Mathematical Statistics I Muzammil Tanveer
Mathematical Statistics I Muzammil Tanveer
Mathematical Statistics I Muzammil Tanveer
Muzammil Tanveer
mtanveer8689@gmail.com
0316-7017457
Dedicated
To
My Honorable Teacher
&
My Parents
1
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Lecture # 01
Statistics:
Statistics is defined as a science which deals with the collection of facts and
data and the drawing conclusions or inference from this data by applying
scientific methods OR It is defined as a science of estimates and probabilities.
Branches: There are two branches of statistics
(i) Inferential Statistics (ii) Descriptive Statistics
Frequency:
The number of times a thing occur in a definite period or interval of time is
called frequency of that thing.
Frequency Distribution:
The process of grouping the data into classes or groups and then determining
the frequency of each class is called the frequency distribution.
Example:
Make a grouped frequency distribution to the weights recorded to the nearest
grams of 60 apples picket out at from a consignment.
106 107 76 82 109 107 115 93
187 95 123 125 111 92 86 70
126 68 130 129 139 119 115 128
100 186 84 99 113 204 111 141
136 123 90 115 98 110 78 185
162 178 140 152 173 146 158 194
148 90 107 181 131 75 184 104
110 80 118 82
Solution: R = Range = Max Value Lowest Value
= 20468 = 136
Class Interval = h =20
2
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Classes Tally Frequency
65—84 IIII IIII 9
85—104 IIII IIII 10
105—124 IIII IIII IIII II 17
125—144 IIII IIII 10
145—164 IIII 5
165—184 IIII 4
185—204 IIII 5
∑ = 60
Measure of central Tendency or Averages:
An average is a single value which is intended to represents a set of data or a
distribution as a whole. It is a less or more central value around which the
observations tends to cluster such a central value is called the measure of central
tendency as it indicates the central position of the distribution. It is also known
as measure of location or measure of position.
Types of Averages: There are five types of averages.
(i) Arithmetic Mean (ii) Geometric Mean
(iii). Harmonic Mean (iv) The Median
(v). The Mode
(i) Arithmetic Mean:
It is defined as the value obtained by dividing the sum of observation by their
⋯ ∑
number. = =
⋯ ∑
For grouped data = = ∑
⋯
∑ = 60 = 7350
3
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Lecture # 2
Geometric Mean: The Geometric mean G of the set n positive values
, ,…, is defined as the nth root of their product.
G= . . …
Taking log on both sides
/
log G = log ( . . … )
= log( . . … )
= [ + + + ⋯+ ] ∑
G = anti-log ∑
log G = [ 45 + 32 + 37 + 46 + 39 + 36 + 41 + 48 + 36]
log G = 1.59856
G = anti-log (1.59586)
G = 39.68
Example: Find G.M of data given below
(Classmarks log G = anti-
Weight (g) log
or midpoint)
16.8498 log ∑
65--84 9 = 74.5 1.8722 ∑
4
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
The Harmonic Mean:
The Harmonic mean of a set of n values is defined as the reciprocal of the
arithmetic mean of the reciprocal of the values.
∑
H.M = A.M =
∑
H.M = = = 19.15
.
Weight (Classmarks 1
(g) or midpoint)
0.12078
65—84 9 = 74.5 0.01342
85—104 10 94.5 0.01058 0.1058
105—124 17 114.5 0.008734 0.165478
125—144 10 134.5 0.007435 0.07435
145—164 5 154.5 0.006472 0.3236
165—184 4 174.5 0.005731 0.022924
185—204 5 194.5 0.005141 0.025705
∑ = 0.547397
∑ = 60
∑
H.M = = = 109.61
∑ .
The Median:
The median is a value which divides an ordered set of data into two equal parts.
Median is the middle value when n is odd and mean of two values when n is
even.
5
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Example: Find median of data 45,32,37,46,39,36,41,48,36
Solution: Here n = 9
Arrange into ascending order
32,36,36,37,39,41,45,46,48 median = 39
Example: Find median of data 4,8,9,10,6,18,24,16,15,19
Solution: Here n = 10
Arrange into ascending order
Median = + − where n = ∑
Median class
30—39 8 29.5—39.5 8
40—49 87 39.5—49.5 8+87 = 95
50—59 190 49.5—59.5 94+190 = 285
60—69 304 59.5—69.5 285+304 = 589
70—79 211 69.5—79.5 589+211 = 800
80—89 85 79.5—89.5 800+85 = 885
90—99 20 89.5—99.5 885+20 = 905
∑ = 905
n = ∑ = 905 n/2 = 905/2 = 452.5 Median class: A class carrying highest frequency
l = 59.5 , c = 285 , h = 10
6
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
The Mode:
The mode is the value in the data which occurs most frequent.
Example: Find mode of 1,4,5,6,1,8,9,1
Solution: mode = 1
Modal class
30—39 8 29.5—39.5 8
40—49 87 39.5—49.5 8+87 = 95
50—59 190 49.5—59.5 94+190 = 285
60—69 304 59.5—69.5 285+304 = 589
70—79 211 69.5—79.5 589+211 = 800
80—89 85 79.5—89.5 800+85 = 885
90—99 20 89.5—99.5 885+20 = 905
= 59.5 , = 211 , = 304 , h = 10 = 190
Mode = + ( ) ( )
×ℎ
= 59.5 + ( ) ( )
× 10 mode = 65 (marks)
7
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Measure of Dispersion:
By Dispersion we mean the extent to which the values are spread from an
average. A quantity which measures this characteristic is called the measure of
dispersion.
