Random Variables
and their Properties
4
4.1 Random Variables
4.2 Distribution Functions
4.3 Discrete Random Variables and Probability Mass Functions
4.4 Continuous Random Variables and Probability Density Functions
4.5 Mathematical Expectations
61
Learning Objectives:
It is expected that you will be able to do the following:
1. Identify different types of random variables and probability
distributions.
2. Calculate probabilities using the probability function and cumulative
distribution function of a random variable.
3. Determine the mean and variance of a random variable.
4. Apply the Chebyshev’s inequality and the law of large numbers.
4.1 Random Variables
A random experiment is defined earlier as any process that generates a
set of outcomes. Usually, the interest is not on the details of each of these
outcomes, but only in some quantitative characteristic or description of the
outcome. For instance, in tossing a fair coin three times, one may be
interested only with the number of heads, so that to each possible outcome in
the sample space may be associated a numerical value of 0, 1, 2, or 3. These
random quantities may be thought of as the possible values a random
variable X may assume, and depends on the outcome of the experiment.
Definition 4.1
Random Variable
For a given probability space
(,A ,P()), a random variable, denoted by X or X() is a function with
domain  and counterdomain or range the real line.
62
Fig. 4.1
The random variable X
A random variable X is a single-valued real function that assigns a
real number called the value of X() to each sample point  of . The
sample space  is termed the domain of the random variable X, and the
collection of all numbers R is termed the range or counterdomain of the
random variable X. Thus the range of X is a certain subset of the set of all
real numbers. X is a function since two or more different sample points
might give the same value of X(), but two different numbers in the range
cannot be assigned to the same sample point.
Since each value of X corresponds to an event or subset of the sample
space, then in set notation, we can define the event (X=x) as
(X=x) = {| X() = x}.
Similarly, the following events can be defined:
(X<x) = {| X()< x}
(X>x) = {| X()> x}.
(x1 < X < x2) = {| x1 < X() < x2}.
Consider the example of tossing a fair coin three times, the sample
space  = {HHH,HHT,HTH,THH,HTT,THT,TTH,TTT} is finite with
equally likely sample points. If X is the random variable which is equal to
the number of heads in the outcome, then the event defined by X = 2 is
63
A = (X = 2) = {| X() = 2} = {HHT, HTH, THH}
and P(A) = P(X = 2) = 3/8.
Also, the event defined by X < 2 is
B = (X<2) = {| X() < 2} = {TTT, TTH, THT, HTT}
and P(B) = P(X<2) = 4/8 = 1/2
Different random variables may be defined on the same sample space.
The choice of the random variable depends on the kind of information
sought. In the same experiment of tossing a fair coin three times, other
random variables may be defined such as Y equal to say 1 if the outcome has
a head and 0 if there is no head and Z equal to the number of tosses until a
head appears. The values of the random variables X, Y and Z are defined in
table 1.
Table 1 Values of random variables X, Y and Z
Sample point
Values of X
Values of Y
HHH
3
1
HHT
2
1
HTH
2
1
THH
2
1
HTT
1
1
THT
1
1
TTH
1
1
TTT
0
0
Values of Z
1
1
1
2
1
2
3
0
Example 1
In an experiment of tossing a coin three times, X, defined as the number
of heads in the sequence is a random variable. Use the notations of
functions to define the values of X
64
Solution
Let (i,j,k) denote the number of head in ith, jth, and kth toss of the
coin.
X(i,j,k) = i+j+k
for
i=0,1; j=0,1; k=0,1
///
Example 2
In an experiment of tossing two dice (one red, one green), the following
are examples of random variables. Define each using the notations of
functions.
a. X=the number of dots on the red die
b. Y=sum of the dots on the upturned faces of the two dice
c. Z=absolute difference of the results of the two dice
Solution
Let (i,j) denote the outcome, where i is the number of dots on the fred
die and j is the number of dots on the blue die.
X(i,j) = i
Y(i,j)=i+j
Z(i,j)=|i-j|
for i=1,2,3,4,5,6
for i=1,2,3,4,5,6 ; j= 1,2,3,4,5,6
for i=1,2,3,4,5,6 ; j= 1,2,3,4,5,6
///
Example 3
In the enrollment of CFY students, the following are random variables:
a. number of units enrolled by a student
b. total amount paid by a student
c. length of time a student was able to finish enrolling
d. weight in kg of a student
e. height in meters of a student
///
4.2 Distribution Functions
The most important function associated with the random variable X is
the cumulative distribution function usually denoted by FX() or F().
65
Definition 4.2
Cumulative Distribution Function The cumulative
distribution function of a random variable X, denoted by Fx(), is
defined to be that function with domain the real line and
counterdomain the interval [0,1] which satisfies
Fx(x) = P(X < x) = P({:X() < x})
for every x  R
Properties of a Cumulative Distribution function:
(i) 0 ≤ F (x) ≤ 1 for all x  R
(ii) F is a monotone, nondecreasing function, that is, if x1 < x2, then
F(x1) ≤ F(x2)
(iii) F is right continuous, that is, lim Fx (x  h)  Fx (x)
0h0
(iv) lim F (x)  0 and lim F (x)  1
x  
x 
Example 4 A fair die is thrown once with the number of dots on the top
face being of interest, so that the sample space
 = {1, 2, . . . , 6} with each element being equally likely to
occur. If we let the random variable X to be this value, then
X() =  and the cumulative distribution function Fx() is given
below:
0
1/6

