### 3. Random Variables

3. Random Variables
Let (, F, P) be a probability model for an experiment,
and X a function that maps every   , to a unique
point x  R, the set of real numbers. Since the outcome 
is not certain, so is the value X ( )  x. Thus if B is some
subset of R, we may want to determine the probability of
“ X ( )  B ”. To determine this probability, we can look at
the set A  X 1 ( B)   that contains all    that maps

into B under the function X.

X ( )
x
A
B
Fig. 3.1
R
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Obviously, if the set A  X 1( B) also belongs to the
associated field F, then it is an event and the probability of
A is well defined; in that case we can say
Probability of theevent" X ( )  B "  P( X 1 ( B)).
(3-1)
However, X 1( B) may not always belong to F for all B, thus
creating difficulties. The notion of random variable (r.v)
makes sure that the inverse mapping always results in an
event so that we are able to determine the probability for
any B  R.
Random Variable (r.v): A finite single valued function X (  )
that maps the set of all experimental outcomes  into the
set of real numbers R is said to be a r.v, if the set  | X ( )  x 
is an event ( F ) for every x in R.
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Alternatively X is said to be a r.v, if X 1( B)  F where B
represents semi-definite intervals of the form {  x  a}
and all other sets that can be constructed from these sets by
performing the set operations of union, intersection and
negation any number of times. The Borel collection B of
such subsets of R is the smallest -field of subsets of R that
includes all semi-infinite intervals of the above form. Thus
if X is a r.v, then
 | X ( )  x    X
x
(3-2)
is an event for every x. What about  a  X  b ,  X  a?
Are they also events ? In fact with b  a since {X  a}
and  X  b  are events,  X  a c   X  a  is an event and
hence  X  a   X  b  {a  X  b} is also an event.
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1


a

 X a 

n


Thus,
Consequently
is an event for every n.

1


a


X

a
  { X  a}
 
n

n 1
(3-3)
is also an event. All events have well defined probability.
Thus the probability of the event  | X ( )  x  must
depend on x. Denote
P | X ( )  x   FX ( x)  0.
(3-4)
The role of the subscript X in (3-4) is only to identify the
actual r.v. FX (x) is said to the Probability Distribution
Function (PDF) associated with the r.v X.
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Distribution Function: Note that a distribution function
g(x) is nondecreasing, right-continuous and satisfies
g ()  1, g ()  0,
(3-5)
i.e., if g(x) is a distribution function, then
(i) g ()  1, g ()  0,
(ii) if x1  x2 , then g ( x1 )  g ( x2 ),
(3-6)
and

g
(
x
)  g ( x), for all x.
(iii)
We need to show that FX (x) defined in (3-4) satisfies all
properties in (3-6). In fact, for any r.v X,
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(i)
FX ()  P | X ( )     P()  1
(3-7)
and
FX ()  P | X ( )     P( )  0.
(3-8)
(ii) If x1  x2 , then the subset (, x1 )  (, x2 ).
Consequently the event  | X ( )  x1   | X ( )  x2 ,
since X ( )  x1 implies X ( )  x2 .As a result
FX ( x1 )  P X ( )  x1   P X ( )  x2   FX ( x2 ),


(3-9)
implying that the probability distribution function is
nonnegative and monotone nondecreasing.
(iii) Let x  xn  xn1    x2  x1, and consider the event
since
Ak   | x  X ( )  xk .
(3-10)
 x  X ( )  xk   X ( )  x  X ( )  xk , (3-11)
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using mutually exclusive property of events we get
P( Ak )  Px  X ( )  xk   FX ( xk )  FX ( x).
But
 Ak 1  Ak  Ak 1 ,

