Transcript Chapter 5 Discrete Random Variables and Probability Distributions © Random Variables A random variable is a variable that takes on numerical values determined by the outcome.

Chapter 5
Discrete Random Variables and
Probability Distributions
©
Random Variables
A random variable is a variable that
takes on numerical values determined
by the outcome of a random
experiment.
Discrete Random Variables
A random variable is discrete if it
can take on no more than a
countable number of values.
Discrete Random Variables
(Examples)
1.
The number of defective items in a sample of
twenty items taken from a large shipment.
2. The number of customers arriving at a checkout counter in an hour.
3. The number of errors detected in a
corporation’s accounts.
4. The number of claims on a medical insurance
policy in a particular year.
Continuous Random
Variables
A random variable is continuous
if it can take any value in an
interval.
Continuous Random Variables
(Examples)
1. The income in a year for a family.
2. The amount of oil imported into the U.S. in a
particular month.
3. The change in the price of a share of IBM
common stock in a month.
4. The time that elapses between the installation
of a new computer and its failure.
5. The percentage of impurity in a batch of
chemicals.
Discrete Probability
Distributions
The probability distribution function (DPF), P(x),
of a discrete random variable expresses the
probability that X takes the value x, as a
function of x. That is
P( x)  P( X  x), for all valuesof x.
Discrete Probability
Distributions
Graph the probability distribution function for
the roll of a single six-sided die.
P(x)
1/6
1
2
3
4
Figure 5.1
5
6
x
Required Properties of Probability
Distribution Functions of Discrete
Random Variables
Let X be a discrete random variable with
probability distribution function, P(x). Then
i. P(x)  0 for any value of x
ii. The individual probabilities sum to 1; that is
 P( x )  1
x
Where the notation indicates summation over
all possible values x.
Cumulative Probability
Function
The cumulative probability function, F(x0), of a
random variable X expresses the probability that
X does not exceed the value x0, as a function of
x0. That is
F ( x0 )  P( X  x0 )
Where the function is evaluated at all values x0
Derived Relationship Between Probability
and Cumulative Probability Function
Let X be a random variable with probability function
P(x) and cumulative probability function F(x0).
Then it can be shown that
F ( x0 )   P( X )
x x0
Where the notation implies that summation is over
all possible values x that are less than or equal to x0.
Derived Properties of Cumulative
Probability Functions for Discrete
Random Variables
Let X be a discrete random variable
with a cumulative probability
function, F(x0). Then we can show
that
i. 0  F(x0)  1 for every number x0
ii. If x0 and x1 are two numbers with x0
< x1, then F(x0)  F(x1)
Expected Value
The expected value, E(X), of a discrete random
variable X is defined
E ( X )   xP( x)
x
Where the notation indicates that summation
extends over all possible values x.
The expected value of a random variable is
called its mean and is denoted x.
Variance and Standard
Deviation
Let X be a discrete random variable. The expectation
of the squared discrepancies about the mean, (X - )2,
is called the variance, denoted 2x and is given by
  E( X   x )   ( x   x ) P( x)
2
x
2
2
x
The standard deviation, x , is the positive square root