There are two types of measure of dispersion
(i) Absolute measure of dispersion
(ii) Relative measure of dispersion
Example: The marks obtained by 9 students are given
45,32,37,46,39, 36,41,48,36
Find the coefficient of dispersion.
Solution: Here the highest marks = = 48
Lowest marks = = 32
Coefficient of dispersion =
= = 0.2
8
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Lecture # 03
Measure of Dispersion:
(i) The Range
(ii) The Variance
(iii) The Standard Deviation
(i) The Range:
The range is defined as the difference between the highest observation to the
lowest observation.
i.e. R= −
Where us the maximum value and is the smallest value/observation.
Example: Find the range from the data
45,32,37,46,39,36,41,48,36
Solution: Highest observation = 48
Lowest observation = 32
R= −
R = 48 − 32 = 16 (marks)
(ii) The Mean Deviation:
The mean deviation of n observation of data is defined as the mean of absolute
deviation or mod deviations of observations from their mean.
n
x
i 1
i x
M .D for mean
n
n
f
i 1
i xi x
M .D n
for group data
f
i 1
i
x
i 1
i median
M.D for median
n
9
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
n
f
i 1
i xi median
M .D n
for group data
f
i 1
i
− | − | − | − |
32 -8 5 -7 7
36 -4 4 -3 3
36 -4 4 -3 3
37 -3 3 -2 2
39 -1 1 0 0
41 1 1 2 2
45 5 5 6 6
46 6 6 7 7
48 8 8 9 9
xi x 40 xi median 39
M .D
x i x
40
4.4(marks )
n 9
M .D
x i median
39
4.3( marks )
n 9
(iii) The Variance: The variance of a set of n observations is defined
as mean of squared deviations of the observations from their mean
n 2
2
i 1
xi x
S
n
10
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
n 2
fi xi x
For group data 2 S2 i 1
n
f
i 1
i
f i 1
i
2
xi 1
i
2
2 xi x x 2
Or S
n
n n
2
x i 2 x xi nx 2
S2 i 1 i 1
n n
xi 2 2 x.nx nx2 x i n
2
S i 1
x i 1
& nx x i
n n i 1
n n
xi 2 nx 2 x i
2
i 1
i 1
x2
n n
n 2
2 n n
x i xi x i
S 2 i 1
i 1 x i 1
n n n
11
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
n 2
2 n
x
i 1
i xi
S i 1
n n
For group data
n 2
2 n
2
fxi 1
i i fi xi
S n
i1n
f
i 1
i fi
i 1
n 2
2 n
fx
i 1
i i f i xi
And S n
i 1n
f
i 1
i fi
i 1
− ( − )
7 -10 100
8 -9 81
10 -7 49
13 -4 16
14 -3 9
19 2 4
20 3 9
25 8 64
26 9 81
28 11 121
∑( − ) = 534
n 2
x i x 534
The Variance S2 i 1
53.4
n 10
x
fx i i
7350
122.5
f i 60
2
The Variance S2
f x x
i i
72960
1216
f i 60
m1
x i x
n
2
m
x x i
2
n
3
m3
x i x
n
4
And m4
x i x
n
13
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Lecture # 04
Probability:
Probability is the measure of uncertainty or measure of degree of belief in a
particular statement or assumption.
Venn Diagram:
A diagram that is understood to represent the set by circular region or by parts
of circular regions or their compliments with respect to a rectangle represent a
space S is called the Venn Diagram.
A B A B
S
S
A B A B
S
S
1 2 3 4 5 6
(T,2)
(T,4)
(T,5)
(T,1)
(T,3)
(T,6)
Random Experiment:
The term experiment means a planned activity which results a set of data.
Trail:
A single performance of an experiment is called Trial.
Outcome:
The results obtained from an experiment or a trial is called outcome.
And experiment which produces different results even through it is repeated a
large number of time under some essential conditions is called random
experiment.
(i) The experiment can be repeated practically or theoretically any
number of times.
(ii) The experiment has always two or more outcome.
(iii) The outcome of each repetition is unpredictable.
Sample Space:
A set consisting of all possible outcomes of a random experiment is called the
sample space.
A = {H,T} for a coin
S = {1,2,3,4,5,6} for a dice
15
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Event:
Any subset of a sample space S is called an event
A = {H} {H,T}
B = {1,4,6} {1,2,3,4,5,6}
Mutually Exclusive Events:
Two events A and B of single experiment are said to be mutually exclusive or
disjoint events iff they cannot occur both together.
Exhaustive Events:
Events are said to be exhaustive if their union make the whole sample space.
A = {H} , B = {T}
AB = {H,T}
Equally likely events:
Two events A and B are said to be equally likely events if they both have the
same chance of occurrence or when one event is as likely to occur as to other.
Permutations:
A permutation is any ordered subset from a set of ‘n’ distinct objects. The
number of ‘r’ objects selected in a definite order from ‘n’ distinct is denoted by
n
Pr and given as
n n!
Pr
n r !
Combinations:
A combination is a set of ‘r’ objects selected without the regard of their order
from the set of ‘n’ objects is denoted by n Cr and given as
n n!
Cr
r ! n r !
Example:
A club consist of four members. How many sample points are in a sample space
when three officer’s president, secretary and treasurer are chosen.
16
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Solution:
Here n = 4 , r =3
n n! 4! 4.3.2.1
Pr = 24
n r ! 4 3 ! 1!
Example:
How many sample points are in a sample space S when a person draws a hand
of 5 cards from a well shuffled pack of 52 playing cards.
Solution:
Here n = 52 , r = 5
n n! 52! 52.51.50.49.48.47!
Cr 2,598,960
r ! n r ! 5! 52 5 ! 5! 47!
