# Unit 1 Probability Theory 1.1 Set Theory Definition : sample space ...

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Unit 1

Probability Theory

1.1 Set Theory

Definition

: sample space, all possible outcomes

Example: tossing a coin,

Example: reaction time to a certain stimulus,

Sample
space
:
may be
c
ountable or
uncountable

C
ountable: put 1
-
1 correspondence with a subset of integers

F
inite elements

countable

I
nfinite elements

countable

or uncountable

Fact: There is only countable sample space since measu
infinite accuracy

Definition
event: any measurable collection of possible outcomes, subset of

I
f
,

occurs if outcome is in the set
.

: probability of an event (rather than a set)

Theorem

: events on

(1).

Commutativity
,

(2).

Associativity

,

(3).

Distributive Laws

(4).

DeMorgan

s Law

,

Example:
(a) If
,
,
.

(b) If
,
.

Definition

(a)

are disjoint if
.

(b)

are pairwise disjoint if
,
.

Definition

are pa
irwise disjoint and
, then

form a
partition of
.

1.2
Basic Probability Theory

: event in
,
,
:

probability of

D
omain of
: all measurable subsets of

D
efinition

(sigma algebra,
-
algebra, Borel field):
collection

of subsets of

satisfies (a)

(b) if

(
closed

under complementation)

(c) if

(closed under countable unions)

Properties
(a)

(b)

Example
: (a)
: trivial
-
algebra

(b) smallest
-
algebra that contains all of the open sets in
=
{all open sets in
}=

(intersection on all possible
-
algebra)

Definition

Kolmogorov Axioms (Axioms of Probability)

Given
, probability function is a function

with domain

s
atisfies

(a)
,

(b)

(c) If
, pairwise disjoint

Exercise: axiom of finite additivity + continuity of
(if
)

Theorem

If
,
: probability

(a)

(b)

(c)

Theorem

(a)

(b)

(c)
, then

Bonferroni

s Inequality

Example: (a)
,

(b)
,
,
, useless but correct

Theorem

(a)
, for any partition

(b)

for any

(Boole

s inequality)

General
versio
n of Bonferroni inequality:

Counting

without

replacement

with

replacement

Ordered

Unordered

Let

be a sequence of sets.
The

set of all points

that belong to

for infinitely many values of

is known as the

limit superior

of the sequence and is
denoted by

or
.

The set of all points that belong to

for all but a finite number of values of

is known as the

limit inferior

of the sequence

and is denoted by

or
. If
, we say that the limit exists and write

for the
common set and call it the
limit set
.

We have
.

If the sequence

is such that
, for
, it is called
nondecreasing
; if
,
, it is called
nonincreasing
. If the sequence

is nondecreasing, or nonincreasin
g, the limit exists and we have

if

is nondecreasing and

if

is nonincreasing.

Theorem
Let

be a nondecreaing sequence of events in
; that is
,
, and
,

Then

.

P
roof
. Let
. Then

.

have

,

and letting
, we see that

.

The
second

term on the right tends to zero as

since the sum

and each summand is
nonnegative
.
T
he re
sult follows.

C
orollary

Let

be a nonincreasing sequence of events in
. Then

.

Proof
.
Consider the nondecreasing sequence of events
. Then

.

It follows from
the

above Theorem that

.

Hence,
.

Example (Bertrand

s Paradox) A chord is drawn at random in
the

unit circle. What is
the probability that the chord is longer than the side of the equilater
al triangle
inscribed in the circle?

Solution 1
. Since the length of a chord is uniquely determined by the position of its
midpoint, choose a point

at random in the circle and draw a line through

and
, the center of the circle. Draw the chord through

perpendicular to the line
.
If

is the length of the chord with

as midpoint,

if and only if

lines inside the circle with center

. Thus
.

Solution 2
.
Because of symmetry, we may fix one endpoint of the chord at some point

and then choose the other endpoint

at random. Let the probability that

lies on an arbitrary arc of the circle be proportional to the length of this arc. Now the
inscribed equilateral
triangle having

as one of

its vertices divides the
circumference into three equal parts. A chord drawn through

will be longer than
the side of the triangle if and only
if the other endpoint

of the chord lies on that
one
-
third of the circumference that is opposite
. It follows that the required
probability is
.

Solution 3
.
Note that the length of a chord is determined uniquely by the distanc
e of
its midpoint from the center of the circle. Due to the symmetry of the circle, we
assume that the midpoint of the chord lies on a fixed radius,
, of the circle. The
probability that the midpoint

lies in a

given segment of the radius through

is
then proportional to the length of this segment. Clearly, the length of the chord will be
longer than the side of the inscribed equilateral triangle if the length of

is

less
than
. It follows that the required probability is
.

Question: What

s happen? Which answer(s) is (are) right?

