ECE 302 Spring Ilya Pollak

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1 ECE 302 Spring 202 Practice problems: Multiple discrete random variables, joint PMFs, conditional PMFs, conditional expectations, functions of random variables Ilya Pollak These problems have been constructed over many years, using many different sources. If you find that some problem is incorrectly attributed, please let me know at Suggested reading: Sections in the recommended text []. Equivalently, Sections (discrete random variables only) in the Leon-Garcia text [3]. Problem. The Curious Case of a 992 Virginia Lottery. (Ilya Pollak.) Before 992, the state of Virginia ran a lottery where a winning six-number combination was selected as six distinct numbers out of 44 numbers. Anybody could buy any number of tickets and, on each ticket, select any set of six numbers. The price of one ticket was $. Thus, in some drawings no one selected the winning combination and therefore no one won the jackpot. On the other hand, there were a few drawings where more than one person happened to select the winning combination. In this case, the jackpot was divided equally among the winners. Whenever there was a drawing with no jackpot winners, the jackpot for the next drawing would increase. This led to a $27M jackpot in the February 5, 992 drawing, payable in 20 yearly installments of $.35M. (a) If all six-number combinations are equally likely, what is the probability that any one particular set of six numbers will be the winning combination? (b) If you want to be guaranteed to win (or share) the jackpot, how much money do you need to spend on the tickets? (c) Suppose you undertake to buy tickets with all possible six-number combinations. Suppose further you estimate, based on historical data, that the chances that you will win the entire jackpot (i.e., that there will not be anyone else with the winning combination) are 0.70, the chances that you will win one-half of the jackpot (i.e., that there will be one other person with the winning combination) are 0.24, and the chances that you will win one-third of the jackpot are Assuming that the jackpot is $27M, what are your expected net winnings? For the sake of simplicity, assume there are no prizes other than the jackpot (i.e., that you do not win anything if you match five or fewer numbers). (d) Under the same assumptions as in Part (c), what is your probability to incur a loss? (e) Suppose that, during your buying spree, you are prevented from completing the entire purchase by the overwhelming logistics of having to buy so many tickets and fill them out by hand. Suppose that in the end you are only able to buy 5 million tickets. What is your probability of a net loss now? What are the net expected winnings?

2 [The remainder of the text in this problem is historical background and intuition behind the questions. It is not essential to read it in order to solve the problem.] An Australian investment fund took the approach outlined in this problem. Even though it may seem like a riskless way of making $27M from a $7M investment, Part (e) illustrates that actually there was some risk. The fund, in fact, did run out of time and was able to only buy about 5 million tickets. Despite this, it turned out that they did buy a winning ticket. Another potential source of risk, which is not easily quantifiable, is touched on in Parts (c) and (d). The probability estimates are based on prior history; however, an event that is not in the historical data could always occur. Even in the case that you are able to get tickets with all the combinations and thus guarantee that you have a winning ticket, it can happen that three or more people besides you happen to have winning tickets. Your gross winnings will then be at most $6.75M, resulting in a loss. In addition, the case of you sharing with two more winners should also be considered a loss because of the fact that the winnings are paid in installments over a 20-year period. Investing the $7,059,052 in any low-risk asset (FDIC-insured savings accounts, certificates of deposit, US saving bonds, etc.) for any annual return exceeding.222% would produce more than $9M at the end of 20 years. The case of a two-way win results in about 3.3% annual return which is still not stellar, considering the risks. As it happens, the syndicate did win the entire jackpot as well as a number of lesser prizes, for a total winnings of about $27.9M. Their outlays were $5M for the tickets they bought, plus some unknown amount for the logistics and legal representation. Another source of risk which is impossible to quantify was litigation. Virginia State Lottery took several weeks to review its legal options but in the end paid up, having concluded that they would not have a strong case in court. They have since changed the lottery rules to restrict block purchases. In addition, they have increased the number of possible combinations, so that now the probability of winning or sharing the jackpot is about in 76 million. There were other risks faced by the fund, also quite difficult to quantify, such as the possibility that Virginia State Lottery could default on its obligations and be unable to pay some part of their winnings. This story was reported by The New York Times [5,6]. Problem 2. On life expectancies and life insurance. (Ilya Pollak.) (a) Let L be a discrete random variable, and let n be a number such that P(L > n) > 0. Show that the following inequality holds: E[L L > n] > n Let m be a number such that P(L m) > 0. Show that the following inequality holds: E[L L m] m (b) Use the total expectation theorem and your results from Part (a) to show that E[L] < E[L L > n] for any random variable L and any number n such that P(L > n) > 0 and P(L n) > 0. 2