Definition of Probability:
If a random experiment produces ‘n’ equally likely and mutually exclusive
events are considered favorable for the occurrence of an even A then probability
of an event is denoted by P(A) and given as
.
P(A) = =
.
(i) P(A) = 0
(ii) P(S) = = 1
(iii) If A and B are two mutually events then P(AB) = P(A)+P(B)
Example:
A card is drawn from an ordinary deck of 52 playing cards. Find the probability
that
(i) Card is a red card
(ii) Card is a diamond
(iii) Card is 10
Solution:
Here n = 52
(i) Let A represents the event that card is a red card.
P(A) = =
17
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
(ii) Let B represent the event that card is a diamond card
P(B) = =
(iii) Let C represent the event that card is 10
P(C) = =
For Information:
18
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Lecture # 05
Example:
A fair coin is tossed three times. What is the probability that at least one head
appears?
Solution:
The sample space for this experiment is
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Thus n(S) = 8
Let A denotes the event that at least one head appears. Then
A = {HHH, HHT, HTH, THH, HTT, THT, TTH}
Therefore, n(A) = 7
n(A) 7
Hence P(A) =
n(S) 8
Example: If two fair dice are thrown, what is the probability of getting
(i) A double six
(ii) A sum of 8 or more dots?
Solution: The sample space for this experiment is
S = {(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2)
(3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4)
(5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}
Thus n(S) = 36
(i) Let A represent the event that a double six appears. Then
A = {6,6} n(A) = 1
n(A) 1
Therefore P(A) =
n(S) 36
(ii) Let B denotes the event that a sum of 8 or more dot.
B = {(2,6) (3,5) (3,6) (4,4) (4,5) (4,6) (5,3) (5,4) (5,5) (5,6) (6,2) (6,3) (6,4)
(6,5) (6,6) }
n(B) 15 5
n(B) = 15 P(A) =
n(S) 36 12
19
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Example: Six white balls and four black balls which are indistinguishable
apart from color are placed in a bag. If six balls are taken from the bag, find the
probability of their being three white and three black.
Solution: Here total balls =10
White balls = 6
Black balls = 4
Let 6 balls are selected at random from ten balls. The possible number if
outcomes in which 6 balls are selected from 10 balls.
10!
n(S) 10C6 210
6!(10 6)!
Let A denote the event that three balls are white and three are black
6 4
n(A)
3 3
6! 4! 6.5.4.3! 4.3!
n(A) 80
3!(6 3)! 3!(4 3)! 3.2.1.3! 3!.1!
n(A) 80 8
Therefore, P(A) =
n(S) 210 21
(i) Let A be the event that all three selected are men. The possible
outcomes for this event are
8 8! 8.7.6.5! n(A) 56 8
n(A) 56 P( A)
3
3!(8 3)! 3.2.1.5! n (S) 455 65
20
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
(ii) Let B represents the event that at least one women is selected. The
total number of ways for this event are
7 8 7 8 7 8
n(B)
1 2 2 1 3 0
7! 8! 7! 8! 7! 8!
n(B)
1!(7 1)! 2!(8 2)! 2!(7 2)! 1!(8 1)! 3!(7 3)! 0!(8 0)!
n(B) 399 sample points
n(B) 399 57
P(B)
n(S) 455 65
Example:
Four items are taken at random from a box of 12 items and inspected. The box
is rejected if more than one item is found to be faulty. If there are three faulty
items in the box, find the probability that the box is accepted.
Solution:
The possible number of ways in which 4 items are selected from 12 items
12 12!
n(S) 495 sample points
4 4!(12 4)!
Let A denotes the event that shows no faulty item or one faulty item
3 9 3 9
n(A)
0 4 1 3
3! 9! 3! 9!
n(A) 378
0!(3 0)! 4!(9 4)! 1!(3 1)! 3!(9 3)!
n(A) 378
Therefore, P(A) 0.76
n(S) 495
21
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Lecture # 06
Laws of Probability:
Theorem: If is the impossible event then P() = 0
Proof: The sure event S and the impossible event the mutually exclusive
events or S=S
P(S) = P(S )
P(S) = P(S) + P() P( A B) P( A) P(B)
P() = 0
Laws of Complement:
Theorem: If A is the complement of an event A then P( A ) = 1P(A)
Proof:
We can write A A = S
Where A and A are both mutually exclusive events.
P A A P(S )
P( A) P( A) 1
n( A) n(S)
P( A) P(S) 1
n( S ) n( S )
P( A) 1 P( A)
Or P( A) 1 P( A)
22
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
n( A) 15
n(A) = 15 P( A)
n( S ) 16
P( A) 1 P( A)
1 15
P( A) 1
16 16
Example:
3
A coin is biased so that the probability that it falls showing tails is
4
(i) Find the probability of obtaining at least one head when the coin is
tossed five times.
(ii) How many times must the coin be tossed so that the probability of
obtaining at least one head is greater than 0.98?