2/6

Fx ( x )  P(X  x)  ...
...

5/6
1

x 1
1  x  2,
2x3
5  x  6,
6 x
Graphically,
66
Fig. 4.2
It can be verified that all properties mentioned above are satisfied by
the F(x) of this example.
///
4.3 Discrete Random Variables
and Probability Mass Functions
A random variable may be discrete or continuous. It is discrete if the
range of X is countable. If a random variable X is discrete, then its
corresponding cumulative distribution function F() is also discrete, and it
has an associated probability mass function f() which gives the probability
of each numerical value that the random variable may assume.
Definition 4.3 Probability Mass Function of a discrete random variable
If X is a discrete random variable with distinct values x1, x2, … xn, …
then the probability mass function of X, denoted by fx() is defined by
P(X  x i )
f x (x)  
0
if x  x i , i  1,2,..., n,...
if x  x i
Properties of a Probability Mass Function Any function fx() with domain
the real line and counterdomain [0,1] is defined to be a probability
mass function if for some countable set x1, x2, … , xn ,,,,
(i)
(ii)
f(xi) > 0 for i = 1,2,…
f(xi) = 0 if x  xi ; I = 1, 2, …
67
(iii)
 f(x
i
)  1 where the summation is over the points x1,x2 , … , xn, …
Theorem 4.1
Let X be a discrete random variable.
determined from fx(), and vice versa.
Proof:
Fx() can be
Suppose fx() is given, then Fx(xi) can be determined by
Fx(x) =
f
x
(x i )
x i x
Conversely, suppose Fx() is given, then fx(xi) can be
determined by
fx(xi) = Fx(xi) – lim Fx(x i - h)
0 h 0
Also, fx(x) = 0 for all x  xi, i = 1,2, . . . so fx(x) is
determined for all real numbers.
Example 5 Determine the corresponding probability mass function fx() of
the cumulative distribution Fx() in example 4.
Solution:
The graph of Fx() shows jumps at xi = 1,2,3,4,5,6. At this
points, by fx(xi) = Fx(xi) – lim Fx(x i - h) , we can obtain fx(xi) = 1/6. For
0 h 0
values of x other than these, the value of fx(x) = 0. Thus,
1/6
f x (x)  
0
x  1,2,3,4,5,6
otherwise
///
Example 6
Consider the example of tossing a fair coin three times where X
denotes the number of heads.
(a) Find and sketch the cumulative distribution function of X.
(b) Find the probability of obtaining at least 2 heads.
68
Solution:
(a) The following table can be constructed for X
X
-1
0
1
2
3
4
5
(X=x)