3 (c) This part refers to the life expectancies found in Table 7 on page 26 of National Vital Statistics Reports for 2006, Vol. 57, No. 4, published April 7, 2009 [4]. It can be viewed at: 4.pdf Looking at the first column, it appears that the life expectancy in the US has been steadily decreasing. For example, somebody born in 2006 had life expectancy A person born in 996 and who therefore was 0 years old in 2006, had remaining life expectancy of 68.4, for a total of A person born in 976 was 30 years old in 2006 and had remaining life expectancy of 49.2, for a total of A person born in 936 was 70 years old in 2006 and had remaining life expectancy of 4.9, for a whopping total of 84.9! Yet, here is a headline from The Washington Post, April 9, 2009 [8]: Life Expectancy Hits New High of Nearly 78. Is the headline wrong? If not, how do you reconcile this headline with the argument given above that seems to show that the life expectancies are declining? (d) Suppose you are a life insurance company and are offering a whole life insurance policy that pays $00,000 to the family of the insured upon the insured s death. Your every client gets charged a yearly premium that remains constant throughout the client s life. This premium is determined by the sex of the client as well as his or her age at the time that they purchase the insurance policy. The premium payments stop when the client dies. The premium is designed so that your total expected profit per customer is $5000. What amount will you charge a male customer who is 40 years old when he purchases the policy? What amount will you charge a female customer who is 20 years old when he purchases the policy? Refer to the same life expectancy table as in Part (c). Assume zero interest rates and inflation, i.e., assume that $ today is worth exactly $ at any time in the future. Problem 3. (Drake [2], Problem 2.06.) A fair four-sided die (with faces labeled 0,, 2, 3) is thrown once to determine how many times a fair coin is to be flipped: if N is the number that results from throwing the die, we flip the coin N times. Let K be the total number of heads resulting from the coin flips. Given K = k, the coin flips are conditionally independent. Determine each of the following probability mass functions for all values of their arguments: (a) p N (n). (b) p K N (k 2). (c) p N K (n 2). (d) p K (k). (e) Also determine the conditional PMF for random variable N, given that the experimental value k of random variable K is an odd number. Problem 4. A pair of fair four-sided dice is thrown once. Each die has faces labeled, 2, 3, and 4. Discrete random variable X is defined to be the product of the down-face values. The down-face values for the two dice are independent. Determine the conditional variance of X 2 given that the sum of the down-face values is greater than the product of the down-face values. 3

4 Problem 5. Evaluate the following summation without too much work: N ( ) N n 2 A n ( A) N n, n where 0 < A < and N > 2. n=0 Problem 6. The joint PMF of discrete random variables X and Y is given by: { Cx 2 y, for x = 5, 4,...,4,5 and y = 0,,...,0. p X,Y (x,y) = 0, otherwise. Here, C is some constant. What is E[XY 3 ]? (Hint: This question admits a short answer/explanation. Do not spend time doing calculations.) Problem 7. Let X and Y be independent random variables. Random variable X has a discrete uniform distribution over the set {, 2, 3}, and Y has a discrete uniform distribution over the set {,3}. Let V = X + Y, and W = X Y. (a) Are V and W independent? Explain without calculations. (b) Find and plot p V (v). Also, determine E[V ] and var(v ). (c) Find and show in a diagram p V,W (v,w). (d) Find E[V W > 0]. (e) Find the conditional variance of W given the event V = 4. (f) Find and plot the conditional PMF p X V (x v), for all values. Problem 8. (Ross [7], Chapter 4, Problem 27.) An insurance company writes a policy to the effect that an amount of money x must be paid if some event A occurs within a year. If the company estimates that A will occur within a year with probability p, what should it charge the customer so that its expected profit will be 0% of x? Problem 9. (Ross [7], Chapter 4, Problem 2.) A total of four buses carrying 48 students from the same school arrive at a football stadium. The buses carry, respectively, 40, 33, 25, and 50 students. One of the students is selected randomly (according to a discrete uniform distribution) from the set of all 48 students. Let X denote the number of students that were on the bus carrying this randomly selected student. A bus driver is selected randomly (also according to a discrete uniform distribution) from the set of the four bus drivers. Let Y denote the number of students on his bus. Compute E[X] and E[Y ]. Problem 0. (Aditya Mohan and Ilya Pollak.) Let n be a non-random, fixed positive integer. Let p be a non-random, fixed positive real number such that p <. Let X,...,X n be independent Bernoulli random variables, with P({X i = }) = p and P({X i = 0}) = p for all i =,...,n. Random variable Z is defined as Z = X + X X n. 4