3
Solution: (i) P(tail or no head) =
4
3 1
P(head) = 1 P( A) 1 P( A)
4 4
The sample space consists of 25 32 sample points. Let A be the event that
at least one head appears and A be the event that no head appears
. A = {TTTTT}
5
3
P( A ) =
4
5
3
P( A) 1 P( A) 1 0.763
4
23
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
(ii). Let the coin be tossed n times to obtained the probability of at least one
head is greater than 0.98 Then
n
3
1 0.98
4
n
3
1 0.98
4
n
3
0.02
4
Taking log on both sides
3
n log log 0.02
4
3 3
Dividing both side by log and reversing the inequality as log is
4 4
negative. We have
log 0.02
n
3
log
4
n 13.6
So n = 14
Probability of subset:
Theorem: If A and B be the two event such that A B then P(A) P(B)
Solution: Let the event B can be written as B A B A where A and
Where P B A 0 B A
24
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Theorem: If A and B be the two events define in a sample space S, then
P A B P( A) P( A B)
A A B A B
Where A B and A B are mutually exclusive events
P(A) P A B A B
A B
S
P(A) P A B P A B
P A B P (A) P A B
A B
A B A B
Theorem:
If A and B are any two events define in a sample space S, then
P A B P ( A) P ( B ) P( A B)
Or “If two events A and B are not mutually exclusive then the probability that
at least one of then occurs, is given by the sum of the separate probabilities of
event A and B minus the probability of the joint event A B ”
Proof:
The event A B can be written as
A B
S
A B A A B where A and A B are two
P A B P A A B
A B
A B A B
P A B P A P A B ___(i)
B A B A B
25
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Where A B and A B are two mutually exclusive events.
P ( B ) P A B A B
P( B) P A B P A B
P A B P( B) P A B
Put in (i)
P A B P ( A) P ( B) P A B
26
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Lecture # 07
Corollary:
If A and B are mutually exclusive events then
P A B P( A) P( B)
Proof:
Since the events A and B are mutually exclusive so
A B
P A B P 0
By addition law P A B P( A) P( B ) P A B
P A B P ( A) P ( B ) 0
P A B P( A) P( B) proved
Alternative:
Let n be the total number of sample points in
A B
a sample space S. m1 be the sample points that S
contains the event A and m2 be the sample points
n A B m1 m2
P A B
n( S ) n
m1 m2
P A B
n n
P A B P( A) P( B)
27
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Corollary:
If A1 , A2 , A3 ,.... Ak are k mutually exclusive events then the probability that one
of them occurs is the sum of the probabilities of the separate event.
Proof:
Since P A B P( A) P( B)
P A1 A2 A3 .... Ak P A1 P A2 P A3 .... P Ak
Corollary:
If A and B are any two events then P A B P( A) P( B)
Proof:
These events are not mutually exclusive. Then by addition law
P A B P ( A) P ( B) P A B
P A B P A B P ( A) P ( B )
P A B P ( A) P ( B)
In general
P A1 A2 A3 .... Ak P A1 P A2 P A3 .... P Ak
28
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Let B be the event that the card is a face card
n(B) = 12
n(B) 12
P B
n(S) 52
Let A B be the event that the card is both club and face card
n( A B ) = 3
3
P( A B ) =
52
The probability is
P A B P ( A) P ( B) P A B
13 12 3 22
52 52 52 52
Example:
An integer is chosen at random from the first 200 +ve integers. What is the
probability that integer chosen is divided by 6 or by 8?
Solution:
The sample space is S = {1,2,3,…..200}
Let A represent the event that integer chosen is divisible by 6.
200
n(A) = 33
6
n( A) 3
P A
n(S) 200
29
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Let A B be the event that integer chosen is divisible by both 6 and8
200
n( A B ) = 8
24
n( A B ) 8
P A B
n(S) 200
The probability is
P A B P ( A) P ( B) P A B
33 25 8 50
200 200 200 200
1
P A B
4
Example:
A pair of dice are thrown. Find the probability of getting a total of either 5 or 11
Solution: The sample space for this experiment is
S = {(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2)
(3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4)
(5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}
n(S) = 36
Let A be the event that a total of 5 occurs
A = {(1,4) , (2,3) , (3,2) , (4,1)}
n(A) = 4
n( A) 4
P A
n(S) 36
30
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
The events A and B are mutually exclusive as a total of 5 and 11 cannot both
occur together
P A B P ( A) P ( B)
4 2 1
P A B
36 36 6
Example:
Three horses A, B & C are in a race A is twice as likely to win as B and B is
twice likely to win as C. What is the probability that A or B wins.
Solution:
According to given condition
P(A) = 2P(B)
P(B) = 2P(C)
Let P be the probability of C
P(C) = P
Then P(B) = 2P(C) = 2P
P(A) = 2P(B) = 2(2P) = 4P
We know that sum of all probabilities is equal to one
4P + 2P + P = 1
7P = 1
1
P=
7
1
P (C )
7
2
P (B)
7
4
P (B)
7
Hence P A B P( A) P( B )
4 2 6
7 7 7
31
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Theorem:
If A, B and C are any three events in a sample space S, then the probability of at
least one of them occurring is given as
P A B C P( A) P(B) P(C) P A B P B C P A C P A B C
Solution:
As A B C = A B C
P A B C P A B C
P A P B C P A B C
P A P B P C P B C P A B C
By distributive law
A B C A B A C
P A B C P A P B P C P B C P A B A C
P A P B P C P B C P A B P A C P A B A C
P A B C P( A) P(B) P(C) P A B P B C P A C P A B C
Example:
A card is drawn at random from a deck of ordinary playing cards. What is the
probability that it is a diamond a face or a king?
Solution: Here n(S) = 52
Let A represents the event that the card is diamond card
n(A) = 13
n( A) 13
P A
n(S) 52
P A B C P( A) P(B) P(C) P A B P B C P A C P A B C
13 12 4 3 4 1 1
52 52 52 52 52 52 52
13 12 4 3 4 1 1
52
22
52
33
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Lecture # 08
Conditional Probability:
The sample space for an experiment must often be changed when some
additional information pertaining to the outcome of the experiment is received.
The effect of such information is to reduce the sample space by excluding some
outcomes as being impossible which before receiving the information where
believed possible. The probabilities associated with such a reduce sample space
are called conditional probabilities.
