V. RANDOM VARIABLES, PROBABILITY DISTRIBUTIONS, EXPECTED VALUE

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1 V. RANDOM VARIABLES, PROBABILITY DISTRIBUTIONS, EXPECTED VALUE A game of chance featured at an amusement park is played as follows: You pay $ to play. A penny a nickel are flipped. You win $ if either heads or tails turn up, you get your dollar back. You lose $ if a head a tail turn up, you do not get your dollar back. Another way to describe this game is: Two coins are flipped. You win $ if either heads or 0 heads turn up; you lose $ if exactly one head turns up. Suppose you can play this game as many times as you like. Would you play over over again? Does either side (i.e., you or the amusement park) have an advantage? If so, what is it? The objective of this section is to provide the mathematical methods to answer these questions. Rom Variable: The probability experiment involved in the game described above is: Flip a penny a nickel. The sample space of equally likely simple events is: S = { HH, HT, TH, TT}. For the purposes of our game, we are not interested in the particular pattern of the coins. Rather we want to know the number of heads that turns up. So, we associate with HH, with HT with TH, 0 with TT. That is, we have defined a function that associates a real number with each of the simple events in S. Given a probability experiment with sample space S. A rom variable X is a function that assigns a numerical value to each simple event in S. Examples 5.:. Flip a penny, a nickel a dime. Let S be the sample space of equally likely simple events let X be the number of heads that turn up. Give the values of X. Sample Space S Number of Heads X(e i ) e : HHH e : HHT e : HTH 7

2 e : THH e 5 : HTT e : THT e 7 : TTH e : TTT 0. In the game of craps, dice are tossed (If it helps you, imagine that one die is red the other is green). The interest in this game is not on the particular number, say i, on the red die the particular number, say j, on the green die that turns up. Instead, we are interested in the some of the two numbers i + j. Thus, to each of the equally likely simple events (i,j) in the sample space, we assign the number i + j. This association defines the rom variable X the sum of the numbers that turn up. Some of the values of X are: X [(,)] =, X[(,)] = 7, X[(,)] = 9, X[(,)] =. Probability Distribution for a Rom Variable: In Example 5.. we defined the rom variable X whose values were 0,,,. We now consider the questions: If we toss coins, (you can think of them as a penny, a nickel a dime), what is the probability that X = 0? That X =? That X =? That X =? The value X = 0 corresponds to the event E 0 = exactly 0 heads turn up ; E 0 = { TTT} n( E ) P ( X = 0) = P( E ) 0 0 = =. n( S) The value X = corresponds to the event E = exactly head turns up ; E = { HTT, THT, TTH} Similarly, n( E ) P ( X = ) = P( E ) = =. n( S) P ( X = ) = P ( X = ) =. These results are summarized in the table below. This table is called a probability distribution for the rom variable X. 7

3 Number of Heads x 0 Probability P(X = x) A graphical representation of this probability distribution is shown in the following figure, called a histogram Probability, P(X=x) / / / / 0 Number of Heads (x) Note from the table that () 0 P( X = x) (P(X = x) is the probability of an event) () P ( X = 0) + P( X = ) + P( X = ) + P( X = ) =. These are the properties that characterize the probability distribution of a rom variable X associated with any probability experiment: 7

4 Let X be a rom variable with values { x, x,..., xk}. The probability function P X = x), x { x, x,..., x }, is a probability distribution of the rom variable X if ( k () For each x x, x,..., x }, 0 P ( X = x) { k () P X x ) + P( X = x ) P( X = x ). ( = k = Examples 5.:. Two dice are rolled X is the rom variable the sum of the numbers that turn up. Find the probability distribution of X.. A box of 0 flashbulbs contains defective bulbs. A rom sample of is selected tested. Let X be the number of defective bulbs in the sample. Find the probability distribution of X. Solutions:. Based on the results of preceding examples, the probability distribution of X is X P(X = x) 5 5. Let X be the rom variable: the number of defective bulbs. Then the values of X are 0,,. The total number of ways to select objects from a collection of 0 without regard to the order of selection is. C 0, The probability of 0 defective bulbs: the number of ways to select good bulbs is C 7, C7, 7 P(0) = = =. C 5 5 0, The probability of defective bulb( good bulb, bad bulb): the number of ways to select good bulb defective bulb is C7, C, 75

5 C7, C, 7 7 P() = = =. C 5 5 0, The probability of defective bulbs: the number of ways to select defective bulbs is C, C, P() = = =. C 5 5 0, The probability distribution of X is: X 0 P(X = x) Expected Value: We continue with our probability experiment of tossing coins letting X be the number of heads that turn up. Suppose we repeat the experiment n times, recording the number of heads that turn up each time. If we then add up the number of heads that occur in each of the n tosses divide by n, we will get the average number of heads per toss. What would we expect the average number of heads per toss to be? According to the probability distribution given in the table above, when n is large 0 heads will occur in of the tosses, head will occur in of the tosses, heads will occur in of the tosses, heads will occur in of the tosses. Therefore, for large n, the total number of heads that will occur is: 0 ( ) n ( ) n + ( ) n + ( ) n = [0( ) + ( ) + ( ) + ( )] n +. Dividing by n we get the average number of heads per toss: 7

