1. 1 Chapter 16 Random Variables Random Variables and Expected Value

2. 2 Betting on Death!  Many people in America have life insurance policies. Although you might not want to think of it this way, when you purchase a life insurance policy, you’re betting that you will die sooner rather than later…  Although this is a bet that people really don’t want to win, it is a bet that they are willing to take just to be sure that their families are financially secure in the event of death.  Most families depend on the income of one or more people in the household. What would happen if that income suddenly disappeared? Life insurance help us handle such disasters.  When you purchase a life insurance policy, it’s in your best interests that the company makes a profit and does well; why do you think that is?

3. 3 Betting on Death! Question:  You purchase a policy that charges only $50 a year. If it pays $10,000 for death and $5000 for a permanent disability, is the company likely to make a profit?  Actuaries at for the company have determined the following probabilities in any given year:  P (Death) = 1/1000  P (Permanently disabled) = 2/1000  P (Healthy) = 997/1000  We’ll come back to this problem later on…

4. 4 Random Variables and Expected Value Random Variable  A Random Variable is a variable whose values are numbers that are determined by an outcome of a random event. values  Random variables are denoted by capital letters, while the values of random variables are denoted with lowercase letters (small letters) mean expected value  The mean of the discrete random variable, X, is also called the expected value of X. Notationally, the expected value of X is denoted by E(X). It is what we expect to happen. The formula for expected value is:

5. 5 Examples  In the experiment of flipping three coins, consider the outcomes and define the random variable X as the number of heads that appear.  The outcomes are {no heads, 1 head, 2 heads, or 3 heads}  X has values in the set: {0, 1, 2, 3   When rolling two dice and finding the sum, determine the outcomes and the random variable Y.  The outcomes are {(1,1), (1,2), (1,3), (1,4), (1,5), etc…}  Y has values in the set: {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}  In our life insurance example, what are the outcomes and random variables Z if we define them as the possible payments.  The outcomes are {die, disabled, healthy}  Z has values in the set: {$10,000, $5000, $0}

6. 6 Back to Betting on Death  So, will the company make a profit for any given year? How much will they make or lose?  These questions are answered by finding the expected value.  The expected Value is: Policyholder Outcome PayoutxProbability P(X = x) Die $10,0001/1000 Disability $50002/1000 Healthy $0997/1000

7. 7 Back to Betting on Death  So, what does this mean? Since each customer pays $50 per year, the company expects to make a profit of $30 per customer per year.  The expected value for the company is a payout, on average, of $20 per customer per year. Since each customer pays $50 per year, the company expects to make a profit of $30 per customer per year. expected average payout  It’s important to note that the insurance company will never really pay anyone $20; it only pays $10,000, $5000, or $0. $20 is the expected average payout given a large number of policy holders based on the LLN.

8. 8 Labor Costs  A car’s air conditioner recently needed to be repaired at the auto shop. The mechanic said that it could for $60 in 75% of the cases by drawing down and recharging the coolant. If that fails, it will cost an additional $140 to replace the unit.  What are the outcomes, random variables, and the probability distribution? OutcomeCostxProbability P(X = x) Quick fix works$60¾ =.75 Replace unit$200¼ =.25

9. 9 Labor Costs  A car’s air conditioner recently needed to be repaired at the auto shop. The mechanic said that it could for $60 in 75% of the cases by drawing down and recharging the coolant. If that fails, it will cost an additional $140 to replace the unit.  What is the expected value of the cost of this repair?

10. 10 Labor Costs  A car’s air conditioner recently needed to be repaired at the auto shop. The mechanic said that it could for $60 in 75% of the cases by drawing down and recharging the coolant. If that fails, it will cost an additional $140 to replace the unit.  What does this mean in context of this problem?  Car owners with this problem will spend an average of $95 to get their car fixed at this auto shop.

11. 11 Got to Love Those Aces  It takes $5 to play a game  From a standard 52 card deck of cards, if you get an ace of hearts, you get $100  If you get any other ace, you get $10  If you get any other heart, you get your $5 back  If you get any other card, you lose  What is the expect value of this game and is it worth it to play this game?

12. 12 Got to Love Those Aces Outcome X = Payout Probability: P(X = x) Ace of Hearts$951/52 =.0192 Other Aces$53/52 =.0577 Other Hearts$012/52 =.2308 Other Cards-$536/52 =.6923  First, you want to determine your possible winnings (let’s include the $5 cost) and probabilities:   Now, we can find the expected value, E(X):   Is this game worth playing?

13. 13 Probability Histogram  We can use histograms to display probability distributions as well as distributions of data.

14. 14 Discrete Versus Continuous  Earlier this year, we covered the notion of discrete and continuous, but it needs more exploration now… A discrete random variable has a countable number of outcomes. In other words, it is possible for you to count and make a list of all of the possible outcomes. Discrete random variables take on only integer values. Suppose, for example, that we flip a coin and count the number of heads. The number of heads results from a random process - flipping a coin. And the number of heads is represented by an integer value.

15. 15 Continuous Random Variable  Continuous random variables average  Continuous random variables, in contrast, can take on any value within a range of values. For example, suppose we flip a coin many times and compute the average number of heads per flip. The average number of heads per flip results from a random process - flipping a coin. And the average number of heads per flip can take on any value between 0 and 1, even a non-integer value. Therefore, the average number of heads per flip is a continuous random variable. A continuous random variables are not countable. In other words, you cannot list every single possible outcome. For example, the amount of water can you put into a 5-gallon container – there are an infinite number of possibilities.

16. 16 Example  Which of the following is a discrete random variable? I. The average height of a randomly selected group of boys. II. The annual number of sweepstakes winners from New York City. III. The number of presidential elections in the 20th century. (A) I only (B) II only (C) III only (D) I and II (E) II and III

17. 17 Solution  The correct answer is B. The annual number of sweepstakes winners is an integer value and it results from a random process; so it is a discrete random variable. The average height of a group of boys could be a non-integer, so it is not a discrete variable. And the number of presidential elections in the 20th century is an integer, but it does not vary and it does not result from a random process; so it is not a random variable.

18. 18 Assignment Chapter 16 Lesson : Random Variables and Expected Value Read : Chapter 16 Problems : 1 – 41 (odd)