of the variance.
Variance
(Alternative Formula)
The variance of a discrete random variable
X can be expressed as
  E( X )  x
2
x
2
2
  x P( x)   x
2
x
2
Expected Value and Variance for
Discrete Random Variable Using
Microsoft Excel
Sales
0
1
2
3
4
5
(Figure 5.4)
P(x)
Mean
0.15
0.00
0.30
0.30
0.20
0.40
0.20
0.60
0.10
0.40
0.05
0.25
1.95
Expected Value = 1.95
Variance
0.5704
0.2708
0.0005
0.2205
0.4203
0.4651
1.9475
Variance = 1.9475
Bernoulli Distribution
A Bernoulli distribution arises from a random
experiment which can give rise to just two possible
outcomes. These outcomes are usually labeled as either
“success” or “failure.” If  denotes the probability of a
success and the probability of a failure is (1 -  ), the
the Bernoulli probability function is
P(0)  (1   ) and P(1)  
Mean and Variance of a
Bernoulli Random Variable
The mean is:
 X  E( X )   xP( x)  (0)(1   )  (1)  
X
And the variance is:
  E[( X   X ) ]   ( x   X ) P( x)
2
X
2
2
X
 (0   ) (1   )  (1   )    (1   )
2
2
Sequences of x Successes in
n Trials
The number of sequences with x successes in n independent
trials is:
n!
C 
x!(n  x)!
n
x
Where n! = n x (x – 1) x (n – 2) x . . . x 1 and 0! = 1.
These C xn sequences are mutually exclusive,
since no two of them can occur at the same time.
Binomial Distribution
Suppose that a random experiment can result in two possible
mutually exclusive and collectively exhaustive outcomes,
“success” and “failure,” and that  is the probability of a success
resulting in a single trial. If n independent trials are carried out,
the distribution of the resulting number of successes “x” is called
the binomial distribution. Its probability distribution function for
the binomial random variable X = x is:
P(x successes in n independent trials)=
n!
x
( n x )
P( x) 
 (1   )
x!(n  x)!
for x = 0, 1, 2 . . . , n
Mean and Variance of a Binomial
Probability Distribution
Let X be the number of successes in n independent
trials, each with probability of success . The x
follows a binomial distribution with mean,
 X  E( X )  n
and variance,
  E[( X   ) ]  n (1   )
2
X
2
Binomial Probabilities
- An Example –
(Example 5.7)
An insurance broker, Shirley Ferguson, has five contracts,
and she believes that for each contract, the probability of
making a sale is 0.40.
What is the probability that she makes at most one sale?
P(at most one sale) = P(X  1) = P(X = 0) + P(X = 1)
= 0.078 + 0.259 = 0.337
5!
P (nosales)  P (0)
(0.4) 0 (0.6) 5  0.078
0!5!
5!
P (1sale)  P (1)
(0.4)1 (0.6) 4  0.259
1!4!
Binomial Probabilities, n = 100,  =0.40
(Figure 5.10)
Sample size
100
Probability of success
0.4
Mean
40
Variance
24
Standard deviation
4.898979
Binomial Probabilities Table
X
36
37
38
39
40
41
42
43
P(X)
0.059141
0.068199
0.075378
0.079888
0.081219
0.079238
0.074207
0.066729
P(<=X)
0.238611
0.30681
0.382188
0.462075
0.543294
0.622533
0.69674
0.763469
P(<X)
0.179469
0.238611
0.30681
0.382188
0.462075
0.543294
0.622533
0.69674
P(>X)
0.761389
0.69319
0.617812
0.537925
0.456706
0.377467
0.30326
0.236531
P(>=X)
0.820531
0.761389
0.69319
0.617812
0.537925
0.456706
0.377467
0.30326
Hypergeometric Distribution
Suppose that a random sample of n objects is chosen from a
group of N objects, S of which are successes. The distribution
of the number of X successes in the sample is called the
hypergeometric distribution. Its probability function is:
C xS CnNxS
P( x) 
N
Cn
S!
( N  S )!

