1. 4.2 (cont.) Expected Value of a Discrete Random Variable
A measure of the “middle” of the values of a random variable

2. Center
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)

3. Mean or Expected Value k = the number of possible values (k=4)
Probability
Great
0.20
Good
0.40 OK
0.25
Economic
Scenario
Profit
($ Millions) 5 1 -4
Lousy
0.15 10
P(X=x4) X x1 x2 x3 x4 P
P(X=x1)
P(X=x2)
P(X=x3)
Mean or Expected Value
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

4. Mean or Expected Value k = the number of outcomes (k=4)
µ = x1·p(x1) + x2·p(x2) + x3·p(x3) xk·p(xk)
Weighted mean
Each outcome is weighted by its probability

5. 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

6. Other Weighted Means GPA A=4, B=3, C=2, D=1, F=0
Course grade: tests 40%, final exam 25%, quizzes 25%, homework 10%
"Average" ticket prices

7. Mean k = the number of outcomes (k=4)
Probability
Great
0.20
Good
0.40 OK
0.25
Economic
Scenario
Profit
($ Millions) 5 1 -4
Lousy
0.15 10
P(X=x4) X x1 x2 x3 x4 P
P(X=x1)
P(X=x2)
P(X=x3)
Mean
k = the number of outcomes (k=4)
µ = x1·p(x1) + x2·p(x2) + x3·p(x3) xk·p(xk)
EXAMPLE

8. Mean k = the number of outcomes (k=4)
Probability
Great
0.20
Good
0.40 OK
0.25
Economic
Scenario
Profit
($ Millions) 5 1 -4
Lousy
0.15 10
P(X=x4) X x1 x2 x3 x4 P
P(X=x1)
P(X=x2)
P(X=x3)
Mean
k = the number of outcomes (k=4)
µ = x1·p(x1) + x2·p(x2) + x3·p(x3) xk·p(xk)
EXAMPLE
µ = 10* * *.25 – 4*.15 = 3.65 ($ mil)

9. Mean k = the number of outcomes (k=4)
µ=3.65
k = the number of outcomes (k=4)
µ = x1·p(x1) + x2·p(x2) + x3·p(x3) xk·p(xk)
EXAMPLE
µ = 10· · · ·.15 = 3.65 ($ mil)

10. Interpretation
E(x) is not the value of the random variable x that you “expect” to observe if you perform the experiment once

11. 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.

12. Example: Green Mountain Lottery
State of Vermont
choose 3 digits from 0 through 9; repeats allowed
win $500
x $0 $500
p(x)
E(x)=$0(.999) + $500(.001) = $.50

13. 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)

14. Example: coin tossing
Suppose a fair coin is tossed 3 times and we let x=the number of heads. Find m=E(x).
First we must find the probability distribution of x.

15. 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

16. Example (cont.)

17. Example (cont.) So the probability distribution of x is: x 0 1 2 3
p(x) 1/8 3/8 3/8 1/8

18. Example So the probability distribution of x is: x 0 1 2 3
p(x) 1/8 3/8 3/8 1/8
Example

19. US Roulette Wheel and Table
American Roulette (The European version has only one 0.)
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

20. 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) 37/38 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 …