Explore the concepts of expectation, standard deviation, variance, and covariance

1. Explore the concepts of expectation, standard deviation, variance, andcovariance It is based on a lecture given by Professor Costis Maglaras at Columbia University.

2. Inventory management problem • Variance Associates (VA) is a commercial aircraft leasing firm. • It has two customers, FearUs and Oops, both of which are major overnight package delivery companies. • Airlines and overnight package companies normally own only a portion of their total fleet of aircraft, and lease the balance on an as-needed basis. • FearUs and Oops have long-term contracts with Variance, under which they have the option to request planes from VA on a monthly basis. • Current Situation: • Each month FearUs can request up to 4 planes for the succeeding month, at a price of $150,000 per month for each plane requested. • Oops can also request up to 4 planes for the succeeding month, at a price of $150,000/plane per month. • Variance owns 8 planes, in order to cover its contracts. • It has had both good and bad years in the past, but cumulatively it is just breaking even. • Out of frustration, the president of Variance Associates hires a consultant. • McBain & Co. assigns Poindexter Harbus to assess the problem and deliver a solution. • The president of VA explains to Poindexter that although VA has great margins on its leased planes, the firm's total performance has been uneven.

3. Analysis • VA has fixed costs of $75,000/plane per month (i.e., whether or not the plane was leased that month), but that all other costs incurred when a plane is in service (e.g., fuel, maintenance, crews) are covered by the customer at the customer's expense. • Examine Variance's data on the demand for planes by each customer. Compute relative frequencies and tabulates the following probabilities: FearUs Demand Oops Demand Demand 0 1 2 3 4 Demand 0 1 2 3 4 Probability 0.2 0.1 0.4 0.1 0.2 Probability 0.1 0.2 0.4 0.2 0.1 • Conclusion made by Poindexter: • VA should expect 4 of their planes to be on the ground at VA each month. • Under VA's current business policies, the expected cash flow (revenues less expenditures) each month are in fact $0. • How did Poindexter arrive at these conclusions? • What is the expected demand from FearUs? From Oops?

4. A New Strategy • To improve Variance's profitability, Poindexter proposes a new, two-tier strategy for VA: • Sell off one plane, and keep only 7 to meet demand. • To satisfy the customers, offer a new policy: If a customer ever asks for a plane and VA does not have one, VA will pay the customer $150,000. • (Actually, VA offers a “get your next plane for free” policy rather than paying the customer directly. But that is more complicated to analyze so let's assume VA pays the customer directly.) • Determine expected profits under the new strategy. • Poindexter needs to determine the probability that total demand would be 8 planes (i.e., a “stock out” event). • To do this, Poindexter multiplies the marginals for each customer's demand to construct a probability table: FearUs Demand 0 1 2 3 4 0 0.02 0.01 0.04 0.01 0.02 0.1 Oops 1 0.04 0.02 0.08 0.02 0.04 0.2 Demand2 0.08 0.04 0.16 0.04 0.08 0.4 3 0.04 0.02 0.08 0.02 0.04 0.2 4 0.02 0.01 0.04 0.01 0.02 0.1 0.2 0.1 0.4 0.1 0.2

5. Analysis probability distribution for total demand: Total Demand 0 1 2 3 4 5 6 7 8 Probability .02 .05 .14 .17 .24 .17 .14 .05 .02 • Under the new strategy, • What is the revenue? • What is the fixed cost? • What is the compensation policy cost numbers? • How to compute expected profits? • What does Poindexter's analysis indicate VA's expected profits will be? • Explain how did Poindexter derive the distribution for total demand. • Why is P(total demand = 1) = 0.05?

6. Is there a better strategy? • The president of VA likes Poindexter's results. But after pondering Poindexter's reasoning for a few minutes, he wonders: • “Why downsize by just one plane if we expect demand to be only four planes?" He reasons, “If that's what we expect, we should just carry four planes. • What is the expected demand number of planes? • Instead of carrying enough planes to handle the maximum demand, we'll carry enough to handle the average demand, and suitably compensate FearUs and Oops with a free plane if we're short. • Will VA see higher expected profits with only 4 planes to meet customer demand? • Why or why not? (Hint: Make another table.) • What is the optimal number of planes for VA to own?