lim Ak   Ak  
k 
(3-12)
and hence
and hence lim P( Ak )  0.
(3-13)
k 
k 1
Thus
lim P( Ak )  lim FX ( xk )  FX ( x )  0.
k 
k 
xk  x  , the right limit of x, and hence
But lim
k 
FX ( x  )  FX ( x),
(3-14)
i.e., FX (x) is right-continuous, justifying all properties of a
distribution function.
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(iv) If
FX ( x0 )  0
for some x0 , then FX ( x)  0, x  x0.
(3-15)
This follows, since FX ( x0 )  P X ( )  x0   0 implies X ( )  x0 
is the null set, and for any x  x0 ,  X ( )  x  will be a subset
of the null set.
(v) P X ( )  x   1  FX ( x).
(3-16)
We have  X ( )  x   X ( )  x   , and since the two events
are mutually exclusive, (16) follows.
(vi) P x1  X ( )  x2  FX ( x2 )  FX ( x1 ), x2  x1.
(3-17)
The events  X ( )  x1  and {x1  X ( )  x2} are mutually
exclusive and their union represents the event  X ( )  x2 .
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(vii) PX ( )  x   FX ( x)  FX ( x ).
(3-18)
Let x1  x   ,   0, and x2  x. From (3-17)
lim P x    X ( )  x   FX ( x )  lim FX ( x   ),
(3-19)
P X ( )  x   FX ( x)  FX ( x  ).
(3-20)
 0
 0
or
According to (3-14), FX ( x0 ), the limit of FX (x) as x  x0
from the right always exists and equals FX ( x0 ).However the
left limit value FX ( x0 ) need not equal FX ( x0 ). Thus FX (x)
need not be continuous from the left. At a discontinuity
point of the distribution, the left and right limits are
different, and from (3-20)
P X ( )  x0  FX ( x0 )  FX ( x0 )  0.
(3-21)
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Thus the only discontinuities of a distribution function FX (x)
are of the jump type, and occur at points x0 where (3-21) is
satisfied. These points can always be enumerated as a
sequence, and moreover they are at most countable in
number.
Example
3.1: X is a r.v such that
Solution:
For
FX (x).
X ( ) Find
c,  
.
so that
x and
c,  X (for
 )  x    ,
FX ( xso
)  that
0,
FX ( x )  1.
(Fig.3.2)
x  c, X ( )  x  ,
FX (x)
1
c
x
Fig. 3.2
Example 3.2: Toss a coin.   H ,T . Suppose the r.v X is
10
such that X (T )  0, X ( H )  1. Find FX (x).
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Solution: For x  0,  X ( )  x    , so that FX ( x)  0.
0  x  1,
x  1,
 X ( )  x    T , so that FX ( x)  P T   1  p,
 X ( )  x    H , T   , so that FX ( x)  1. (Fig. 3.3)
FX (x)
1
q
1
x
Fig.3.3
•X is said to be a continuous-type r.v if its distribution
function FX (x) is continuous. In that case FX ( x  )  FX ( x) for
all x, and from (3-21) we get PX  x  0.
•If FX (x) is constant except for a finite number of jump
discontinuities(piece-wise constant; step-type), then X is
said to be a discrete-type r.v. If xi is such a discontinuity
point, then from (3-21)
(3-22) 11
pi  PX  xi   FX ( xi )  FX ( xi ).
PILLAI
From Fig.3.2, at a point of discontinuity we get
P X  c   FX (c)  FX (c )  1  0  1.
and from Fig.3.3,
P X  0   FX (0)  FX (0 )  q  0  q.
Example:3.3 A fair coin is tossed twice, and let the r.v X
represent the number of heads. Find FX (x ).
Solution: In this case    HH , HT , TH , TT , and
X ( HH )  2, X ( HT )  1, X (TH )  1, X (TT )  0.
x  0,
X ( )  x  
 FX ( x )  0,
0  x  1, X ( )  x   TT   FX ( x )  P TT   P (T ) P (T ) 
1
,
4
1  x  2, X ( )  x   TT , HT , TH   FX ( x )  P TT , HT , TH  
x  2,
X ( )  x    FX ( x)  1. (Fig. 3.4)
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3
,
4
From Fig.3.4, PX  1 FX (1)  FX (1 )  3 / 4  1 / 4  1 / 2.
FX (x)
1
3/ 4
1/ 4
1
x
2
Fig. 3.4
Probability density function (p.d.f)
The derivative of the distribution function FX (x) is called
the probability density function f X (x) of the r.v X. Thus

f X ( x) 
dFX ( x )
.
dx
(3-23)
Since
dFX ( x )
F ( x  x )  FX ( x )
 lim X
 0,
x 0
dx
x
from the monotone-nondecreasing nature of FX ( x ),
(3-24)
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it follows that f X ( x)  0 for all x. f X (x) will be a
continuous function, if X is a continuous type r.v.
However, if X is a discrete type r.v as in (3-22), then its
f (x)
p.d.f has the general form (Fig. 3.5)
pi
X
f X ( x)   pi ( x  xi ),
(3-25)
i
xi
x
Fig. 3.5
where xi represent the jump-discontinuity points in FX (x).
As Fig. 3.5 shows f X (x) represents a collection of positive
discrete masses, and it is known as the probability mass
function (p.m.f ) in the discrete case. From (3-23), we
also obtain by integration
FX ( x)  
x