6 ( ) ( ) + ( ) + ( ) = = This number is called the expected value of the rom variable X. Note that the expected value of X is not necessarily a value that will occur in a single trial of the experiment (In this particular case, you can not get.5 heads if you flip coins). The expected value is the average of what occurs over a large number of trials of the experiment. In general, let X be a rom variable with probability distribution given by the table X x x x k P(X = x) p p p k The expected value of X, denoted by E(X), is given by the formula E X ) = x p + x p x k p k (. Examples 5.:. An experiment consists of rolling a single die. Let the rom variable X be the number that turns up. What is the expected value of X?. Two dice are rolled X is the rom variable the sum of the numbers that turn up. Find E(X).. A box of 0 flashbulbs contains defective bulbs. A rom sample of is selected tested. Let X be the number of defective bulbs in the sample. Find E(X). Solutions:. Let X = the number that turns up. The probability distribution of X is Therefore, X 5 P(X = x) 77

7 EX ( ) = + = =.5.. From Example 5.., EX ( ) = = = 7. From Example 5.., EX ( ) = = = Expected Value of a Game: Now we use the expected value of a rom variable to find the expected value of a game of chance. We return to the example that began this section. You pay $. A penny a nickel are flipped. You win $ if either heads or tails turn up, your $ is returned; otherwise you lose $ your $ is not returned. The sample space is: S = {0 heads,head, heads}. This time our rom variable X is the amount you win if 0 heads, head, or heads turn up. The probability distribution for X, called a payoff table, is: Number of Heads 0 X $ -$ $ Probability p The expected value of the game is: ( ) + ( $) ( ) + $( ) $ $ = $ = 50 E ( X ) = $ =. 7

8 If you play this game over over again, you will lose an average of 50 per game. The amusement park has the advantage in this game; the park wins an average of 50 each time the game is played. A game is fair if only if the expected value E ( X ) = 0. That is, neither side has an advantage. Our amusement park game is not fair. So this raises the question: For the game to be fair, how much should you lose, in addition to the $ you paid to play, if exactly one head turns up? Denote that amount by u. Then the payoff table is: Number of Heads 0 X $ -$(u + ) $ Probability p Since the game is fair, we have Solving this equation for u we get, ( ) + [ $( + )] ( ) + $( ) = 0 E ( X ) = $ u. u = 0 which gives u =. Thus, the game will be fair if it is modified so that we lose $ if exactly one head turns up. Examples 5.:. A roulette wheel has equally likely spaced slots numbered 00, 0,,,,. A player bets $ on one of the numbers wins $5 ( gets the $ back) if the ball comes to rest on the chosen number; otherwise the player loses the $. the rom variable X assigns the value $5 to having the ball l on the chosen number $ to having the ball l on any other number. What is the expected value of roulette?. One thous tickets are sold at $ each for a charity raffle. Tickets are to be drawn at rom cash prizes are to be awarded as follows: prize of $00, prizes of $50, 5 prizes of $0. What is the expected value of this raffle if you buy ticket? 79

9 . In the original Texas Lotto game you paid $ to select numbers from the set {,,,..., 50}. (a) What is your expected value if you buy ticket the jackpot is $ million? (b) What is your expected value if you buy ticket the jackpot is $0 million? (c) At what jackpot level is Texas Lotto a fair game? Solutions:. The payoff table for roulette is: X $5 $ p 7 The expected value of the game is: 7 ( ) + ( ) ( ) = $ = $ 5 E ( X ) = You have a: = 0.00 chance of winning $99 [ $00 $ (the cost of the ticket)] = 0.00 chance of winning $9 = chance of winning $9 99 = 0.99 chance of winning $. 000 Therefore, the payoff table is: X $99 $9 $9 $ p

10 The expected value is: E ( X ) = 99(.00) + 9(.00) + 9(.005) 0.99 = 0.5. Thus E( X ) = $0.5 = 5.. You are choosing numbers from the set of 50 numbers. This can be done in C 50, different ways (the order of the selection is immaterial). Therefore, the probability of having the winning ticket is / C 50, = /5,90, (a) The payoff table is: X $,999,999 -$ p E ( X ) = (,999,999) = $0.75 (b) The payoff table is X $9,999,999 -$ p E ( X ) = (9,999,999) = $.5 (c) Let N be the jackpot level at which the game is fair. Then E ( X ) = N( ) = N = 5,7, Exercises:

Expected Value 7/7/2006

Expected Value 7/7/2006 Definition Let X be a numerically-valued discrete random variable with sample space Ω and distribution function m(x). The expected value E(X) is defined by E(X) = x Ω x m(x), provided