{TTT}
{HTT,THT,TTH}
{HHT,HTH,THH}
{HHH}


fx(x)
0
1/8
3/8
3/8
1/8
0
0
Fx(x)=P(X<x)
0
1/8
4/8
7/8
1
1
1
Fig. 4.3
Solution (b) P(at least 2) = P(X=2 0r X=3)
= P(X=2) + P(X=3)
= f(2) + f(3)
= (3/8) + (1/8)
= 1/2
Example 7 Let X be a discrete random variable whose possible values are
0,3, and 5. Given the probability mass function of X, find the
cumulative distribution function of X.
69
if x  0
1 / 6

f(x) = 1/2
1/3

x3
x 5
Solution
By definition, F(x) = P(X< x).
First, determine F(x) for values of x < 0. Since none of the
possible values of X are less than 0, then
F(x) = P(X< x) = 0,
for x<0.
Next, determine F(x) for values 0 < x < 3. Since 0 is the only
value of X that is less or equal to any value x in this
interval, then
F(x) = P(X<x) = P(X = 0) = f(0) = 1/6 for 0 < x < 3
Next, determine F(x) for values 3 < x < 5. Since 0 and 3 are the
only values of X that is less or equal to any value x this interval,
then
F(x) = P(X<x) = P(X=0) + P(X=3)
= f(0) + f(3) = 1/6 + 1/2
= 2/3
for 3 < x < 5
To cover the entire set of real numbers, for the interval 5 < x,
since 0, 3, and 5 are the only values of X that is less or equal to
any value x in this interval, then
F(x) = P(X<x) = P(X=0) + P(X=3) + P(X=5)
= f(0) + f(3) + f(5) = 1/6 + 1/2 + 1/3
=1
for 5<x
To conclude, the F(x) for - < x <  is as follows:
0
1 / 6
F(x) = 
2/3
1
for x  0
0 x  3
3 x5
5 x
70
4.4 Continuous Random Variables and Probability
Density Functions
A random variable X is called continuous if there exists a function
x
fx() such that Fx(x) =
f
x
( u )du for every real number x. The cumulative

distribution function Fx() of a continuous random variable is also
continuous.
Definition 4.4 Probability Density Function of a continuous Random
Variable. If X is a continuous random variable, the function fx() in
x
Fx(x) =
f
x
( u )du is called the probability density function of X.

Properties of a Probability Density Function
Any function f() with
domain the real line and counterdomain [0,) is defined to be a
probability density function if
(i) f(x) > 0 for all x

(ii)
 f(x)dx
1

Example 8 (a) Verify if fx(x) =
x 1
I [ 2 , 4 ] ( x ) is a probability density function
8
and (b) solve for Fx(3).
Solution for (a):
(ii)
It can be verified that for 2<x<4, fx(x) =
x 1
>0
8
and by definition fx(x) = 0 if x[2,4], thus property 1 is
satisfied.

(iii)
x 1
 8 dx =
=
x 1
1
2 8 dx = 8 2 (x  1)dx
1 164
 2  4  2 = 1
8
4
4
=
4
1 x2
( 2  x)
8
2
71
Thus, fx(x) =
x 1
I [ 2 , 4 ] ( x ) is a probability density function.
8
Solution for (b):
x 1
dx =
8

3
Fx(3) = 
=
3
1 x2
x 1
=
(

x
)
dx
2
2 8
8
2
1 94
 2  3  2 = 7/16
8
3
///
Example 9 Let X be a continuous random variable with probability density
function
2
x for 1  x  2
f(x) =  3
0
elsewhere
Find the cumulative distribution function of X.
Solution
x
By definition: F(x) =
f
x
( u )du

x
First, for x < 1,
Next, for 1<x<2
F(x) =
F(x) =
 f( u)du =
x
 0  du = 0


x
1
x

1
2
 f( u)du =  0  du +  3 udu

2  x  1


3  2 
2
=0+
x2 1
=
3
x
Then for x>2
1
2
x
2
F(x) =  f( u )du =  0  du +  udu +  0  du
3


