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Random Variable and Probability Distribution
Random Variable
Definition: If S is a sample space with a
probability measure and X is a real-valued
function defined over the elements of sample
space S, then X is called a random variable.
[ X(s) = x ]
Random Variable
and
Probability Distributions
Random variables are usually denoted by
capital letter.
X, Y, Z, ...
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Why Random Variable?
Discrete Random Variable
Example: (Toss a balanced coin)
X = 1, if Head occurs, and
X = 0, if Tail occurs.
Probability
Bar Chart
1/2
P(Head) = P(X = 1) = P(1) = .5
P(Tail) = P(X = 0) = P(0) = .5
Probability mass function:
f(x) = P(X = x) =.5 , if x = 0, 1,
and 0 elsewhere.
0
Lower case of letters are usually for denoting
a element or a value of the random variable. 2
1
• A simple mathematical notation to
describe an event. e.g.: X < 3, X = 0, ...
• Mathematical function can be used to
model the distribution through the use of
random variable. e.g.: Binomial, Poisson,
Normal, …
• Types of random variables:
– Discrete
– Continuous
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Probability Mass Function
(or Probability Distribution)
The probability mass function (p.m.f.) f(x) of a
discrete random variable X is a function that
satisfies the following properties:
a) 0 ≤ f(x) ≤ 1
b) SxS f(x) = 1 (Total probability equals 1.)
Discrete Distributions
Random Variables of
the Discrete Type
(Random variables that assume
discrete values by chance.)
* Some places use p(x) instead of f(x).
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Random Variable and Probability Distribution
Uniform Distribution
Checking Distribution
(Uniform Distribution)
If random variable X has a Discrete Uniform
Distribution over first m integers, x = 1, 2, …, m,
then X has a p.m.f. f(x) = 1 .
m
7
X=x
-100
f (x) = P(X = x) = 1/6, if x = 1, 2, 3, 4, 5, or 6,
and 0 elsewhere.
Actual Data
1
2
3
4
5
6
8
Is it a profitable insurance
premium?
f (x)
.1
.75
-1
.25
.50
11
?
.65
.25
Probability line chart
Random Variables of
the Continuous Type
-100 -1 11
Probability Mass Function
Distribution: f (-100) = .1
f (-1) = .25,
f (11) = .65,
.65 , if

