P. Solution: assume P (A) = 0 then P̃ (A) = A Z(w)dP (w) = 0
(iii) Show that there is a set A for which P̃(A) = 0 and P(A) > 0. In other words P̃
and P are not equvalent.
Take A = [0, 1/2). P (A) = 1/2. P̃ (A) = 0.
3