Let us consider a die-throwing experiment
The sample space is S = {1,2,3,4,5,6} n(S) = 6
Let A be the event that only 6 occur. A = {6}
If there is a condition that 6 occur only when the die shows an even number of
dots. Let B be the event that only even number occur. B = { 2,4,6}
1
P(die shows 6 / die shows even numbers) =
3
B nnABB
P A
B PPABB
P A where P(B) 0
34
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
(i)
0 P AB 1
P S B P(B)
(ii) P S 1
S B B, P S B 1
B P ( B) P( B)
A1 A2 P A1 P A2
(iii) P
B B
B
Example:
Two coins are tossed. What is the conditional probability that two heads result,
given that there is at least one head?
Solution:
The sample space for this experiment is
S = {HH, HT, TH, TT} n(S) = 4
Let A represent the event that two heads appears
A = {HH} n(A) = 1
n(A) 1
P(A) =
n( S ) 4
P A B
We have to calculate P A B P B
______(i)
A B HH n A B 1
1
1
P A 4
B 3 3
4
35
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Question:
A man tosses two fair dice. What is the conditional probability that the sum of
the two dice will be 7 given that
(i) The sum is odd
(ii) The sum is greater than 6
(iii) The two dice had the same outcome
Solution:
The sample space for this experiment is
S = {(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2)
(3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4)
(5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}
n(S) = 36
Let A be the event that the sum is 7
A = {(1,6) , (2,5) , (3,4) , (4,3), (5,2), (6,1)}
n(A) = 6
n( A) 6 1
P A
n(S) 36 6
36
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Let D be the event that two dice has same outcome
A = {(1,1) , (2,2) , (3,3) , (4,4), (5,5), (6,6)}
n(A) = 6
n(D) 6 1
P D
n(S) 36 6
n A B 6
n A B 6 1
P A B
n S 36 6
n A C6
n A C 6 1
P A C
n S 36 6
A D
P A D 0
1
P A B 6 1
P A
B P B
1 3
2
1
P A C 2
C
P A
P C
6
7 7
12
P A D 0
D
P A
P D
1
0
37
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Example: What is the probability that a randomly selected poker hand,
contain exactly 3 aces, given that it contains at least 2 aces.
Solution: Here total cards = 52
Total no. of ways in which 5 cards are randomly selected from 52 playing cards
52 52! 52
= C5 2,598,960
5! 52 5! 5
Let A be the event that exactly 3 aces appear
4 48
n(A)
3 2
4 48
n A 3 2
P A
n S 52
5
Let B be the event that at least 2 aces appear
4 48
n A B
3 2
4 48
n A B 3 2
P A B
n S 52
5
4 48 4 48 4 48
n B 2 3 3 2 4 1
P B
n S 52
5
38
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
4 48 52
P A B 3 2 5
P
A
B
P B
52
.
4 48 4 48 4 48
5 2 3 3 2 4 1
4512
B 108336
P A 0.0416
A
P A B P A . P B provided P A 0
P B . P A provided P B 0
B
Proof: By definition of conditional probability
P A B
B
P A
P B
, P(B) 0
B
P A B P B P A , P(B) 0
A PPAAB
P B , P(A) 0
A
P A B P A P B , P(A) 0
Alternative proof:
Let n be the total number of sample points of an experiment. Let m1 be the
number of sample points for the occurrence of event A and m2 be the sample
points for the occurrence of event B. Let m3 be the sample point for the
occurrence of event A B
m3
i.e. P A B
n
m3
The fraction may be written as
n
39
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
m3 m3 m1 m1
. But P A
n m1 n n
m3
And
m1
= conditional probability of B given that A has occurred i.e. P B
A
Hence A
P A B P A P B
Similarly, P A B P B P A
B
Corollary: In the case of event A,B and C
A PC A B
P A B C P A P B
Proof: Let D A B
A BC D C
P A B C P D C
P D P C D
P A B P C
A B
A PC A B
P A B C P A P B proved
Example: A box contains 15 items 4 of which are defective and 11 are good.
Two items are selected what is the probability that the first is god and second is
defective.
n( A) 11
Solution: Let A be the event that the first item is good P A
n(S) 15
n(B) 4
Let B be the event that second item is defective P B
n(S) 14
11 4 44
We have to find P A B
P A B P A P B A
15 14 210
0.16
40
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Lecture # 09
Example: Two cards are dealt from a pack of ordinary playing cards. Find the
probability that the second card dealt is a heart.
Solution: Let H1 be the event that first card dealt is a heart and H2 be the event
that the second card dealt is a heart. Then
P (second card is heart = P (first card is heart not heart) + P (first card is not
heart and second card is heart)
P H2 P H1 H2 P H1 H2
P H1 P H2 / H1 P H1 P H2 / H1
13 12 39 13 13 13 39
. .
52 51 52 51
P H1
52
, P H1 1
52 52
1 13 4 13 17 1
17 68 68 68 4
Example: Box A contain 5 green and 7 red balls. Box B contain 3 green, 3 red
and 6 yellow balls. A box is selected at random and a ball is drawn at random
from it. What is the probability that the ball drawing is green.
Solution: Let G denote the event that green ball is drawn
P(Green ball) = P(Box A is selected and green ball is drawn)
+ P(Box B is selected and green ball is drawn)
P A B P B G
A P B PG B
P A P G
1 5 1 3
. .
2 12 2 12
5 3 53 8 1
24 24 24 24 3
41
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Example: An urn contains 10 white and 3 black balls another urn contains 3
white and 5 black balls. Two balls are transferred from first urn and placed in
the second and then one ball is taken from the latter. What is the probability that
it is a white ball?