1
2
= 0+1+0
=1
72
Therefore, the cumulative distribution function of X is
for x  1
0
 2
x 1
F(x) = 
 3
1
1 x  2
x2
///
Rules for finding probabilities for Continuous Random Variables
The basic formula for finding probabilities for continuous random
x
variables is the statement P(X<x) =
 f(t)dt .
It can be expanded to include

the following rules:
1. P(X=x) = 0 for - < x < 
x
2. P(X<x) = P(X< x) =
for - < x < 
 f(t)dt

b
3. P(a<X<b) = P(a<X<b) = P(a<X<b) = P(a<X<b) =
 f(t)dt
for a<b
a
a
In number 1, P(X = a) = P(a< X< a) =  f(t)dt = 0.
a
Example 10
Let X be a continuous random variable with probability
density function
2
x for 1  x  2
f(x) =  3
0
elsewhere
Find (a) P(X<1.5)
(b) P(X>1.5)
(c) P(1.5<X<1.8)
(d) P(X>4)
Solution:
1.5
2
2 x2
(a) P(X<1.5) =  x dx 
3
3 2
1
1.5
1
x2
=
3
1 .5
=
1
(1.5) 2  12
= 0.4167
3
(b) P(X>1.5) = 1 - P(X<1.5) = 0.5833
73
x2
(c) P(1.5<X<1.8) = P(X<1.8) - P(X<1.5) =
3
1 .8
- 0.4167
1
= 0.7467 - 0.4167 = 0.3300
(d) P(X>4) = 0
///
4.5 Mathematical Expectations
4.5.1 Mean and Variance of a Random Variable
To achieve reasonable simplicity, it is often necessary to describe
probability distributions rather summarily by a few “typical values.” Among
these typical values used to summarize a probability distribution, the mean
and the variance are most often used. The mean is a measure of central
tendency or average, while the variance is a measure of dispersion of the
values of the random variable. These two measures do not uniquely identify
a probability distribution, nevertheless, they are simple and useful
summaries of the probability distribution of the random variable X.
Definition 4.5
Expected value of a Random Variable
Let X be a
random variable. The expected value or the mean of X, denoted by x
or E[X] is defined by
(i)
(ii)
x f (x )
E[X] =  xf (x )dx
E[X] =
i x
i


x
if X is discrete with values x1,x2,…,xi, …
if X is continuous with probability
density function fx(x).
Example 11 The mean of the random variable X, the number of heads in
tossing a fair coin three times is
x = E[X] =
x f
i x
1
3
3
1
12
= 1.5
( x i ) = 0 +1 + 2 +3 =
8
8
8
8
8
///
74
Example 12
The mean of the continuous random variable X with
x 1
I [ 2 , 4 ] ( x ) is
8
4
4
x 1
1
= x
dx =  ( x 2  x )dx
8
82
2
probability density function fx(x) =
x = E[X]
=



xf x ( x )dx
4
1  x3 x2 
1 64  8 16  4 

=    = 
 = 3.083
2
8 3
8 3
2 
2
///
Theorem 4.2
Properties of Expected Value
(1) E[c] = c , c is a constant
Proof
There is only one value for the random variable c.
Therefore, E[c] = cP(c = c) = c1 = c
(2) E[cX] = cE[X] , c is a constant
Proof
E[cX] =  cx f X ( x )dx = c  x f X ( x )dx = cE[X]
(3) E[aX + b] = aE[X] + b, a and b are constants
Proof
E[aX + b] =  (ax  b)f X ( x )dx
=  ax f X ( x )dx +  b f X ( x )dx
= a  x f X ( x )dx + b f X ( x )dx
= aE[X] + b
Definition 4.6
Variance of a Random Variable Let X be a random
variable and let x be E[X]. The variance of X, denoted by  x2 or
Var[X] is defined by
 (x - 
Var[X] =  (x - 
(i) Var[X] =
(ii)
i
) 2 f x (x i )
x
) 2 f x (x)dx if X is continuous with probability density


