chapter 16 expected value of a discrete random variable

CHAPTER 16Expected 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)

µ = x1·p(x1) + x2·p(x2) + x3·p(x3) + ... + xk·p(xk)

Weighted mean

Mean orExpectedValue

= x P(X = xi ii=1

k

)

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)


Weighted meanEach outcome is weighted by its probability


= x P(X = xi ii=1

k

)

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

Other Weighted MeansGPA A=4, B=3, C=2, D=1, F=0

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

Five 3-hour courses: 2 A's (6 hrs), 1 B (3 hrs), 2 C's (6 hrs)

4 * 6 3*3 2 * 6 45GPA: 3.0

15 15



EXAMPLE

Mean

= x P(X = xi ii=1

k

)

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)



EXAMPLE

µ = 10*.20 + 5*.40 + 1*.25 – 4*.15 = 3.65 ($ mil)

Mean

= x P(X = xi ii=1

k

)

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)

-4 -2 0 2 4 6 8 10 12

Profit

Probability

Lousy

OK

Good

Great

.05

.10

.15

.40

.20

.25

.30

.35



EXAMPLE

µ = 10·.20 + 5·.40 + 1·.25 - 4·.15 = 3.65 ($ mil)

Mean

= x P(X = xi ii=1

k

)

µ=3.65

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

Green Mountain Lottery (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: expected number of heads in 3 tosses of fair coin

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: expected number of heads in 3 tosses of fair coin (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


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


So the probability distribution of x is:

x 0 1 2 3p(x) 1/8 3/8 3/8 1/8


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 …

Expected Value, Surprise Onside Kicks

http://www.advancednflstats.com/ The change in expected points for the kicking team: successful 1.9; fail -1.4.

X=change in expected points for kicking team when attempting surprise onside kick

What values of p make surprise onside kicks a good strategy?

X 1.9 -1.4

p(x) p 1-p

Expected change should be greater than 0

(1.9) ( 1.4)(1 ) 0

3.3 1.4 0

3.3 1.4

0.424

p p

p

p

p

http://www.advancednflstats.com/

Two More Examples1. X = # of games played in a randomly

selected World SeriesPossible values of X are x=4, 5, 6, 7

2. Y=score on 13th hole (par 5) at Augusta National golf course for a randomly selected golfer on day 1 of 2011 Masters

y=3, 4, 5, 6, 7

Probability Distribution Of Number of Games Played in Randomly Selected World Series

Estimate based on results from 1946 to 2010.

x 4 5 6 7

p(x) 12/65=0.185

12/65=0.185

14/65=0.215

27/65=0.415

Probability Histogram

Probability Distribution Of Score on 13th hole (par 5) at Augusta National Golf Course on Day 1 of 2011 Masters

y 3 4 5 6 7

p(x) 0.040 0.414 0.465 0.051 0.030

Probability Histogram

E(X)= µ =4(0.185)+5(0.185)+6(0.215)+7(0.414)=5.86 games

E(Y)= µ=3(.04)+4(0.414)+5(0.465)+6(0.051)+7(0.03)=4.617 strokes


k

i ii=1

( ) = x P(X=x )E X

x 4 5 6 7

p(x) 12/65=0.185

12/65=0.185

14/65=0.215

27/65=0.415

y 3 4 5 6 7

p(x) 0.040 0.414 0.465 0.051 0.030

E(X)= µ =4(0.185)+5(0.185)+6(0.215)+7(0.414)

=5.86 games

Mean or Expected Value

µ=5.86

chapter 16 expected value of a discrete random variable

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