Solution: Urn I white = 10 , black = 3
Urn II white = 3 , black = 5
Let A be the event that two ball are transferred from urn I to urn II. Then A can
be occur in these possible ways
A1 = 2 white balls
A2 = 1 white and 1 black ball
A1 = 2 black balls
10 3
2 0
45
P A1
13 78
2
10 3
1 1
30
P A2
13 78
2
10 3
0 2
3
P A3
13 78
2
Situation of urn II when two balls are transferred.
(i) 5 white 5 black (when 2 white ball drawn)
(ii) 4 white 6 black (when 1 white 1 black ball drawn)
(iii) 3 white 7 black (when 2 black ball drawn)
Let w be the event that white ball drawn from urn II when 2 balls transferred
from urn I
42
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
P w P w A1 P w A2 P w A3
P A1 P w P A2 P w P A3 P w
A1 A2 A3
45 5 30 4 3 3
. . .
78 10 78 10 78 10
59
0.4538
130
Example: A card is drawn at random from a deck of ordinary playing cards.
What is the probability that it is a diamond, a face card or a king?
13
Solution: Let A be the event that card is diamond P A
52
12
Let B be the event that card is a face card P B
52
4
Let C be the event that card is king card P C
52
13 1 1
P A C P A P C .
A 52 13 52
13 3 1 1
P A B C P A P B P C . .
A A B 52 13 3 52
43
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Example: Three urns of same appearance are given as follows
Urn A contain 5 red and 7 white balls
Urn B contain 4 red and 3 white balls
Urn C contain 3 red and 4 white balls
An urn is selected at random and a ball is drawn from the urn
(i) What is the probability that the ball drawn is red?
(ii) If the ball drawn is red, what is the probability that it came from urn
A.
Solution (i) : Let R be the event that red ball is drawn
P R P A R P B R P C R
A P B P R B PC P R C
P A P R
1 5 14 13
. . .
3 12 3 7 3 7
0.4722
(ii). Probability of red ball from urn A is
P A R
P A
R P R
___(i)
A
P A R P A P R
1 5 5
. 0.1389 put in (i)
3 12 36
R 0.1389
P A
0.4722
0.2941
i.e. B P A
P A and P B A P B
44
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
It follows that two event A and B are independent iff
P A B P A P B
If the two events A and B are mutually exclusive then P(A) P(B) = 0
Which is true when either P(A) = 0 or P(B) = 0.
The events A and B are defined to be dependent if P A B P A P B
(iii). Now P A B P A P B A 14 . 32 16
Also B
P A B P B P A
1
P A B 6 1
P B P B
P A B
1
2
3
45
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Lecture # 10
Example: Two fair dice one red and one green are thrown. Let A denote the
event that the red die shows an even number and B the event that the green die
shows 5 or 6. Show that the event A and B are independent.
Solution: The sample space for this experiment is
S = {(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2)
(3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4)
(5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}
n(S) = 36
Let A be the event that the red die shows an even number
A = {(2,1) , (2,2) , (2,3) , (2,4), (2,5), (2,6),(4,1) ,(4,2) ,(4,3),(4,4),(4,5),(4,6),
(6,1),(6,2),(6,3)(6,4),(6,5),(6,6)}
n(A) = 18
n( A) 18 1
P A
n(S) 36 2
The event A B is
A B = {(2,5),(2,6),(4,5),(4,6),(6,5),(6,6)}
n A B 6
n A B 6 1
P A B
nS 36 6
46
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Example: Let A be the event that a family has children of both sexes and B be
the event that a family has at most one boy. If a family is known to have
(i) Three children then show that A and B are independent events
(ii) Four children then show that A and B are dependent event.
Solution (i): Let b denote the boy and g denote the girl
The sample space is S = {bbb , bbg , bgb , gbb , bgg , gbg , ggb ,ggg}
n(S) = 8
Let A be the event that children of both sexes
A = { bbg, bgb ,gbb , bgg , gbg, ggb}
n(A) = 6
n( A) 6 3
P A
n(S) 18 4
The event A B is
A B = {bgg , gbg , ggb}
n A B 3
n A B 3
P A B
nS 8
3 1 3
Now P A .P B P A B . hence Aand B areindependent
4 2 8
(ii). The sample space is S = {bbbb , bbbg , bbgb , bgbb ,gbbb, bbgg , bgbg ,
bggb ,gbbg, gbgb, ggbb, bggg, gbgg, ggbg ,gggb ,gggg}
n(S) = 16
47
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Let A be the event that children of both sexes
A = {bbbg , bbgb , bgbb ,gbbb, bbgg , bgbg , bggb ,gbbg, gbgb, ggbb, bggg,
gbgg, ggbg ,gggb }
n(A) = 14
n( A) 14 7
P A
n(S) 16 8
The event A B is
A B = {bggg , gbgg , ggbg, gggb}
n A B 4
n A B 4 2
P A B
nS 18 9
7 5 2
Now P A . P B P A B . hence Aand B are dependent
8 16 9
Theorem: If A and B are two independent events then P A B P A P B
As P A B P A P B A
P A B P A P B
A A B A B
P A P A B A B
P A P A B P A B
P A B P A P A B
P A B P A P A P B P A B P A P B
P A B P(A) 1 P B
P A B P(A) P B
(ii). Similarly,
P B P B A P B A
P A B P B P A B
P A B P B P A P B P A B P A P B
P A B P(B) 1 P A
P A B P(B) P A
A B A B
P A B P A B
P A B
1 P A P B P A B P A B P A P B P A B
49
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
1 P A P B P A P B P A B P A P B
1 P A P B 1 P A
1 P A 1 P B
P A P B
n A1 n A2 4
4 1
P A1 P A2
36 9
50
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Then event B can be occur in the following way
B1 = { a total of 11 occur in first throw}
n B1 n B2 2
2 1
P B1 P B2
36 18
Therefore, P A B P A1 B1 P A2 B2
P A B P A1 P B1 P A2 P B2
1 1 1 1 1 1 1
P A B . .