if X is discrete with values x1, x2,…, xi, …
x
function fx(x).
75
Theorem 4.3
Properties of the Variance
(1) Var[X] = E[X2] – (E[X])2
Proof
Var[X] = E[(X-)2] = E[X2-2X + 2]
= E[X2] - 2E[X] + 2
= E[X2] - 22 + 2
= E[X2] - 2 = E[X2] – (E[X]) 2
(2) Var[aX + b] = a2Var[X]
Proof
Var[aX + b] = E[(aX+b)2] – ( E[aX + b])2
= E[a2X2 + 2abX + b2] – (aE[X] + b)2
= a2E[X2] +2abE[X] + b2 – [a2(E[X])2+ 2ab(E[X]) + b2]
= a2(E[X2] – (E[X])2)
= a2Var[X]
Definition 4.7 Standard Deviation of a Random Variable. If X is a
random variable, the standard deviation of X, denoted by  x , is
defined as the positive square root of the variance:  x = + var[ X]
Example 13 The variance of the random variable X, the number of heads in
tossing a fair coin three times is
-  x ) 2 f x (x i ) =  (x i - 1.5) 2 f x (x i )
1
3
3
1
= (0-1.5)2  +(1-1.5)2 + (2-1.5)2 +(3-1.5)2
8
8
8
8
 x2 = var[X] =
 (x
i
= 0.75
and its standard deviation is  x = 0.75 = 0.866
///
Example 14
Find the variance and standard deviation of the
continuous random variable X with probability density function
fx(x) =
x 1
I[ 2,4 ] ( x ) .
8
76
Solution: Recall that the mean of this random variable is E[X] =
3.083.
E[X2] =



x 2 f x ( x )dx
x 1
x3  x2
=x 
dx = 
dx
8
8
2
2
4
4
2
4
1  x4 x3 
1 256  16 64  8 

=    = 

3
8 4
8
4
3 
2
=
1
56 
 60   =9.833
8
3
Therefore Var[X] = E[X2] – (E[X]) 2
= 9.833 - (3.083)2 = 0.3281
///
Often, we are interested not only in the expected value of a random
variable but also in the expected value of a function of a random variable.
For example, if X denotes the number of items sold, our interest might be on
expected value of the gross sale (25X), where the number 25 is the selling
price per item. Here, the gross sale is a function of the random variable X.
Definition 4.8 Functions of a Random Variable Let X be a random
variable defined on the sample space . Let g be a real valued
function defined for values x that are possible values of X. The
random variable Y = g(X) is defined on  such that for each   ,
and for a given y,
(Y< y) = [g(X) < y] = (X  Dy)
where DY is a subset of Rx , and DY = {x | g(x) < y}.
If Y = g(X) is a function of a random variable X, then Y is also a
random variable since it provides a numerical value y = g(x) for each
possible outcome in the sample space, since every outcome also defines a
numerical value x for X. If X is discrete, then Y is also discrete and its
probability mass function can be calculated using the probability mass
77
function of X. To obtain fY(y) for any y, the probabilities of all values of x
such that g(x) = y is added, that is,
fY(y) =
f
X
{x | g(x)  y}
(x) .
Similarly, the cumulative distribution function of Y can be calculated as
Fy(y) = P(Y< y) = P[g(X) < y] =
f
X
{x | g(x)  y}
(x)
Example 15 Let X be a random variable with probability mass function
given by
1/7
if x is an integer in the range [-3,3],

otherwise
fx(x) = 
0
Define Y = |X|. (a) Determine the values of Y
(b) Determine the probability mass functions fY(y).
Solution
The possible values of Y are y = 0,1,2,3.
There is only one value of X that corresponds to y = 0, namely x = 0.
Thus
fY(0) = fX(0) = 1/7.
Also, there are two values of X that corresponds to each value of
y=1,2,3. For instance,
fY(2) = fX(-2) + fX(2) = 2/7
Thus the probability mass function for Y is
1/7
fY(y) = 2/7
0

