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.
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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 CnNxS
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 )
XY
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 bY
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