9 18 18 9 162 162 81
Example: The probability that a man will be alive in 25 years 3/5 and the
probability that his wife will be alive in 25 years 2/3. Find the probability that
(i) Both will be alive
(ii) Only the man will be alive
(iii) Only the wife will be alive
(iv) At least one will be alive
(v) Neither will be alive in 25 years
Solution: Let A be the event that the man will be alive and B be the event that
wife will be alive in 25 years
3 2
P A , P B
5 3
(i) Both will be alive i.e P A B
Since A and B are independent then
3 2 2
P A B P A .P B .
5 3 5
(ii)
Only the man will be alive i.e P A B
3 2 3 1 1
P A B P A .P B P A . 1 P B 1 .
5 3 5 3 5
51
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
(iii)
Only the wife will be alive i.e P A B
2 2 2 2 4
P A B P A .P B 1 P A P B 1 . .
3 3 5 3 15
(iv) At least one will be alive i.e P A B
3 2 2 9 10 4 13
P A B
5 3 5 15 15
(v)
Neither will be alive i.e P A B
Since A and B are independent
P A B P A .P B
P A B 1 P A 1 P B
3 2
P A B 1 1
5 3
2 1 2
P A B .
5 3 15
52
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Lecture # 11
Bayes Theorem: If the events A1 , A2 ,...A k from a partition of sample space S
that is the event Ai are mutually exclusive and their union is S and if B is any
other event of S such that it can occur only if one of the Ai occurs then for any i
P Ai P B / Ai
P Ai / B k
for i 1, 2,..., k
P A PB / A
i 1
i i
And P Ai B P B P Ai / B ___(ii )
P Ai P B/ Ai P B P Ai / B by (i ) & (ii )
P Ai P B/ A i
P Ai / B (iii )
P B
A1 B A2 B .... Ak B
P B P A1 B A2 B .... Ak B
P B P A1 P B / A1 P A2 P B / A2 ... P Ak P B / Ak
k
P B P Ai P B / Ai
i 1
P Ai P B/ A i
Put in (iii) P Ai / B k
P A PB / A
i 1
i i
53
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Example: In a bolt factory machines A, B and C manufacture 25 ,35 and 40
percent of the total output respectively. Of their outputs 5,4,2 percent
respectively are defective bolts. A bolt is selected at random and found to be
defective. What is the probability that the bolt came from machine A,B,C?
Solution:
P A 25% 0.25 , P B 35% 0.35 , P C 40% 0.40
By Bayes Theorem
P A P E/ A
P A E 3
i A, B, C
P A P E/ A
i 1
i i
P A P E/ A
P A P E A P B P E B P C P E C
54
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Example: An urn contains four balls which are known to be either
(i) All white
(ii) Two white and two black
A ball is drawn at random and is found to be white. What is the probability that
all balls are white?
Solution: Let A1 be the assumption that all the balls are white and A2 be the
assumption that two balls are white and two black. Then
1
P A1 P A2
2
Let B be the event that ball drawn is white
2
4
C1 C1 . 2C0 2 1
P B / A1 4
1 , P B / A1 4
C1 C1 4 2
By Bayes theorem
P A1 P B/ A1
P A1 / B
P A1 P B/ A1 P A2 P B/ A 2
1
1 1 4 2
P A1 / B 2 .
1 1 1 2 3 3
1
2 2 2
P A2 P B/ A 2
P A2 / B
P A1 P B/ A1 P A2 P B/ A 2
1 1
1 4 1
P A2 / B 2 2 .
1 1 1 4 3 3
1
2 2 2
Hence the first assumption is preferred.
55
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Lecture # 12
Random Variable:
A numerical quantity whose value is determined by the outcome of some
random experiment is called random variable. It is also known as chance
variable/ stochastic variable or variate.
The random variables are denoted by capital Latin letter such as X,Y,Z. While
the values taken by them are represented by small letter such as x,y,z.
Distribution function:
The distribution function of a random variable X is denoted by F(x) = P(X x).
The function F(x) gives the probability of the event that X can take the value
less than or equal to the specified value of x. The distribution function is
abbreviated to d.f and is also called the cumulative distribution (cdf) as it is the
cumulative probability function of the X from the smallest up to special value of
x. Since F(x) is probability, it is quite obvious that
F(∞) = P() = 0 and F(+∞) = P(s) = 1
Let a and b be two real numbers such that a < b then
F(b) F(a) = P(Xb)P(Xa)
= P(a<Xb)
Which is non-negative and hence F(x) is non-decreasing function of x.
Again Lim F x h F x , i.e. the function F(x) is continuous on the right at
h 0
each value of X.
A d.f F(x) thus has following properties
(i) F(∞) = 0 and F(+∞) = 1
(ii) F(x) is non-decreasing function of x i.e. F(x1) F(x2) if x1 x2
(iii) F(x) is continuous at least on the right of each x.
Discrete random variable and its probability distribution:
A random variable X is said to be discrete if it can assume values which are
finite or countably infinite i.e x1 , x2 ,.......xn ,... are probability points or jump
point.
56
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
F(x)
Distribution function 1.0
F x f xi
…
0.5
X
0
x1 x2 x 3 .......xi ...
0 X
x1 x2 x3 .....xi xi 1...
