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Dr. Hugh Blanton
ENTC 4307/ENTC 5307
Random Variables
Random Variables
• Many random phenomena have outcomes that
are real numbers,
• e.g., the voltage, v(t) at time, t, across a noisy
resistor, number of people on a New York to
Chicago train, etc.
• In engineering, technology, and science; we are
generally interested in numerical outcomes.
• Even when the universal set, S, in not numerical,
we may apply a mapping to convert the outcomes
to real numbers.
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Definition of a Random Variable:
• A random variable is a number
labeling the outcomes of a
probabilistic experiments.
• X can be considered to be a function
that maps all the elements in S into
points on the real line or some parts
thereof.
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X:SR
Universal Set, S
X(.)
Mapping
Domain
Range
R (Real numbers)
Conditions:
The mapping is single-valued.
The set {X  x} is an event. This is the set of random
variable X taking values equal or less than x in a trial
chance experiment, E.
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Basic Definitions
• Discrete Random Variable: A random
variable that has a countable number
of elements in the range.
• Continuous Random Variable: A
random variable that has an
uncountably infinite number of
elements in the range.
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Random Variables
• The mapping (function) that assigns a
number to each outcome is called a
random variable.
• If the random variable is denoted by
X, then the distribution function F(xo)
is defined by
F ( xo )  Pr{ X  xo }
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Example 1:
Suppose you match coins with a friend, winning $1 if two
coins match and losing $1 if the coins do not match.
Example 1: S={HH, HT, TH, TT}
s1
s2
s3
s4
Random Variable: X(s1) = X(s4) = +1
X(s2 ) = X(s3) = -1
Thus,
X
1
-1
-1
1
S
HH
HT
TH
TT
 Single-valued mapping
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In this case, a random variable takes on only a finite
number of values (+1, -1), satisfying property c.
If we let x = 0.6, then X  0.6, if s = HT or TH, i.e., the
event {HT, TH}. Thus x = 0.6 determines an event.
Let x = -10, the {X  -10} = Ø
Let x > 1, then {X  x} = S
Thus, for every x, we have an event and b is satisfied.
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Basic Definitions
• Discrete Random Variable: A random variable
that has a countable number of elements in the
range.
• Continuous Random Variable: A random
variable that has an uncountably infinite number
of elements in the range.
• Probability Assignment: There are two standard
forms for probability assignment either using
Cumulative Distribution Function (CDF) or
Probability Distribution Function (PDF).
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Cumulative Distribution Function (CDF)
Let X : a random variable with a particular value, x, then,
FX(x) = Pr[X  x]
Thus, the CDF is the probability of event {X  x}, i.e., the
random variable, X, takes on a value equal to or less
than x.
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Example 2
Experiment: Observing the parity bit in a word in
computer memory.
Bit “ON”  X = 1
Bit “OFF”  X = 0
The OFF state has a probability q and thus the ON state
has a probability of (1-q).
Sample space, S = {OFF, ON}
Plot FX(x)
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Example 2
(1) For x  0, event X  x    FX ( x)  0
(2) For 0  x  1, event X  x is equivalent to the event
OFF
Thus, FX ( x)  PX  0  q
(3) For x  1, event X  x  OFF , ON   S
Thus, FX ( x)  1
FX(x)
Prob. of event
{X=1}
Prob. of event
{X=0}
q
q
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Example 3
Determine CDF for a single toss of a die.
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Example 3
S  1,2,3,4,5,6
For x  1
FX ( x)  Pr[ X  x]  0
1 x  2
FX ( x)  Pr[ X  x]  1/ 6
2 x3
FX ( x)  Pr[ X  x]  2 / 6
5 x 6
x6
FX ( x)  Pr[ X  x]  5 / 6
FX ( x)  Pr[ X  x]  1
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1
FX(x)
1/6
1
6
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Example 4
A random variable has a PDF given by
FX(x) = 0
= 1-e-2x
- < x  0
0<x
Find the probability that X > 0.5.
Find the probability that X  0.25
Find the probability that 0.3  X  0.7
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Example 4
(a) PrX  0.5  1  Pr[ X  0.5]  1  FX (0.5)
1
 1  (1  e )  0.3679
0.5


