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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 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 i1 (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 …