Example: Find the probability distribution and distribution function for the
number of heads when 3 balanced coins are tossed. Construct a probability
histogram and graph of the distribution.
Solution: The sample space for three coins is
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Let X be a random variable that denote the number of heads. Then X can take
the value 0,1,2,3.
The corresponding probability are
f(0) = P(X=0) = P[{TTT}] = 1/8
f(1) = P(X=1) = P[{HTT,THT,TTH}] = 3/8
f(2) = P(X=2) = P[{HHT,HTH,THH}] = 3/8
f(3) = P(X=3) = P[{HHH}] = 1/8
57
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
X 0 1 2 3
f(x) 1/8 3/8 3/8 1/8
Now we find distribution function
For x < 0 F(x) = P(X x)
F(0) = 0
For 0 x < 1 we have P(X < x) = P(X = 0) = 1/8
For 1 x < 2 , we have P(X < x) = P(X = 0) + P (X = 1)
= 1/8 + 3/8 = 4/8
For 2 x < 3 , we have P(X < x) = P(X = 0) + P (X = 1)+P(X = 2)
= 1/8+3/8+3/8 = 7/8
Finally For x 3 P(X < x) = P(X = 0) + P (X = 1)+P(X = 2)+P(X=3)
= 1/8+3/8+3/8+1/8 = 8/8 = 1
Hence the desired distribution function is
0 for x 0
1
for 0 x 1
8
4
F x for 1 x 2
8
7
8 for 2 x 3
F(x)
1 for x 3
Graph of distribution
Histogram
1
F(x)
7/8
5/8 6/8
5/8
4/8
4/8
3/8
3/8
2/8
2/8
1/8 1/8
X X
0 1 2 3 0 1 3
2
58
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Lecture # 13
Example: (a) Find the probability distribution of the sum of the dots when
two fair dice are thrown.
(b) Use the probability distribution to find the probabilities of obtaining
(i) a sum of 8 or 11
(ii) a sum is greater than 8
(iii). a sum that is greater than 5 but less than or equal to 10
Solution: The sample space for this experiment is
S = {(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2)
(3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4)
(5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}
(a) Let X denote the random variable that denote the sum of dots which
appear on the dice. Then X cab take the value 2,3,4,5,….,12
The corresponding probabilities are
f(2) = P(X=2)=P[{(1,1)}] = 1/36
f(3) = P(X=3)=P[{(1,2),(2,1)}] = 2/36
f(4) = P(X=4)=P[{(1,3),(2,2),(3,1)}] = 3/36
Similarly, f(5) = 4/36 , f(6) = 5/36 , f(7) = 6/36, f(8) = 5/36 , f(9) =4/36 ,
f(10)=3/36 , f(11) = 2/36 , f(12) = 1/36
xi 2 3 4 5 6 7 8 9 10 11 12
f(xi) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36
59
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
= P(X=6)+P(X=7)+ P(X=8)+P(X=9)+ P(X=10)
= 5/36 + 6/36 +5/36+4/36+ 3/36 = 23/36
Continuous Random variable and its probability Density
function:
A random variable X is defined to be continuous if it can assume every possible
value in an interval [a,b] , a < b, where a and b may be ∞ and +∞ respectively.
A random variable X is defined to be continuous if its distribution function F(x)
is continuous and differentiable everywhere except at the isolated points in a
given range.
F(x)
The graph of F(x) has no jumps.
Let the derivative of F(x) be denoted by f(x) 1
d
i.e. F x f x
dx
Since F(x) is a non-decreasing function of x, we have
(i) f(x) 0 0 X
F(a) F(b)
(ii) F(x) = f x dx , x
(iii) The probability that X can take the value in the interval [c,d] , c < d is
given by
P(c < x d) = F(d) F(c)
d c
= f x dx f x dx
d
= f x dx f x dx
c
60
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
d
= f x dx f x dx
c
d
= f x dx
c
Example: (a) Find the value of k so that the function f(x) defined as follow
may be a density function
f(x) = kx ,0x2
f(x) = 0 , otherwise
(b). Find also the probability that both of two sample value will exceed 1
(c) Compute the distribution function F(x).
Solution: (a) The function f(x) will be a density function, if
(i) f(x)
(ii) f x dx 1
f x dx 1
a 2
f x dx f x dx f x dx 1
0 2
2
0 kx dx 0 1
0
2
x2 4 1
k 1 k 1 k
2 0 2 2
x
for 0 x 2
Hence f ( x) 2
0 otherwise
61
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
2
(b). P X 1 f x dx
1
2 2
x x2
dx
1
2 4 1
4 1 3
4 4 4
(c) To compute distribution function, we find
x
F x P X x f x dx for any x
for x 0
x x
F x f x dx 0dx 0
x 0 x
for 0 x 2 F x f x dx f x dx f x dx
0
x 2
x x
0 dx
0
2 4
for 2 x
x 0 2
F x f x dx f x dx f x dx f x dx
0 2
2
x
0 dx 0
0
2
2
x2 4
1
4 0 4
0 for x 0
2
x
F x for 0 x 2
2
1 for 2 x or x 2
62
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer
Mathematical expectation of a random variable:
Let X be a discrete random variable having possible values x1 , x2 ,.....xn ,... with
the corresponding probabilities f x1 , f x2 ,..... f xn ,... such that f x 1
Then the mathematical expectation or the expectation or the expected value of
X, denoted by E(X) is defined as
E X x1 f x1 x2 f x2 .... xn f xn ,...
= xi f xi , provided it converge absolutely.
i 1
i.e. x f x dx is finite.
63
Collected by: Muhammad Saleem Composed by: Muzammil Tanveer