4.2 (cont.) Expected Value
of a Discrete Random
Variable
A measure of the “middle”
of the values of a random
variable
Probability
Center
Good
.40
.35
OK
.30
Gre at
.25
Lousy
.20
.15
.10
.05
-4
-2
0
2
4
6
8
10
12
Profit
The mean of the probability distribution is
the expected value of X, denoted E(X)
E(X) is also denoted by the Greek letter µ
(mu)
Economic
Scenario
Mean or
Expected
Value
Profit X
($ Millions)
Probability P
Great
x1 10
P(X=x1) 0.20
Good
x2 5
P(X=x2) 0.40
OK
x3 1
P(X=x3) 0.25
Lousy
x4 -4
P(X=x4) 0.15
k = the number of possible values (k=4)
k
E ( x)   =
x
i
 P (X = x i )
i= 1
E(x)= µ = x1·p(x1) + x2·p(x2) + x3·p(x3) +
... + xk·p(xk)
Weighted mean
Sample Mean
Mean or
Expected
Value
X
=
n
X

i
i = 1
n
x +x +x +...+x
n
X= 1 2 3
n
1
1
1
1
= x + x + x +...+ x
n 1 n 2 n 3
n n
k = the number of outcomes (k=4)
k
E ( x)   =
x
i
 P (X = x i )
i= 1
µ = x1·p(x1) + x2·p(x2) + x3·p(x3) + ... +
xk·p(xk)
Weighted mean
Each outcome is weighted by its probability
Other Weighted Means
Stock Market: The Dow Jones
Industrial Average
The “Dow” consists of 30 companies (the 30
companies in the “Dow” change periodically)
To compute the Dow Jones Industrial
Average, a weight proportional to the
company’s “size” is assigned to each
company’s stock price
Economic
Scenario
Mean
Profit X
($ Millions)
Probability P
Great
x1 10
P(X=x1) 0.20
Good
x2 5
P(X=x2) 0.40
OK
x3 1
P(X=x3) 0.25
Lousy
x4 -4
P(X=x4) 0.15
k = the number of outcomes (k=4)
k
E ( x)   =
x
i
 P (X = x i )
i= 1
µ = x1·p(x1) + x2·p(x2) + x3·p(x3) + ... +
xk·p(xk)
EXAMPLE
Economic
Scenario
Mean
Profit X
($ Millions)
Probability P
Great
x1 10
P(X=x1) 0.20
Good
x2 5
P(X=x2) 0.40
OK
x3 1
P(X=x3) 0.25
Lousy
x4 -4
P(X=x4) 0.15
k = the number of outcomes (k=4)
k
E ( x)   =
x
i
 P (X = x i )
i= 1
µ = x1·p(x1) + x2·p(x2) + x3·p(x3) + ... +
xk·p(xk)
EXAMPLE
µ = 10*.20 + 5*.40 + 1*.25 – 4*.15 = 3.65
($ mil)
Probability
Good
.40
.35
OK
.30
Gre at
.25
Mean
Lousy
.20
.15
.10
.05
-4
-2
0
2
4
6
8
10
12
Profit
µ=3.65
k = the number of outcomes (k=4)
k
E ( x)   =
x
i
 P (X = x i )
i= 1
µ = x1·p(x1) + x2·p(x2) + x3·p(x3) + ... +
xk·p(xk)
EXAMPLE
µ = 10·.20 + 5·.40 + 1·.25 - 4·.15 = 3.65
($ mil)
Interpretation
E(x) is not the value of the random
variable x that you “expect” to observe if
you perform the experiment once
Interpretation
E(x) is a “long run” average; if you
perform the experiment many times and
observe the random variable x each time,
then the average x of these observed xvalues will get closer to E(x) as you
observe more and more values of the
random variable x.
Example: Green Mountain
Lottery
State of Vermont
choose 3 digits from 0 through 9; repeats
allowed
win $500
x
$0
$500
p(x)
.999
.001
E(x)=$0(.999) + $500(.001) = $.50
Example (cont.)
E(x)=$.50
On average, each ticket wins $.50.
Important for Vermont to know
E(x) is not necessarily a possible value of
the random variable (values of x are $0
and $500)
Example: coin tossing
Suppose a fair coin is tossed 3 times and
we let x=the number of heads. Find
E(x).
First we must find the probability
distribution of x.
Example (cont.)
Possible values of x: 0, 1, 2, 3.
p(1)?
An outcome where x = 1: THT
P(THT)? (½)(½)(½)=1/8
How many ways can we get 1 head in 3
tosses? 3C1=3
Example (cont.)
p (0)  3 C 0 
p (1)  3 C 1 
  
0
1
2
1
2
 
1
1
2
1
2
3
2

p (2)  3 C 2 
1
2
  
p (3)  3 C 3 
1
2
  
2
3
1
2
1
2

1
8
3
8
1

3
8
0

1
8
Example (cont.)
So the probability distribution of x is:
x
p(x)
0
1/8
1
3/8
2
3/8
3
1/8
 So the probability distribution of x is:
Example
x
p(x)
0
1/8
1
3/8
E(x) (or μ ) is
E(x)
4
  x  p(x )
i
i
i1
 (0  1 )  ( 1  3 )  (2  3 )  (3  1 )
8
8
8
8
 12  1.5
8
2
3/8
3
1/8
US Roulette Wheel
and Table
 The roulette wheel has
alternating black and
red slots numbered 1
through 36.
 There are also 2 green
slots numbered 0 and
00.
 A bet on any one of
the 38 numbers (1-36,
0, or 00) pays odds of
35:1; that is . . .
 If you bet $1 on the
winning number, you
receive $36, so your
winnings are $35
American Roulette 0 - 00
(The European version has
only one 0.)
US Roulette Wheel: Expected Value of a
$1 bet on a single number
Let x be your winnings resulting from a $1 bet
on a single number; x has 2 possible values
x
p(x)
-1
37/38
35
1/38
E(x)= -1(37/38)+35(1/38)= -.05
So on average the house wins 5 cents on every
such bet. A “fair” game would have E(x)=0.
The roulette wheels are spinning 24/7, winning
big $$ for the house, resulting in …