f ( x)  .25 , if
 .1 , if

x  11
x  -1
x  -1009
Density Curve
Percent
Continuous Distributions
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Density Curve
f(x)
Percent
A smooth curve that fit
the distribution
f(x)
A smooth curve that fit
the distribution
Density
function, f (x)
Test scores
10 20 30 40 50 60 70 80 90 100 110
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Probability
Density
Function, f (x)
Shows the
probability
density for all
values of x.
Test scores
10 20 30 40 50 60 70 80 90 100 110
Probability density is not probability!
Use a mathematical model to describe the variable.
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Random Variable and Probability Distribution
Meaning of Area Under Curve
Continuous Distribution
Probability Density Function (p.d.f.) of a random
variable X of continuous type with a space S is an
integrable function, f(x), that satisfying the following
conditions:
Example: What percentage of the
distribution is in between 72 and 86?
f(x)
1. f (x)  0, x  S,
2. S f(x) dx = 1, (Total area under curve is 1.)
b
3. For a and b in S, P ( a  X  b )   f ( x ) dx
P(72  X  86)
a
f(x)
72 86
a b
x
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Meaning of Area Under Curve
X (Height)
P(X = 72) = 0, density is not probability.
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Uniform Distribution
Example: What percentage of the
distribution is in between 72 and 86?
P(72  X  86) = P(72 < X < 86)
f(x) = P(72  X <86) = P(72 < X  86)
The continuous random variable X has a
uniform distribution if its p.d.f. is equal
to a constant on its sample space, S. If tS
is the interval [a, b], then its p.d.f. is
f ( x) 
1
,
b-a
a  x  b.
It is usually denoted as U(a, b).
72 86
X (Height)
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f(x)
1
b-a
0
a
b
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Uniform Distribution
Example: An index score is uniformly
distributed between 4 to 8. What
percentage of the distribution is in
between 4 and 7?
Mean and Standard Deviation
of a Random Variable
¼ x (7 - 4) = 3/4
f(x)
(Mathematical Expectation)
1/4
4
5
6
7
8
X (Index)
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Random Variable and Probability Distribution
Expected Value
Expected Value
Empirical study: Play the game 1000
times, if head turns you win $1,
otherwise, you win $0. On average, how
much money do you win per game?
Outcome
Frequency
x
Empirical study: Play the game 1000
times, if head turns you win $1,
otherwise, you win $0. On average, how
much money do you win per game?
Relative Frequency
Average
f (x)
Head($1)
500
.5
Tail ($0)
500
.5
Total
1000
1
= (1 · 500 + 0 · 500) / 1000
= 500 / 1000 = .5
or
f (x)
= 1 · 500/1000 + 0 · 500/1000
= 1 · .5 + 0 ·.5 = .5
x
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Mathematical Expectation
Expected Value
Empirical study: Play the game 1000
times, if head turns you win $1,
otherwise, you win $0. On average, how
much money do you win per game?
If f (x) is the p.m.f. of a discrete random
variable X, with sample space S, the mean
(mathematical expectation) of X is
Outcome, x
Relative
Frequency, f(x)
Product, x·f (x)
mx = E[X] = SS x ·f(x)
Head($1)
.5
1 · .5
Tail ($0)
.5
0 · .5
and the mean (mathematical expectation)
of u(X) is
Total
1.0
.5
E[u(X)] = SS u(x) ·f(x)
S x·f (x)
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What is the probability
distribution of rolling a die?
Probability mass function:
(Uniform Distribution)
f (x) = P(X = x) = 1/6, if x = 1, 2, 3, 4, 5, or 6,
and 0 elsewhere.
x
f(x)
1
1/6
mx = E[X] = ?
2
3
4
5
6
1/6
1/6
1/6
1/6
1/6
1
1
1
E[ X ]  1  2   3 
6
6
6
1
1
1 21
 4  5  6 
 3.5
6
6
6 6
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Is it a profitable insurance
premium?
x
f(x)
f(x)
x·f(x)
f(x)
-100
-100
.1
.1
-100 ·.1
-1
-1
.25
.25
-1·.25
.75
11
.65
11·.65
.50
11
.65
Probability line chart
.25
-100 -1 11
The mean of the distribution is
E[X] = (-100) · .1+ (-1) · .25+ 11 · .65 = -3.1
(Weighted by probabilities.)
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Random Variable and Probability Distribution
Example
Example
Let random variable X have the p.m.f.
f (x) = 0.25 , x  S, where x = -1, 0, 1, 2,
find E[(X)].
Let random variable X have the p.m.f.
f (x) = 0.25 , x  S, where x = -1, 0, 1, 2.
Let u(x) = 3x, find E[u(X)].
Sol:
E[X] = 0.5
Sol:
E[X] = -1 ·f(-1) + 0 ·f(0) + 1·f(1) + 2 ·f(2)
E[u(X)]= u(-1)·f(-1) + u(0)·f(0) + u(1)·f(1) + u(2)·f(2)
= -1 ·.25 + 0 ·.25 + 1 ·.25 + 2 ·.25
=3(-1)·f(-1) + 3·0·f(0) + 3·1·f(1) + 3·2·f(2)
= 0.5
= -3 ·.25 + 0 ·.25 + 3 ·.25 + 6 ·.25
= 1.5
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Example
Let random variable X have the p.m.f.
f (x) = 0.25 , x  S, where x = -1, 0, 1, 2,
find E[X 2].
The Mean, Variance and
Standard Deviation of
Random Variable
Sol:
E[X2] = (-1)2·f(-1) + (0)2·f(0) + (1)2·f(1) + (2)2·f(2)
= 1·f(-1) + 0 ·f(0) + 1·f(1) + 4 ·f(2)
= 1 ·.25 + 0 ·.25 + 1 ·.25 + 4 ·.25
= 1.5
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Mean and Variance of a
Random Variable
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Measure of Center for a
Continuous Distribution
If f (x) is the p.m.f. of a discrete random
variable X, with space S, then the mean of
the random variable is
The mean value (expected value) of a
continuous random variable (distribution)
X, denoted by mX or E[X] or just m is
defined as
m = E[X] = SS x ·f(x)
and the variance of the random variable is

m X   x  f ( x ) dx
s 2 = E[(X - m)2] = SS (x - m)2 ·f(x)
-
Standard Deviation = s = ? E [( X - m ) 2 ]
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Random Variable and Probability Distribution
Measure of Spread for a
Continuous Distribution
A Short Cut formula for
Variance of a Random Variable
The variance of a continuous random
variable (distribution) X, denoted by sX2
or just s2 (or Var[X]) is defined as
s 2 = E[X 2] - m 2

s 2  E [( X - m ) 2 ]   ( x - m ) 2  f ( x ) dx
-
The standard deviation of X is
s  s 2  E [( X - m ) 2 ]
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Example
Example: Variance
x
Let random variable X have the p.m.f.
f (x) = 0.25 , x  S, where x = -1, 0, 1, 2,
find the variance of X
f(x)
x·
x·f(x)
f(x)
.75
-100
.1
-100 ·.1
.50
-1
.25
-1·.25
.25
11
.65
11·.65
Sol:
Probability line chart
-100 -1 11
E[X] = 0.5
E[X2] = 1.5
E[X] = (-100) · .1+ (-1) · .25+ 11 · .65 = -3.1
E[X2] = (-100)2 · .1+ (-1) 2 · .25+ 112 · .65 = 1078.9
Variance, s 2 = E[X 2] – (E[X])2
= 1078.9 – (– 3.1)2 = 1069.29
s2 = E[X2] - m2
= 1.5 – 0.52 = 1.5 – 0.25 = 1.25
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Linear Function of a
Random Variable
Example
Let random variable X have the p.m.f.
f (x) = 0.25, where x = -1, 0, 1, 2,
find the mean and variance of Y = 3X + 2
Let X be a random variable with mean mX and
variance sX 2, and let random variable Y be a
linear function of the X, and Y = aX + b, then
Sol:
mY  E[Y ]  E[aX  b]  a  E[ X ]  b  am X  b
mX = 0.5, sX2 = 1.25 from earlier example
mY = 3 x 0.5 + 2 = 3.5
s Y2  E[(Y - mY ) 2 ]  E[( aX  b - am X - b) 2 ]
sY2 = 32 x 1.25 = 11.25
 E[a 2 ( X - m X ) 2 ]  a 2s X2
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What is the standard deviation?
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