if
y0
y  1,2,3
otherwise
///
78
Definition 4.9 Mathematical Expectation of a Function of a Random
Variable
The mathematical expectation of the random variable Y = g(X) is the
quantity
(i)
E(Y) = E[g(X)] =  g(x i )f X ( x i )
if X is discrete
i
(ii)
E(Y) = E[g(X)] =



g(x)f x (x)dx
if X is continuous
provided that the series converges absolutely (for the discrete case)
and the integral is defined (for the continuous case).
Theorem 4.4
Properties of Expected Value of a Function of a
Random Variable
(1) E[cg(X)] = cE[g(X)] , c is a constant
Proof
E[cg(X)] =  c  g(x )f X ( x )dx = c  g(x )f X ( x )dx = cE[g(X)]
(2) E[g1(X) + g2(X)] = E[g1(X)] + E[g2(X)]
E[g1(X) + g2(X)] =  [g1 (x )  g 2 (x )]f X ( x )dx
Proof
=  g1 (x )f X ( x )dx +  g 2 (x )f X ( x )dx
= E[g1(X)] + E[g2(X)]
Example 16 Let X be a random variable with E[X] = 1 and E[X(X-1)] = 4.
Find (a) Var[X] and (b) Var[2-3X].
Solution
Given:
E[X] = 1
E[X(X-1)] = E[X2-X] = E[X2] – E[X] = 4
Thus E[X2] = 4+E[X] = 4+1 = 5
79
(a) Var[X] = E[X2] – (E[X])2 = 5 – 12 = 4
(b) Var[2-3X] = Var[2] +(-3)2Var[X]
= 0 + 9(4)
= 36
///
4.5.2 The Chebyshev’s Inequality
Theorem 4.5 Chebyshev’s Inequality Let X be a random variable with an
expected value E[X] and variance Var[X]. Then for any constant c>0,
E[ X 2 ]
.
c2
P(|X| > c) <
Proof: For the discrete case:
Using the definition of variance of the random variable X,
Var[X] = E[(X – E[X])2]
= E[X2] – (E[X])2
Observe that
E[X2] =
 X f (x)
2
x
where summation is taken over all the values of X.
Since
 X f (x) =  X f (x )
2
2
{ x :|x | c}
x
it follows that
 X f (x)
2
>
>
 c f (x)
= c2
2
2
{x :|x | c}
2
{ x :|x | c}
2
 X f (x)
{ x :|x | c}
Now,
 X f (x)
 X f (x)
{ x :|x | c}
x
and
+
 c f (x) .
2
{x :|x | c}
 f (x ) = c2 P(|X|>c)
{x :|x | c}
Therefore
E[X2] > c2 P(|X|>c)
or equivalently
E[ X 2 ]
P(|X|>c) <
.
c2
80
Corollary 4.6 Let X be a random variable with finite mean x and finite
variance  x2 . For constants d > 0 and k > 0 we have
 2x
d2
(a)
P(|X - x| > d) <
(b)
P(|X - x| > k  x ) < 1 -
1
k2
Proof: Part of the exercise.
Example 17
Let X be a random variable denoting the measurements
in degree Celsius of the temperature of a material using a new device.
Suppose the mean of X is x and its standard deviation is x = 0.5, use
the Chebyshev’s inequality to estimate the probability that the new
device will provide a measurement that is within 1 degree of the
actual temperature x of the material.
Solution: Given: x = 0.5 and d = 1
We are asked to estimate P(|X - x| < 1)
By Corollary 4.6 we have
P(|X - x| > d) <
 2x
d2
From which we can obtain that
1 - P(|X - x| < 1) <
 2x
d2
It follows that
 2x
1- 2 <
d
P(|X - x| < 1) > 1-(0.5)2
P(|X - x| < 1) > 0.75
Thus
4.5.3
P(|X - x| < 1)
///
Law of Large Numbers
Theorem 4.7 The Law of Large Numbers Given a random experiment
with sample space . Let X be a random variable defined on  with
expected value  X and variance  2X . If the random experiment is
repeated n times under identical conditions and xi is the value of the
random variable X at the ith performance of the random experiment,
and let X =
X 1  X 2  ...  X n
, then for any  > 0
n
81
lim P (| X -  X |  )  1
n
The law of large numbers can be used to answer some practical
questions. It implies that if n is large, there is a great probability that the
value of X will be approximately equal to the mathematical expectation
E[X].
X is a random variable with mean  X and variance
 2X
n
as will be
discussed in chapter 7. We will use these properties of X and the
Chebyshev inequality to prove the Law of Large Numbers.
Proof:
Using the properties of X and applying the Chebyshev
inequality (corollary 4.6), we obtain
P(| X -  X | > ) <
 2X
.
n 2
Equivalently, this is
 2X
n 2
2
P(| X -  X | < ) > 1 - X2
n
1 - P(| X -  X | < ) <
Taking limits on both sides as n approaches infinity will give us the
result
lim P (| X -  X |  )  1
n
Activities:
1. Use the notation of functions to define the random variables
enumerated in Example 3. Indicate your assumptions on the possible
values of these random variables.
2. An urn contains three balls numbered 1,2, and 3. A ball is drawn
from the urn, and then a coin is tossed as the number shown on the
drawn ball. Use a table of values to define X as the number of heads
in the tosses.
82
3. Two dice are rolled. Let X be the sum of the values of the two dice.
(a) Using a table of values, define the random variable X.
(b) Give the probability mass function of X in tabular form.
(c) Include the cumulative probability distribution of X in the same
table.
(d) Use your answers in (b) and (c) to find the probability of
obtaining (i)
a sum less than 9;
(ii) a sum that is greater than 4 but less than 11;
(iii) a sum that is at least 3.
4. Check whether the following functions satisfy the conditions of a
discrete probability mass function. Give the reasons for your answers.
(a) f(x) = 1-x
(b) f(x) = 1/x
(c) f(x) = 1/5
for x = 0,1/2,3/2
for x = 1,2,3,4
for x = -4,2,6,8,9
5. Let X be a discrete random variable whose cumulative distribution
function is
0
1/6