(b) Pr X  0.25  FX (0.25)  (1  e )  0.3935
(c) Pr0.3  X  0.7  FX (0.7)  FX (0.3)
 (1  e 1.4 )  (1  e 0.6 )  0.3022
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1
FX(x)
x
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Example 5
A random variable has PDF given by:
FX(x) = A(1-e-(x-1))
=0
1< x < 
-<x1
Find A for a valid CDF
FX(x) = ?
Pr[2 < X < ] = ?
Pr[1 < X  3] = ?
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Example 5
(a) Since FX() = 1,
 A [1
– e-]  A = 1
(b) FX(2) = [1 – e-1] = 0.6321
Pr[2 < X <  ] = FX() - FX(2) = 1 - 0.6321 = 0.3679
(c) Pr[1 < X  3 ] = FX(3) - FX(1)
= (1 – e-2) - (1 – e0) = 0.8647
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CDF or Discrete Random Variable:
A discrete random variable , X, taking on one of the
countable set of possible values x1, x2,  with
probability Pr[X = xk],  k[1,N] forming a stair-step CDF
with amplitude of each step being Pr[X = xk], k = 1, 2,
. Thus,
N
FX ( x) 
where,
 Pr[ X  x ]u( x  x )
k
k 1
k
x0
x0
1
u ( x)  
0
Or more compactly,
N
FX ( x) 
 Prx u( x  x )
k
k
k 1
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Example 6
A bus arrives at random in (0, T], i.e., 0 < t  T. Let X be
a random variable representing time of arrival, then
clearly,
FX(t) = 0
FX(T) = 1
for t  0
impossible event
certain event
Bus is uniformly likely to come at any time within (0,T].
t0
0
FX(t)
Then

FX (t )  t / T
1

0t T
t T
1
0
T
t
A continuous random variable has a continuous CDF.
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Probability Density Function (PDF)
A PDF is defined as
dFX ( x)
f x ( x) 
dx
Properties of PDF: If fX(x) exists, then
x
(1)
FX ( x) 

f X ( )d
i.e., CDF

(2) Pr[a  x  b]  FX (b)  FX (a)
b



a
f X ( )d 

b
f X ( )d 

Dr. Blanton - ENTC 4307 - Random Variables

f X ( )d
a
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(3) If a = - and b = , then



f X ( )d  FX ()  FX ()  1
(4) f X ( x)  0
decreasing
  x  
since CDF is non-
From (2), the probability that X takes on values
between x and x + x is
Pr[ x  X  x  x]  FX ( x  x)  FX ( x)
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x  x
FX ( x  x)  FX ( x) 

f X ( )d  f X ( x)x
x
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Generalization
For discrete random variables, the PDF has a general
N
form of
f X ( x) 
 Pr( x ) ( x  x )
k
k
k 1
Example 8: For a random variable, X, we have
 Ax(1  x)
f X ( x)  
0
0  x 1
otherwise
(a)Find A so that this function is a valid PDF.
(b) Find Pr[1/2  x 1].
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Example 8

(a)

1

f X ( x)dx  1  Ax(1  x)dx  1

0
1
1

A ( x  x 2 )dx  A ( x)dx  x 2 dx 


0
0
0

1


1
x
x 
A A A
 A       1
 2 3 0 2 3 6
2
A6

3
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Example 8 (cont.)
1
1

(b) Pr   x  1  6 x(1  x)dx
2
 1

2
1
1
 1

2
3 1

 
  6x 6x 
2
2

 6 ( x  x )dx   6 xdx  x dx   

3 1
1
 1
  2
1
2
2
 2
  2




 3x  2 x
2

3 1
1
2

3
2
1 1
 3   2   1 
4
8
2 2
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