Download Introduction to Probability and Statistics Eleventh Edition

Random Variables
•A random variable is a numerical measure of the
outcome from a probability experiment, so its value is
determined by chance. Random variables are
denoted X, Y, Z etc.
•A discrete random variable is a random variable
that has either a finite number of possible values or
a countable number of possible values.
•A continuous random variable is a random
variable that has an infinite number of possible
values that is not countable.
Example
• Discrete random variable with a finite number of
values
Let X = number of TV sets sold at the store in one
day where x can take on 5 values (0, 1, 2, 3, 4)
• Discrete random variable with an infinite sequence
of values
Let X = number of customers arriving in one day
where x can take on the values 0, 1, 2, . . .
We can count the customers arriving, but there is no
finite upper limit on the number that might arrive.
• The waiting time of a customer in a queueContinuous.
• We use capital letter , like X, to denote the random
variable and use small letter to list the possible
values of the random variable.
• Example. A single die is cast, X represent the
number of pips showing on the die and the
possible values of X are x=1,2,3,4,5,6.
A probability distribution provides the possible
values of the random variable and their corresponding
probabilities. A probability distribution can be in the
form of a table, graph or mathematical formula.
The table below shows the probability distribution for
the random variable X, where X represents the
number of DVDs a person rents from a video store
during a single visit.
Probability Distributions for
Discrete Random Variables
(Probability Mass Function (PMF)).
• The probability distribution is defined by a
probability function, denoted by p(x), which
provides the probability for each value of the
random variable.
• The probability distribution for discrete random
variable is called Probability Mass Function
(PMF).
p x i   P X  x i

Properties for
Discrete Random Variables
• The properties for a discrete probability function
(PMF) are:
p( x)  P( X  x)
0  p( x)  1 x
 p( x)  1
all x
• Cumulative Distribution Function (CDF)
F ( x)  P( X  x)
F (b)  P ( X  b) 
b
 p( x)
y  
F ()  0
F ( )  1
Example
n
Using past data on TV sales (below left), a tabular
representation of the probability distribution for TV
sales (below right) was developed.
Units Sold
0
1
2
3
4
Number
of Days
80
50
40
10
20
200
X
0
1
2
3
4
p(x)
.40
.25
.20
.05
.10
1.00
• Graphical Representation of the Probability Distribution
.50
p(x)
.40
.30
.20
.10
0
1
2
3 4
Values of Random Variable X (TV sales)
Example
• Random Variable: Grades of the students
Student ID
1
2
3
4
5
6
7
8
9
10
Grade
3
2
3
1
2
3
1
3
2
2
Probability Mass Function
2
p 1  P  X  1 
 0.2
10
PMF
4
p  2  P  X  2 
 0.4
10
4
p  3  P  X  3 
 0.4
10
Grade
Example
• Random Variable: Grades of the students
Student ID
1
2
3
4
5
6
7
8
9
10
Grade
3
2
3
1
2
3
1
3
2
2
Probability Mass Function
 p x
i
CDF
  p 1  p  2  p 3  1
i
Cumulative Distribution Function
p  X  x   x
p  X  2  x
i
i
x
2
p  X  3   x
i
p (x i )
p (x i )  p 1  p  2   0.2  0.4  0.6
2
p (x i )  p 1  p  2   p  3  1
Grade
Example
• Toss a fair coin three times and
define X = number of heads.
x
HHH
1/8
3
HHT
1/8
2
HTH
1/8
2
THH
1/8
2
HTT
1/8
1
THT
1/8
1
TTH
1/8
1
TTT
1/8
0
P(X = 0) =
P(X = 1) =
P(X = 2) =
P(X = 3) =
1/8
3/8
3/8
1/8
X
0
1
2
3
p(x)
1/8
3/8
3/8
1/8
Probability
Histogram for x
Expected Value and Variance
• The expected value, or mean, of a random variable
is a measure of its central location.
– Expected value of a discrete random variable:
n
E  X      xi p  xi 
11
• The variance summarizes the variability in the
values of a random variable.
– Variance of a discrete random variable:
Var  X   
2
n
 E  X      ( xi   ) 2 . p  xi 
2
i 1
 EX
2
 EX 
2
n
  xi2 p  xi    2
i 1
Expected Value and Variance

