Lec 2 discrete random variable

9/28/2018
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SBE 304: Bio-Statistics
Discrete Random Variables
and their Probability
Distributions
Dr. Ayman Eldeib
Systems & Biomedical
Engineering Department
Fall 2018
SBE 304: DRV - PMF
Outline
 Introduction
 What is a discrete random variable?
 Discrete Probability Distribution
 Probability Mass Function (PMF)
 Cumulative Distribution Function
 The Mean and Variance of a Discrete Random Variable
 Bernoulli Trial
 Binomial, Negative Binomial, Uniform, and Poisson
Distribution
 Geometric and Hypergeometric Random Variable
 The finite population correction factor
 Joint Probability Distribution

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SBE 304: DRV - PMF
A Random Variable and
its Probability Distribution
A variable whose value is uncertain and
dependent on chance is called a random
variable.
All the possible values of a random
variable and their associated probability
values constitute a probability
distribution.
SBE 304: DRV - PMF
A random variable is a function that assigns a
real number to each outcome in the sample
space of a random experiment.
A random variable is a real valued function defined
on the sample space of an experiment. Associated
with each random variable is a Probability Density
Function (PDF) for the random variable.
Cont.

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SBE 304: DRV - PMF
Suppose you toss a fair coin 2 times and observe the outcomes.
You then define the random variable X to be the number of heads
observed in the 2 coin tosses. This is a valid random variable,
because it is a function assigning real numbers to outcomes, as
follows:
Outcome (in ) HH HT TH TT
Value of X 2 1 1 0
The range of the random variable X is R(X) = {0; 1; 2}. Notice that,
although the 4 outcomes in are equally likely (each having
probability 1=4), the values of X are not equally likely to occur.
Cont.
SBE 304: DRV - PMF
A Discrete/Continuous
Random Variable
 A discrete random variable is a random variable with a
finite (or countable infinite) range. It is one that can
assume only distinct (whole number, 0, 1, 2, 3, 4, 5, 6,
etc. ) Values.
Examples: Response with Yes - No values, number of
transmitted bits received in error, year (2010), etc.
 A continuous random variable is a random variable with
an interval (either finite or infinite) of real numbers for its
range. It is one that can assume any value over a
continuous range of possibilities.
Examples: temperature, weight, electrical current, length,
pressure, time, voltage, etc.

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SBE 304: DRV - PMF
Discrete Probability Distribution
The probability distribution function p(x) (pX(x))
for a discrete random variable X is a function
assigning probabilities to the elements of its
range R(X)
If we adopt the notation that large letters (like X) are
used to stand for random variables, and corresponding
small letters (like x) are used to stand for realized
values (i.e., elements of the range of) these random
variables, we see that
p(x) = P(X = x)
SBE 304: DRV - PMF
x p(x) = P(X = x)
2 ¼
1 ½
0 ¼
Outcome (in Ω) HH HT TH TT
Value of X 2 1 1 0
R(X) = {0; 1; 2}
Cont.

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SBE 304: DRV - PMF
x p(x) = P(X = x)
1 ½
0 ½
R(X) = {0; 1}
Suppose you throw a fair die, and code
the outcomes as in the table below
Outcome (in Ω) 1 2 3 4 5 6
Value of X 1 0 1 0 1 0
Cont.
SBE 304: DRV - PMF
For a discrete random variable X with possible
values x1, x2, …, xn, a probability mass function p(x)
is a function such that,
0 ≤ p(x) ≤ 1 for all x
p(x) = 1
p(x) = P(X = x)
∑
n
x = 1
Probability Mass Function (PMF)

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SBE 304: DRV - PMF
Calculation of PMF
To calculate the PMF of a random variable X
with a probability density function p(x):
1. Collect all possible outcomes in Ω which
are mapped by X to x, i.e. {X = x}.
2. Add their probabilities to obtain p(x).
SBE 304: DRV - PMF
Cumulative Distribution Function
The cumulative distribution function of a
discrete random variable X, denoted as F(x),
satisfies the following properties:
(1) F(x) = P( X ≤ x) = p(xi)
(2) 0 ≤ F(x) ≤ 1
(3) If x ≤ y, then F(x) ≤ F(y)
∑
xi ≤ x

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SBE 304: DRV - PMF
Suppose that a day’s production of 850
manufactured parts contains 50 parts that do not
conform to customer requirements. Two parts are
selected at random, without replacement, from the
batch. Let the random variable X equal the number
of nonconforming parts in the sample.
What is the cumulative distribution function of X?
Cont.Cumulative Distribution Function
SBE 304: DRV - PMF
Total # of parts = 850
Nonconforming parts = 50
Two parts are selected at random, without
replacement
random variable X equal the number of
nonconforming parts in the sample.
F(0) = P(X≤0) = 0.886
P(X=0) = 800/850 * 799/849 = 0.886
P(X=1) = 2*800/850 * 50/849 = 0.111
P(X=2) = 50/850 * 49/849 = 0.003
F(1) = P(X≤1) = 0.886 + 0.111 = 0.997
F(2) = P(X≤2) = 1
Cont.Cumulative Distribution Function