Example: Consider sampling

items from

items,

with replacement. The
outcomes in the ordered and unordered sample spaces are these.

Unordered

{1,1}

{2,2}

{3,3}

{1,2}

{1,3}

{2,3}

Probability

1/6

1/6

1/6

1/6

1/6

1/6

Ordered

(1,1)

(2,2)

(3,3)

(1,2), (2,1)

(1,3), (3,1)

(2,3), (3,2)

Probability

1/9

1/9

1/9

2/9

2/9

2/9

Which one is correct?

Hint: The confusion arises because the phrase

with replacement

will typically be
interpreted with the sequential kind of sampling, leading to assigning a probability 2/9
to the event {1, 3}.

1.3
Conditional Proba
bility and Independence

Definition

C
onditional

probability of

given

is

,
provided
.

Remark: (a) In the above definition,

becomes the samp
le space and
.
All events are calibrated with respect to
.

(b) If

then

and
.
D
isjoint is not
the same as independent.

Definition

and

are independent if
.

(or
)

Example: Three prisoners,
,
, and
, are on death row. The governor de
cides to
pardon one of the three and chooses at random the prisoner to pardon. He informs the
warden of his choice but requests that the name be kept secret for a few days.

The
next day,

tries to get the warden to tell him who had
been pardoned. The
warden refuses.

or

will be executed. The warden
thinks for a while, then tells

that

is to be execu
ted.

Warden

s reasoning:

Each prisoner has a 1/3 chance of being pardoned. Clearly,
either

or

must be executed, so I have given

wil
l be pardoned.

s reasoning:

Given that

will be executed, then either

or

will be
pardoned. My chance of being pardoned has risen to 1/2.

Which one is correct?

Bayes

Rule

: partition of sample space,
: any set,
.

Example:
When coded messages are sent, there are sometimes errors in transmission.
In particular, Morse code uses

dots

and

dashes

, which are known to occur in the
proportion of 3:4. This means that for any given symbol,

and
.

Suppose there is interference on the transmission line, and with probability 1/8 a dot
s a dash, and vice versa. If we receive a dot, can we be sure
that a dot was sent?

Theorem

If

then (a)
, (b)
, (c)
.

Definition
: mutuall
y independent if any subcollection

then

.

1.4
Random Variable

Definition

Define
new sample space
.

: random
variable
,
,
, where
: induced probability
function on

in terms of original

by

,

and

satisfies the Kolmogorov Axioms.

Exam
ple:
Tossing three coins,

S =

{HHH,

HHT,

HTH,

THH,

TTH,

THT,

HTT,

TTT}

X :

3

2

2

2

1

1

1

0

Therefore,
, and

.

1.5 Distribution Functions

W
ith every random variable
, we associate a function called the cumulative
distribution function of
.

Definition

The cumulative distribution function or cdf of a random variable
,
denoted by
, is d
efined by
, for all
.

Example
:
Tossing three coins,
, the
corresponding
c
.
d
.
f
.

is

,

where
: (a) is a step function

(b) is defined for all
, not just in

(c) jumps at

, size of jump

(d)

for
;

for

(e) is right
-
continuous (is left
-
continuous if
)

Theorem

is a c.d.f.

(a)
,
.

(b)
: non
-
decreasing

(c)
: right
-
continuous

Example: Tossing a coin until a head appears. Define a random variable
:

# of
tosses required to get a head. Then

,
,
.

The c.d.f. of the random variable

is

,
.

It i
s easy to check that

satisfies the three conditions of c.d.f.

Example:
A continuous c.d.f. (of logistic distribution) is
, which
satisfies the three conditions of c.d.f.

Definition

(a)

is continuous if

is continuous.

(b)

is discrete if

is a step
function
.

Definition

and

are identical distributed if
,
.

Example:

Tossing a fair coin three times. Let
: # of tail. Then

,
.

But for each sample point
,
.

Theorem

and

are identical distributed
,
.

1.6 Density and Mass Function

Definition

The probability mass function (p.m.f.) of a discrete random variable

is

for all
.

Example:
For the geometric distribution,
we have the p.m.f.

.

A
nd
size of jump in c.d.f. at
,

,

.

Fact: For continuous random variable (a)
,
.

(b)
. Using the Fundamental Theorem of Calculus, if

is continuous, we have

.

Definition

The probability density function or pdf,
, of a continuous random
variable

is the function that satisfies

for all
.

Notation: (a)
,

is distributed as
.

(b)
,

and

have the same distribution.

Fact: For continuous,
.

Example:
For the logistic distribution

,

we have
, and

.

Theorem

: pdf (or pmf) of a random variable if and only if

(a)
,
.

(b)

(pmf) or

(pdf).

Fact: For any nonnegative function with finite positive integral (or sum) can be turned
into a pdf (or pmf)

then
.