E (aX  b)  aE ( X )  b
Ex. Given that X is random variable whose mean = 4,
find the mean of 3X+5.
Solution. E(3X+5)= 3 E(X)+E(5)= 3x4+5=17

V (aX  b)  a 2V ( X )
Ex. Given that X is random variable whose variance = 2,
find the variance of 3X+5.
Solution. V(3X+5)= 9 V(X)= 9x2=18
Example
n
Using past data on TV sales (below left), a tabular
representation of the probability distribution for TV
sales (below right) was developed.
Units Sold
0
1
2
3
4
Number
of Days
80
50
40
10
20
200
X
0
1
2
3
4
Find the mean and variance.
p(x)
.40
.25
.20
.05
.10
1.00
Example:
• Variance and Standard Deviation of a Discrete Random Variable
x
p(x)
xp(x)
0
1
2
3
4
.40
.25
.20
.05
.10
.00
.25
.40
.15
.40
1.20
x 2p(x)
.00
.25
.80
.45
1.6
3.1
n
E  X      xi p  xi   1.20
n
Var  X      x p  xi   
2
i 1
2
i
2
 3.1  1.20  1.66
2
11
 standard deviation is
1.66 =1.2884
Example
• Toss a fair coin 3 times and
record x the number of heads.
X
p(x)
xp(x)
(x-)2p(x)
0
1/8
0
(-1.5)2(1/8)
1
3/8
3/8
(-0.5)2(3/8)
2
3/8
6/8
(0.5)2(3/8)
3
1/8
3/8
(1.5)2(1/8)
12
   xp( x)   1.5
8
  ( x   ) p( x)
2
2
 2  .28125  .09375  .09375  .28125  .75
  .75  .688
Alternative Solution (Suggested):

x
p(x)
x.p(x)
x2p(x)
0
1/8
0
0
1
3/8
3/8
3/8
2
3/8
3/4
3/2
3
1/8
3/8
9/8
12
   xp( x)   1.5
8


Var  X      x p  xi    xi p  xi  
i 1
 i 1

 3  1.52  0.75
n
2
n
2
i
2
  .75  .688
Example
• The probability distribution for X the
number of heads in tossing 3 fair coins.
•
•
•
•

Shape?
Outliers?
Center?
Spread?
Symmetric;
mound-shaped
None
 = 1.5
 = .688
Some important Differentiation and Integration
Formulas
d
 c.dx  c.x
(c )  0
dx
 dx  x
d
( x)  1
n 1
x
dx
n
x
dx 
; n  1

d
n 1
( x n )  nx n 1
x
x
dx
e
dx

e

d
(e x )  e x
1 x
x
dx
 a dx  ln a a
d
( a x )  a x ln a
 ln( x)dx  x ln x  x
dx
1
d
1
dx  ln x
ln( x ) 

x
dx
x
Notes about Continuous RV
• A continuous random variable can assume any
value in an interval on the real line or in a collection
of intervals.
• It is not relevant to talk about the probability of the
random variable assuming a particular value.
• Instead, we talk about the probability of the
random variable assuming a value within a given
interval.
Probability Distributions for
Continuous Random Variables
(Probability Density Function (PDF)).
• The probability distribution is defined by a
probability function, denoted by f(x), which
provides the probability for each value of the
random variable.
• The probability distribution for continuous
random variable is called Probability
Density Function (PDF).
Properties for
Continuous Random Variables
The properties for a
continuous probability
function (PDF) are:
f x 
0  f ( x)  1

 f ( x) dx 1

f x  
d
F x 
dx
Cumulative Distribution Function
(CDF)
F  x   P X  x  
x
 f ( x) dx

b
F  x   P(a  x  b)   f ( x) dx
a
Example
– Let X be a random variable with range [0,2] and pdf
defined by f(x)=1/2 for all x between 0 and 2 and f(x)=0
for all other values of x. Note that since the integral of
zero is zero we get