x!( S  x)! (n  x)!( N  S  n  x)!

N!
n!( N  n)!
Where x can take integer values ranging from the larger of 0 and
[n-(N-S)] to the smaller of n and S.
Poisson Probability Distribution
1)
2)
3)
Assume that an interval is divided into a very large number
of subintervals so that the probability of the occurrence of
an event in any subinterval is very small. The assumptions
of a Poisson probability distribution are:
The probability of an occurrence of an event is constant
for all subintervals.
There can be no more than one occurrence in each
subinterval.
Occurrences are independent; that is, the number of
occurrences in any non-overlapping intervals in
independent of one another.
Poisson Probability Distribution
The random variable X is said to follow the Poisson
probability distribution if it has the probability
function:
e   x
P( x) 
, for x  0,1,2,...
x!
where
1.
P(x) = the probability of x successes over a
given period of time or space, given 
2.

= the expected number of successes per
time or space unit;  > 0
3.
e
= 2.71828 (the base for natural
logarithms)
Poisson Probability Distribution
• The mean and variance of the Poisson
probability distribution are:
x  E( X )   and   E[(X  ) ]  
2
x
2
Partial Poisson Probabilities for  = 0.03
Obtained Using Microsoft Excel PHStat
(Figure 5.14)
Poisson Probabilities Table
X
P(X)
0
0.970446
1
0.029113
2
0.000437
3
0.000004
4
0.000000
P(<=X)
0.970446
0.999559
0.999996
1.000000
1.000000
P(<X)
0.000000
0.970446
0.999559
0.999996
1.000000
P(>X)
0.029554
0.000441
0.000004
0.000000
0.000000
P(>=X)
1.000000
0.029554
0.000441
0.000004
0.000000
Poisson Approximation to
the Binomial Distribution
Let X be the number of successes resulting from n
independent trials, each with a probability of success, . The
distribution of the number of successes X is binomial, with
mean n. If the number of trials n is large and n is of only
moderate size (preferably n  7), this distribution can be
approximated by the Poisson distribution with  = n. The
probability function of the approximating distribution is then:
P( x) 
e
 n
(n )
, for x  0,1,2,...
x!
x
Joint Probability Functions
Let X and Y be a pair of discrete random
variables. Their joint probability function
expresses the probability that X takes the
specific value x and simultaneously Y takes the
value y, as a function of x and y. The notation
used is P(x, y) so,
P( x, y)  P( X  x  Y  y)
Joint Probability Functions
Let X and Y be a pair of jointly distributed
random variables. In this context the probability
function of the random variable X is called its
marginal probability function and is obtained by
summing the joint probabilities over all possible
values; that is,
P( x)   P( x, y)
y
Similarly, the marginal probability function of
the random variable Y is
P( y)  P( x, y)

x
Properties of Joint Probability
Functions
•
Let X and Y be discrete random variables with
joint probability function P(x,y). Then
1. P(x,y)  0 for any pair of values x and y
2. The sum of the joint probabilities P(x, y) over all
possible values must be 1.
Conditional Probability Functions
Let X and Y be a pair of jointly distributed discrete random
variables. The conditional probability function of the random
variable Y, given that the random variable X takes the value x,
expresses the probability that Y takes the value y, as a function
of y, when the value x is specified for X. This is denoted
P(y|x), and so by the definition of conditional probability:
P ( x, y )
P( y | x) 
P( x)
Similarly, the conditional probability function of X, given Y = y is:
P ( x, y )
P( x | y ) 
P( y )
Independence of Jointly
Distributed Random Variables
The jointly distributed random variables X and Y
are said to be independent if and only if their joint
probability function is the product of their marginal
probability functions, that is, if and only if
P( x, y)  P( x) P( y) for all possible pairsof values x and y.
And k random variables are independent if and only
if
P( x , x ,, x )  P( x ) P( x )P( x )
1
2
k
1
2
k
Expected Value Function of Jointly
Distributed Random Variables
Let X and Y be a pair of discrete random
variables with joint probability function P(x,
y). The expectation of any function g(x, y) of
these random variables is defined as:
E[ g ( X , Y )]   g ( x, y) P( x, y)
x
y
Stock Returns, Marginal Probability,
Mean, Variance
(Example 5.16)
Y Return
X
Return
0%
0%
5%
10%
15%
0.0625
0.0625
0.0625
0.0625
5%
0.0625
0.0625
0.0625
0.0625
10%
0.0625
0.0625
0.0625
0.0625
15%
0.0625
0.0625
0.0625
0.0625
Table 5.6
Covariance
Let X be a random variable with mean X , and let Y be a
random variable with mean, Y . The expected value of (X X )(Y - Y ) is called the covariance between X and Y, denoted
Cov(X, Y).
For discrete random variables
Cov( X , Y )  E[( X   X )(Y  Y )]   ( x   x )( y   y ) P( x, y)
x
y
An equivalent expression is
Cov( X , Y )  E( XY )   x  y   xyP( x, y)   x  y
x
y
Correlation
Let X and Y be jointly distributed random
variables. The correlation between X and Y is:
  Corr( X , Y ) 
Cov( X , Y )
 XY
Covariance and Statistical
Independence
If two random variables are statistically
independent, the covariance between
them is 0. However, the converse is not
necessarily true.
Portfolio Analysis
The random variable X is the price for stock
A and the random variable Y is the price for
stock B. The market value, W, for the
portfolio is given by the linear function,
W  aX  bY
Where, a, is the number of shares of stock A
and, b, is the number of shares of stock B.
Portfolio Analysis
The mean value for W is,
The variance for W is,
W  E[W ]  E[aX  bY ]
 a X  bY
  a   b   2abCov( X , Y )
2
W
2
2
X
2
2
Y
or using the correlation,
  a   b   2abCorr( X , Y ) X  Y
2
W
2
2
X
2
2
Y