f x (u )du.
(3-26)
Since FX ()  1, (3-26) yields



f x ( x )dx  1,
(3-27)
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which justifies its name as the density function. Further,
from (3-26), we also get (Fig. 3.6b)
P x1  X ( )  x2
  FX ( x2 )  FX ( x1 )  x
x2
f X ( x)dx.
(3-28)
1
Thus the area under f X (x) in the interval
the probability in (3-28).
FX (x)
represents
f X (x)
1
x1 x2
( x1 , x2 )
x
(a)
x
x1 x2
(b)
Fig. 3.6
Often, r.vs are referred by their specific density functions both in the continuous and discrete cases - and in what
follows we shall list a number of them in each category.
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Continuous-type random variables
1. Normal (Gaussian): X is said to be normal or Gaussian
r.v, if
f X ( x) 
1
2
2
e ( x   )
2
/ 2 2
(3-29)
.
This is a bell shaped curve, symmetric around the
parameter  , and its distribution function is given by

FX ( x) 
x

1
2 2
e
( y   ) 2 / 2 2
x

dy  G
,
  
(3-30)
where G( x)   1 e dy is often tabulated. Since f X (x)
2
depends on two parameters  and  2 , the notation X  N ( , 2 )
will be used to represent (3-29).
f (x)
x
 y2 / 2

X

Fig. 3.7
x
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2. Uniform: X  U (a, b), a  b, if (Fig. 3.8)
1


, a  x  b,
f X ( x)   b  a

 0, ot herwise.
(3.31)
3. Exponential: X   ( ) if (Fig. 3.9)
1

 e x /  , x  0,
f X ( x)   

 0, ot herwise.
(3-32)
f X (x)
f X (x)
1
ba
a
Fig. 3.8
b
x
x
Fig. 3.9
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4. Gamma: X  G( ,  ) if (  0,   0) (Fig. 3.10)
f X (x)
 1
 x
x / 

e
, x  0,
f X ( x )   ( )  

0, ot herwise.

If   n an integer
(3-33)
x
Fig. 3.10
(n)  (n  1)!.
f X (x )
5. Beta: X   (a, b) if (a  0, b  0) (Fig. 3.11)
0
x
1
Fig. 3.11
1

a 1
b 1
x
(
1

x
)
,
0

x

1
,

f X ( x )    ( a , b)
(3-34)

0,
ot
herwise.

where the Beta function  (a, b) is defined as
1
 (a, b)   u a 1 (1  u)b1 du.
0
(3-35)
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6. Chi-Square: X   2 (n), if (Fig. 3.12)
1

x n / 21e  x / 2 , x  0,
 n/2
f X ( x )   2 ( n / 2 )
(3-36)

0,
ot herwise.

f X (x )
x
Fig. 3.12
Note that  2 (n) is the same as Gamma (n / 2, 2).
7. Rayleigh: X  R( 2 ), if (Fig. 3.13)
2
2
x

 2 e x / 2 , x  0,
f X ( x )  

 0, ot herwise.
f X (x)
(3-37)
x
Fig. 3.13
8. Nakagami – m distribution:
 2  m  m 2 m 1  mx 2 / 
x e
, x0



f X ( x )   ( m )   

0
otherwise

(3-38)
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f X (x )
9. Cauchy: X  C( ,  ), if (Fig. 3.14)
f X ( x) 
 /
  (x  )
2
2
,    x  .
x

(3-39)
Fig. 3.14
10. Laplace: (Fig. 3.15)
f X ( x) 
1 |x|/ 
e
,    x   .
2
(3-40)
11. Student’s t-distribution with n degrees of freedom (Fig 3.16)
f T (t ) 
( n  1) / 2  
t 
1  
n
n ( n / 2) 
2
f X ( x)
 ( n 1) / 2
,    t  .
fT ( t )
x
Fig. 3.15
(3-41)
t
Fig. 3.16
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12. Fisher’s F-distribution
{( m  n ) / 2} m m / 2 n n / 2
z m / 2 1
, z0

(mn) / 2
f z ( z)  
( m / 2) ( n / 2)
( n  mz )
(3-42)