2
1
f ( x)dx   1/ 2dx  x  1  0  1
0
2 0
2
– That is, as with all continuous pdfs, the total area under
the curve is 1. We might use this random variable to
model the position at which a two-meter with length of
rope breaks when put under tension, assuming “every
point is equally likely”. Then the probability the break
occurs in the last half-meter of the rope is
P(3/ 2  X  2)  
2
3/ 2
2
1
f ( x)dx   1/ 2dx  x  1/ 4
3/ 2
2 3/ 2
2
Example
– Let Y be a random variable whose range is the
nonnegative and whose pdf is defined by
1
f y 
e
750

y
750
The random variable Y might be a reasonable choice to model
the lifetime in hours of a standard light bulb with average life
750 hours. To find the probability a bulb lasts under 500 hours,
you calculate
P(0  Y  500)  
500
0
1  x / 750
 x / 750 500
2/3
e
dx  e
 e  1  0.487
0
750
Expected Value and Variance
• The expected value, or mean, of a random variable is a
measure of its central location.
– Expected value of a continuous random variable:
EX    

 x f x  dx

• The variance summarizes the variability in the values of a
random variable.
– Variance of a discrete random variable:
 
Var  X    2  E  X     E X 2  E  X 
2
2



2
2
  x    . f x dx    x . f x dx    2

 


Discrete versus Continuous Random Variables
Discrete RV
Continuous RV
Infinite Sample Space
e.g. [0,1], [2.1, 5.3]
Finite Sample Space
e.g. {0, 1, 2, 3}
Probability Mass Function (PMF)
Probability Density Function (PDF)
f x 
p( xi )  P( X  xi )
1. 0  p( x)  1 x
2.
 p ( x)  1
all x
Cumulative Distribution Function (CDF)
F ( x)  P( X  x) 
b
 p( x)
y  
p X  x 
F  x   P X  x  
x
 f ( x) dx

b
F  x   P(a  x  b)   f ( x) dx
a
Example

We assume that with average waiting time of one customer is 2
minutes
1 x / 2
 e , x0
f ( x)   2
0,
otherwise
PDF: f (time)
time
Example
• Probability that the customer waits exactly 3 minutes is:
1 3  x /2
P (x  3)  P (3  x  3)  3 e dx  0
2
• Probability that the customer waits between 2 and 3
minutes is:
1 3  x /2
P (2  x  3)   e dx  0.145
2 2
• The Probability that the customer waits less than 2
minutes
2
P(0  X  2)   e
0
x/ 2
1
dx  1  e  0.632
Example
• Probability that the customer waits exactly 3 minutes is:
1 3  x /2
P (x  3)  P (3  x  3)  3 e dx  0
2
• Probability that the customer waits between 2 and 3 minutes is:
1 3  x /2
P (2  x  3)   e dx  0.145
2 2
P(2  X  3)  F (3)  F (2)  (1  e(3 / 2) )  (1  e1 )  0.145
CDF
• The Probability that the customer waits less than 2 minutes
P (0  X  2)  F (2)  F (0)  F (2)  1  e 1  0.632
CDF
Expected Value and Variance
A continuous variable X has a probability density function
f ( x)  cx ;0  x  1
2
where c is constant. Find (i) the value of c (ii)
(iii) P ( X  .75) (iv) P (.25  X  .75)
v) compute mean and variance of X.
P ( X  .25)
Key Concepts
V. Discrete Random Variables and Probability
Distributions
1. Random variables, discrete and continuous
2. Properties of probability distributions
0  p( x)  1 and  p( x)  1
3. Mean or expected value of a discrete random
variable: Mean :    xp( x)
4. Variance and standard deviation of a discrete
random variable: Variance :  2  ( x   )2 p( x)
Standard deviation :    2