CHAPTER 16 Expected Value of a Discrete Random Variable

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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

.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

Probability

Great 0.20

Good 0.40

OK 0.25

EconomicScenario

Profit($ Millions)

-4Lousy 0.15

P(X=x4)

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

Sample MeanSample Mean

1 + ... +

x + ... + 3

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

EXAMPLE

= x P(X = xi ii=1

Probability

Great 0.20

Good 0.40

OK 0.25

EconomicScenario

Profit($ Millions)

-4Lousy 0.15

P(X=x4)

P(X=x1)

P(X=x2)

P(X=x3)

EXAMPLE

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

= x P(X = xi ii=1

Probability

Great 0.20

Good 0.40

OK 0.25

EconomicScenario

Profit($ Millions)

-4Lousy 0.15

P(X=x4)

P(X=x1)

P(X=x2)

P(X=x3)

-4 -2 0 2 4 6 8 10 12

Profit

Probability

EXAMPLE

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

= x P(X = xi ii=1

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

So the probability distribution of x is:

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

)81(3)

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

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

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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