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SBE 304: DRV - PMF
The Mean of a Discrete
Random Variable
The mean or expected value of the discrete
random variable X, denoted as µ or E(X), is the
sum of each value of the variable times its
corresponding probability,
µ = E(X) = x * p(x)∑
x
SBE 304: DRV - PMF
The Algebra of Expected Values
For any random variables X, Y, and constants a
and b, the following results hold:
E(a) = a
E(aX) = aE(X)
E(X + Y) = E(X) + E(Y)

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SBE 304: DRV - PMF
The Mean of a Discrete
Random Variable
µ = E(X) = x * p(x) = 0 * 0.6561 + 1 * 0.2916 + 2 * 0.0486 +
3 * 0.0036 + 4 * 0.0001 = 0.4
∑
x
There is a chance that a bit transmitted through a digital
transmission channel is received in error. Let X equal the number
of bits in error in the next four bits transmitted. The possible
values for X are {0,1,2,3,4}. Suppose that the probabilities are
P(X=0) = 0.6561
P(X=1) = 0.2916
P(X=2) = 0.0486
P(X=3) = 0.0036
P(X=4) = 0.0001
Cont.
SBE 304: DRV - PMF
The Variance and standard deviation
of a Discrete Random Variable
The variance of the discrete random variable X,
denoted as σ2 or V(X), is the expected value of
(X - µ)2
σ2 = V(X) = E[(X - µ)2] = (x - µ)2 * p(x) = x2 * p(x) - µ2∑
x
∑
x
The standard deviation of the discrete random
variable X, denoted as σ = σ2

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SBE 304: DRV - PMF
The Algebra of Variance
For any random variables X, Y, and constants a
and b, the following results hold:
V(X) = E(X2) - µ2
V(aX + b) = a2V(X)
V(X + Y) = V(X) + V(Y)
SBE 304: DRV - PMF
The Expected value of a Function of a
Discrete Random Variable
Let X be a discrete random variable with probability mass
function p(x) and g(X) be a real-valued function of X.
Then the expected value of g(X) is given by
E(g(X)) = g(x) * p(x)∑
x
In the special case that g(X) = aX +b for any constants a
and b, Then the expected value of g(X) is given by
E(g(X)) = a * E(X) + b

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SBE 304: DRV - PMF
X is the number of bits in error in the next four bits
transmitted. What is the expected value of the square
of the number of bits in error? Now, g(X) = X2.
Therefore,
E(g(X)) = g(x) * p(x)
= 02 * 0.6561 + 12 * 0.2916 + 22 * 0.0486 +
32 * 0.0036 + 42 * 0.0001 = 0.52
∑
x
Cont.
The Expected value of a Function of a
SBE 304: DRV - PMF
Bernoulli Trial
A trial with only two possible outcomes is used so
frequently as a building block of a random
experiment that it is called a Bernoulli trial. It is
usually assumed that the trials that constitute the
random experiment are independent. This implies
that the outcome from one trial has no effect on the
outcome to be obtained from any other trial.
Furthermore, it is often reasonable to assume that
the probability of a success in each trial is constant.

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SBE 304: DRV - PMF
Discrete Uniform Distribution
A random variable X has a discrete uniform
distribution if each of the n values in its range, say,
x1, x2, …,xn, has equal probability. Then,
p(xi) = 1/n
SBE 304: DRV - PMF
k = 0, 1, 2, …..

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SBE 304: DRV - PMF
Binomial Distribution
Usually, not all patients show up for their doctor appointments, a
doctor office gives 35 appointments each day while the time and
medical supplies are exactly enough for 30 patients only. The
probability that a patient does not show up is 0.10, and the
patients behave independently.
Example
1.What is the probability that every patient who shows up can see the doctor?
2.What is the probability that the doctor office closes early?
Let X denote the patients with appointment that do not show up,
X is binomial with n = 35 and p = 0.1
P(X ≥ 5) = 1 – P(X ≤ 4) = 0.269
P(X > 5) = 1 – P(X ≤ 5) = 0.13
SBE 304: DRV - PMF
Binomial Distribution
The probability that a patient recovers from a stomach disease is 0.8.
Suppose 20 people are known to have this disease. What is the probability
that:
Example
1. at most 16 recover?
2. at least 14 but not more than 18 recover?
( ) ( ) ( ) ( )18........15141814 =++=+==≤≤ YPYPYPYP = 0.844
Let Y be the number of patients who recovered from the stomach
disease among the 20.
Let Y be the number of patients who recovered from the stomach
disease among the 20. Y is a binomial RV
( ) yny
y
CyYP −
== 2.08.020
20,...,2,1,0=y
( ) ( ) ( ) ( ) ( ) ( ){ }20......17116........1016 =++=−==++=+==≤ YPYPYPYPYPYP = 0.589