0
otherwise

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Discrete-type random variables
1. Bernoulli: X takes the values (0,1), and
P( X  0)  q,
P( X  1)  p.
(3-43)
2. Binomial: X  B(n, p), if (Fig. 3.17)
 n  k n k
P( X  k )  
, k  0,1,2,, n.
k 
p q
 
(3-44)
3. Poisson: X  P( ) , if (Fig. 3.18)
P( X  k )  e 
k
k!
, k  0,1,2,, .
(3-45)
P( X  k )
P( X  k )
k
12
n
Fig. 3.17
Fig. 3.18
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4. Hypergeometric:
P( X  k ) 
m
 
k 
 
 N m 


 n k 


,
N 
 
n 
 
max(0, m  n  N )  k  min( m, n )
(3-46)
5. Geometric: X  g ( p ) if
P( X  k )  pqk , k  0,1,2,, ,
q  1  p.
(3-47)
6. Negative Binomial: X ~ NB (r, p), if
 k  1 r k  r
P( X  k )  
p q ,

 r 1
k  r, r  1,
.
(3-48)
7. Discrete-Uniform:
P( X  k ) 
1
, k  1,2,, N .
N
(3-49)
We conclude this lecture with a general distribution due 23
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to Polya that includes both binomial and hypergeometric as
special cases.
Polya’s distribution: A box contains a white balls and b black
balls. A ball is drawn at random, and it is replaced along with
c balls of the same color. If X represents the number of white
balls drawn in n such draws, X  0, 1, 2, , n, find the
probability mass function of X.
Solution: Consider the specific sequence of draws where k
white balls are first drawn, followed by n – k black balls. The
probability of drawing k successive white balls is given by
pW 
a
ac
a  2c
a  b a  b  c a  b  2c
a  (k  1)c
a  b  (k  1)c
Similarly the probability of drawing k white balls
(3-50)
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followed by n – k black balls is given by
pk  p w
b
bc
a  b  kc a  b  ( k  1)c
k 1
  a ab icic
i 0
n  k 1

j 0
b  (n  k  1)c
a  b  (n  1)c
b  jc
.
a b( j  k )c
(3-51)
Interestingly, pk in (3-51) also represents the probability of
drawing k white balls and (n – k) black balls in any other
specific order (i.e., The same set of numerator and
denominator terms in (3-51) contribute to all other sequences
n
as well.) But there are  k  such distinct mutually exclusive
 
sequences and summing over all of them, we obtain the Polya
distribution (probability of getting k white balls in n draws)
to be
k 1
n  k 1
P( X  k ) 
n
 
k 
 
pk 
n
 
k 
  i 0

a ic
a bic

j 0
b jc
,
a b( j  k )c
k  0,1,2,
, n.
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(3-52)
PILLAI
Both binomial distribution as well as the hypergeometric
distribution are special cases of (3-52).
For example if draws are done with replacement, then c = 0
and (3-52) simplifies to the binomial distribution
P( X  k ) 
where
n
 
k 
 
a
p
,
ab
p k qn k ,
k  0,1,2,
,n
(3-53)
b
q
 1  p.
ab
Similarly if the draws are conducted without replacement,
Then c = – 1 in (3-52), and it gives
P( X  k ) 
n
 
k 
 
a(a  1)(a  2) (a  k  1)
b(b  1) (b  n  k  1)
(a  b)(a  b  1) (a  b  k  1) (a  b  k ) (a  b  n  1)
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PILLAI
n!
a !(a  b  k )!
b!(a  b  n )!
P( X  k ) 

k !(n  k )! (a  k )!(a  b)! (b  n  k )!(a  b  k )!
a  b 

  
 k   n k 
  

 a b 


 n 


(3-54)
which represents the hypergeometric distribution. Finally
c = +1 gives (replacements are doubled)
P( X  k ) 
=
n
 
k 
 
( a  k  1)! ( a  b  1)! (b  n  k  1)! ( a  b  k  1)!
( a  1)! ( a  b  k  1)!
(b  1)! ( a  b  n  1)!
 a  k 1  b  n  k 1

 


 

k
n

k

 

 a  b  n 1




n


.
(3-55)
we shall refer to (3-55) as Polya’s +1 distribution. the general
Polya distribution in (3-52) has been used to study the spread
of contagious diseases (epidemic modeling).
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PILLAI