4.2 (cont.) expected value of a discrete random variable a measure of the “middle” of the values...
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4.2 (cont.) Expected Value of a Discrete Random Variable
A measure of the “middle” of the values of a random variable
-4 -2 0 2 4 6 8 10 12
Profit
Probability
Lousy
OK
Good
Great
.05
.10
.15
.40
.20
.25
.30
.35Center
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)
k = the number of possible values (k=4)
E(x)= µ = x1·p(x1) + x2·p(x2) + x3·p(x3) + ... + xk·p(xk)
Weighted mean
Mean orExpectedValue
k
i ii=1
( ) = x P(X=x )E x
Probability
Great 0.20
Good 0.40
OK 0.25
EconomicScenario
Profit($ Millions)
5
1
-4Lousy 0.15
10
P(X=x4)
X
x1
x2
x3
x4
P
P(X=x1)
P(X=x2)
P(X=x3)
k = the number of outcomes (k=4)
µ = x1·p(x1) + x2·p(x2) + x3·p(x3) + ... + xk·p(xk)
Weighted meanEach outcome is weighted by its probability
Mean orExpectedValue
Sample MeanSample Mean
n
n
1=ii
X
= X
nx
n
1 + ... +
3x
n
1 +
2x
n
1 +
1x
n
1 =
nn
x + ... + 3
x + 2
x + 1
x = X
k
i ii=1
( ) = x P(X=x )E x
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
k = the number of outcomes (k=4)
µ = x1·p(x1) + x2·p(x2) + x3·p(x3) + ... + xk·p(xk)
EXAMPLE
Mean
Probability
Great 0.20
Good 0.40
OK 0.25
EconomicScenario
Profit($ Millions)
5
1
-4Lousy 0.15
10
P(X=x4)
X
x1
x2
x3
x4
P
P(X=x1)
P(X=x2)
P(X=x3)
k
i ii=1
( ) = x P(X=x )E x
k = the number of outcomes (k=4)
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)
Mean
Probability
Great 0.20
Good 0.40
OK 0.25
EconomicScenario
Profit($ Millions)
5
1
-4Lousy 0.15
10
P(X=x4)
X
x1
x2
x3
x4
P
P(X=x1)
P(X=x2)
P(X=x3)
k
i ii=1
( ) = x P(X=x )E x
-4 -2 0 2 4 6 8 10 12
Profit
Probability
Lousy
OK
Good
Great
.05
.10
.15
.40
.20
.25
.30
.35
k = the number of outcomes (k=4)
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)
Mean
µ=3.65
k
i ii=1
( ) = x P(X=x )E x
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 x-values will get closer to E(x) as you observe more and more values of the random variable x.
Example: Green Mountain Lottery
State of Vermontchoose 3 digits from 0 through 9;
repeats allowedwin $500
x $0 $500p(x) .999 .001
E(x)=$0(.999) + $500(.001) = $.50
Example (cont.)
E(x)=$.50On average, each ticket wins $.50.Important for Vermont to knowE(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 (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: THTP(THT)? (½)(½)(½)=1/8How many ways can we get 1 head
in 3 tosses? 3C1=3
Example (cont.)
0 31 1 13 0 2 2 8
1 2 31 13 1 2 2 8
2 1 31 13 2 2 2 8
3 01 1 13 3 2 2 8
(0)
(1)
(2)
(3)
p C
p C
p C
p C
Example (cont.)
So the probability distribution of x is:
x 0 1 2 3p(x) 1/8 3/8 3/8 1/8
Example
1.58
12
)81(3)
83(2)
831()
81(0
4
1i)
ip(x
ixE(x)
is )μ (orE(x)
So the probability distribution of x is:
x 0 1 2 3p(x) 1/8 3/8 3/8 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 -1 35p(x) 37/38 1/38
E(x)= -1(37/38)+35(1/38)= -.05So 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 …