5 It is given that np(z = 0) = P(Z = ). Find the value of p. Problem : Expected utility theory and St. Petersburg paradox. (Ilya Pollak.) Recall that, in the St. Petersburg paradox problem, a fair coin gets independently flipped until the first H comes up. If N is the total number of flips up to and including the first H, the casino pays the player 2 N dollars. The paradox is in the fact that, while the player s expected winnings are infinite, most people would not pay much for the privilege of playing this game. An explanation provided by G. Cramer and D. Bernoulli is that, rather than maximizing the expected value of their wealth, people try to maximize the expected utility of wealth. Both Cramer and Bernoulli furthermore argued that any person s marginal utility of wealth is a decreasing function of wealth: an additional $ is less important to you when you are wealthy than when you are poor. Bernoulli proposed a logarithmic function as a model for utility. Let U be the utility function, and W wealth. We assume that U(W) = log 2 W for any W > 0. Thus, if a person with initial wealth w 0 plays one coin-flipping game and gets charged a fee of f < w 0, his wealth W after the game will be W = w 0 f + 2 N. His utility after the game will be U(W) = log 2 (w 0 f + 2 N ). Before playing the game, however, he does not know N and therefore does not know W and U(W). He does, however, know that N is a geometric random variable with parameter p = /2. To decide whether or not to play, he compares the expected utility of wealth after playing, E[U(W)], with the utility of current wealth, U(w 0 ) = log 2 w 0. If the expected utility E[U(W)] exceeds log 2 w 0, he will play. If the expected utility E[U(W)] is smaller than log 2 w 0, he will not play. If the expected utility E[U(W)] is equal to log 2 w 0, he will be indifferent between playing and not playing. His expected utility from playing is: E[U(W)] = 2 n log 2(w 0 f + 2 n ) () n= (a) Let m be the smallest positive integer that is greater than or equal to log 2 (w 0 f). Let m m be an integer. Show that 2 n log 2(w 0 f + 2 n ) m m n=m (b) With m defined as in Part (a), and assuming m 2, show that m n= 2 n log 2(w 0 f + 2 n ) (m + )( 2 m+ ) Marginal utility is the first derivative of utility with respect to wealth. 5

6 (c) Show that E[U(W)] is finite and nonnegative. Solution. By assumption, w 0 f > 0. Therefore, w 0 f + 2 n > 2 for any integer n, which means that log 2 (w 0 f + 2 n ) > for n. Therefore, each term in the summation () is nonnegative, and E[U(W)] > 0. On the other hand, it follows from Parts (b) and (c) that { m+2 if m = E[U(W)] 2 m+2 m + (m + ) m+ = m + if m 2 2 m 2 m 2 m m + 2 m (d) Write a Matlab function that, given an initial wealth w 0, a fee f, and a precision parameter d, approximately computes the expected utility E[U(W)] with error not exceeding d. In other words, the output of your function must be a number V such that V E[U(W)] d (i) First, your function should compute m as follows: if (w0 <= f) error( The fee must be smaller than the initial wealth ); m_star = ceil(log2(w0-f)); if (m_star < ) m_star = ; (ii) Then use the result from Part (a) to find an m such that n=m According to Part (a), any m m such that will do. One way of doing this is: m = m_star; while ((m+2)/2^(m-) >= d) m = m+; (iii) Finally, compute V = 2 n log 2(w 0 f + 2 n ) d m n= m m d 2 n log 2(w 0 f + 2 n ) 6

7 Since, according to Part (c), E[U(W)] V > 0 and E[U(W)] V = 2 n log 2(w 0 f + 2 n ) d, n=m V approximates E[U(W)] to within d. Here is a possible implementation: V=0; for n=:m- V = V + 2^(-n)*log2(w0-f+2^n); (e) Write a Matlab function that, given an initial wealth w 0 and a precision parameter d, determines the maximum fee f which a person with initial wealth w 0 would be willing to pay for playing one game. Assume that the fee f is an integer number of dollars. The precision parameter d is to be used for approximately computing the expected utility, as in Part (d). (f) Using precision parameters d = 0 5 and d 2 = 0 0, calculate the maximum acceptable fees for: (i) w 0 = $0; (ii) w 0 = $00; (iii) w 0 = $000; (iv) w 0 = $0000; (v) w 0 = $,000,000. Are the answers the same for the two precision parameters? References [] D.P. Bertsekas and J.N. Tsitsiklis. Introduction to Probability, 2nd Edition, Athena Scientific, Belmont, Massachusetts, [2] A. Drake. Fundamentals of Applied Probability Theory, McGraw-Hill, Inc., 988. [3] A. Leon-Garcia. Probability and Random Processes for Electrical Engineering, 2nd Edition, Addison Wesley Longman, 994. [4] National Vital Statistics Reports, vol. 57, no. 4, p. 26, pdf. [5] Group Invests $5 Million to Hedge Bets in Lottery. The New York Times, February 25, [6] Lottery Will Pay Disputed Jackpot. The New York Times, March, [7] S. Ross. A First Course in Probability, 6th Edition, Prentice-Hall, [8] Life Expectancy Hits New High of Nearly 78. The Washington Post, August 9,