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SBE 304: DRV - PMF
Negative Binomial Distribution
A generalization of a geometric distribution in which the
random variable is the number of Bernoulli trials required
to obtain r successes results in the negative binomial
distribution.
X is a negative binomial random variable with
parameters 0 < p < 1 and r 1, 2 3, ….. , and
p(x) = ( ) (1 – p)x-r * pr where x = r, r+1, r+2,…..
x – 1
r – 1
µ = E(X) = r/p and σ2 = V(X) = r(1 – p)/p2
SBE 304: DRV - PMF
Hypergeometric Random Variable
A set of N objects contains; K objects classified as successes;
N - K objects classified as failures
A sample of size n objects is selected randomly (without
replacement) from the N objects, where K ≤ N and n ≤ N.
Let the random variable X denote the number of successes in
the sample. Then X is a hypergeometric random variable
and
( ) where p = K/N
( )
p(x) =
( )___________
( )
N – K
n - x
K
x
N
n
x = max{0, n + K - N} to min {k,n}
µ = E(X) = np and σ2 = V(X) = np(1 – p) N – n
N – 1
____

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SBE 304: DRV - PMF
A medical supply manufacture ships one of its products in lots of 20. Testing
to determine whether an item is defective is costly, and hence the tester
samples the production rather than using a 100% inspection plan. A
sampling plan, constructed to minimize the number of defectives shipped to
hospitals, calls for sampling five items from each lot and rejecting the lot if
more than one defective is observed. If a lot contains four defectives,
Example
1.What is the probability that it will be rejected?
2.What is the expected number of defective in the sample of size 5?
3. What is the variance of the number of defectives in the sample of
size 5?
Let Y equal the number of defects in the sample.
Then, N = 20, k = 4, and n = 5 [Hypergeometric Random Variable]
A lot will be rejected if Y = 2, 3, 4. Then
P(rejecting a lot) = P(Y ≥ 2) = P(2) + P(3) + P(4) = 1 – (P(0) + P(1)) = 0.248
µ = (5) (4)/20 = 1
σ2 = 0.632
Cont.Hypergeometric Random Variable
SBE 304: DRV - PMF
A hospital storage room contains ten dialysis machines, four
of which are defective. A nurse selects five of the machines at
random, thinking all are in working condition. What is the
probability that all five of the machines are not defective?
Example
Let Y be the number of non-defective machines found in the 5 selected ones,
Y ~ hyper-geometric dist., N = 10, k = 6, n = 5
0238.0
42
1
252
6
5
10
0
4
5
6
)5( ===


















==YP
Cont.Hypergeometric Random Variable

9/28/2018
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SBE 304: DRV - PMF
The finite population correction factor
The term in the variance of a hypergeometric random variable
( )N – 1
______N – n
is called the finite population correction factor
SBE 304: DRV - PMF
Joint Probability Distribution
In the study of probability, given two random
variables X and Y that are defined on the same
probability space, the joint distribution for X and Y
defines the probability of events defined in terms of
both X and Y. In the case of only two random
variables, this is called a bivariate distribution, but
the concept generalizes to any number of random
variables, giving a multivariate distribution.

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SBE 304: DRV - PMF
The joint probability mass function of the discrete
random variables X and Y, denoted as fXY (x, y), satisfies
0 ≤ fXY (x, y) ≤ 1 for all x and y
fXY (x, y) = 1
fXY (x, y) = P(X = x, Y = y)
∑
y
∑
x
Cont.
SBE 304: DRV - PMF
If X and Y are discrete random variables with joint probability
mass function fXY (x, y), then the marginal probability
mass functions of X and Y are
where Rx denotes the set of all points in the range of (X, Y) for
which X=x and Ry denotes the set of all points in the range of
(X, Y) for which Y=y
fX (x) = P(X = x) = fXY (x, y)∑
Rxand
fY (y) = P(Y = y) = fXY (x, y)∑
Ry
Cont.

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SBE 304: DRV - PMF
If the marginal probability distribution of X has the probability
mass function fX (x), then
where R denotes the set of all points in the range of (X, Y)
E(X) = μX = xfX (x) = xfXY (x, y)
and
V(X) = σX
2 = (x - μX)2fXY (x, y)
∑
R
∑
x
∑
R
Cont.
SBE 304: DRV - PMF
Given discrete random variables X and Y with joint probability
mass function fXY (x, y), the conditional probability mass
function of Y given X = x is
fY|x (y) = fXY (x, y) / fX (x) for fX (x) > 0
Cont.

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SBE 304: DRV - PMF
Cont.
For discrete random variables X and Y, if any one of the following
properties is true, the others are also true, and X and Y are
independent.
1. fXY (x, y) = fX (x) fY (y) for all x and y
2. fY|x (y) = fY (y) for fX (x) > 0
3. fX|y (y) = fX (x) for fY (y) > 0
4. P(X Є A, Y Є B) = P(X Є A)P(Y Є B)
for any sets A and B in the range of
X and Y, respectively
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Lec 2 discrete random variable

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Lec 2 discrete random variable