F(x) = 1/4
1/2

1
for x  - 5
-5 x 3
3 x  7
7  x  10
10  x
(a) Find P(X< 6), P( - 3 < x < 6), P(X = -5), P(X =6)
(b) Find the probability mass function of X.
6. A softdrink dispenser is programmed to pour 16 ounces of soda into
plastic glasses. Sometimes it pours more, and sometimes less. The
actual amount in ounces over or under 16 is a random variable X
whose probability density function is
83
3
f(x) =  4
(1  x 2 )

0
for - 1  x  1
otherwise
Find the probability that a glass of soda poured by this machine will
be
(a) at least 0.5 ounces overfilled
(b) between 0.2 to 0.5 underfilled
(c) neither underfilled nor more than 0.5 ounces overfilled.
(d) neither overfilled nor less than 0.2 underfilled
7. Let X be a continuous random variable with probability density
function
1
x for 1  x  3
f(x) =  4

elsewhere
0
Find (a) the cumulative distribution function of X.
(b) P(X<1.2)
(c) P(X>1.2)
(d) P(1.2<X<1.8)
(e) P(X>3)
8. Let X be a continuous random variable with probability density
function
 4 x
f(x) = 4x
0

for - 1/2  x  0
0  x  1/2
elsewhere
Find the cumulative distribution function of X.
9. Let X be a discrete random variable with probability function
3 / 7
f(x) = 2/7
2/7

Find (a) E[X]
if x  4
x 8
x7
(b) E[X2]
(c) Var[X] and (d) E(X2 – 2X + 7)
84
10. Let X be a continuous random variable with probability density
function given by
2 x
if 0  x  1

otherwise
f(x) = 
0
If g1(x) = x + 5, g2(x) = x2 and g3(x) = 1-x,
Find E[g1(x) +2g2(x) -3 g3(x) -5]
11. Prove Corollary 4.6
85