7. Risk in Cash Flow • Variance Associates' accountant, Flim Flambert, now joins the fray. • Flim is concerned with the risk in VA's monthly cash flow position under the old and new strategies. • To get a measure of this risk, Flim computes the variance in monthly income. He then points out a problem in Poindexter's analysis, and announces “I can't reconcile Poindexter's probabilities with our past cash flows. We have more variance in our cash flows each month than Poindexter's figures imply.” • Flim justifies his conclusion with the actual relative frequencies of historical monthly cash flow amounts for VA: Probability Distribution of Historical Monthly Cash Flow ($ in 000s) Cash Flow ($600) ($450) ($300) ($150) $0 $150 $300 $450 $600 Probability 0.08 0.09 0.08 0.10 0.23 0.14 0.19 0.03 0.06 • To calculate the variance in monthly cash flow under the old policy of keeping 8 planes, he makes the following table: Cash Flow ($600) ($450) ($300) ($150) $0 $150 $300 $450 $600 E(Cash Flow) 0 0 0 0 0 0 0 0 0 Squared Dev 360,000 202,500 90,000 22,500 0 22,500 90,000 202,500 360,000 Sq Dev * Prob 28,800 18,225 7,200 2,250 0 3,150 17,100 6,075 21,600

8. Cash Flow • What is the variance and standard deviation in monthly cash flow (under VA's old policy), based on Flim's cash flow probabilities? • What is the variance and standard deviation in monthly cash flow (under VA's old policy), using Poindexter's probability distribution for total demand? (Hint: Make another table.) • To reconcile the discrepancy between their variance estimates, Flim and Poindexter decide to go back to square one. • Using Variance's data on the number of planes demanded by each customer each month, Flim tabulates the relative frequencies of all the joint events and obtains the following probability table: ... FearUs Demand 0 1 2 3 4 0 1 2 3 4 Oops Demand

9. Demand Pattern • What can we infer from the table about the demand patterns of the two customers? • To quantify the association between FearUs' and Oops' demand for aircraft, Flim calculates the covariance between them. • Calculate the covariance between FearUs' demand and Oops' demand. (Hint: To do this, you will need to construct a table something like the following, where X = demand from Oops, and Y = demand from FearUs: • In their analyses, Poindexter and Flim used different probability tables for the joint distribution of FearUs' and Oops' demands. • In doing so, they arrived at the same number for the expected value of total profit under the 8 plane policy. They did not arrive at the same number for the variance of total profit, however. • Why not?

10. Optimum Proposal • With the new probability table, Flim and Poindexter re-analyze Poindexter's original proposal to improve VA's profitability. • Use Flim's new probability table to find the optimal number of planes to own, if VA compensates customers with $150k/plane for shortages? • What is What are VA's expected profits under this strategy? • Are expected profits higher with covarying demands? • Does it matter what the sign of the covariance is?

11. Summary questions: • What is the optimal level of capacity for a firm facing uncertain demand: enough to cover maximum demand, average demand, or something else? • Why - what are we trading over here? • What information do we need to answer this for a particular firm? • If a firm has fixed costs and capacity constraints, which is better, to have customers whose demands positively covary or whose demands negatively covary? • Why? • What does this imply about the firm's capacity requirements and, given that, capital utilization rates?

12. Probability: the study of randomness

13. Randomness • Coin tossing. • A phenomenon is random • if individual outcomes are uncertain but there is a regular distribution of outcomes in a large number of repetitions.

14. Probability • The probability of any outcome of a random phenomenon is • long term relative frequency, i.e. • the proportion of the times the outcome would occur in a very long series of repetitions. (empirical) • Trials need to be independent. • Computer simulation is a good tool to study random behavior. • The uses of probability • Begins with gambling. • Now applied to analyze data in astronomy, mortality data, traffic flow, telephone interchange, genetics, epidemics, investment...

15. Probability Terms • Random Experiment: A process leading to at least 2 possible outcomes with uncertainty as to which will occur. • Event: An event is a subset of all possible outcomes of an experiment. • Intersection of Events: Let A and B be two events. Then the intersection of the two events, denoted A B, is the event that both A and B occur. • Union of Events: The union of the two events, denoted A B, is the event that A or B (or both) occurs. • Complement: Let A be an event. The complement of A (denoted ) is the event that A does not occur. • Mutually Exclusive Events: A and B are said to be mutually exclusive if at most one of the events A and B can occur. • Basic Outcomes: The simple indecomposable possible results of an experiment. One and exactly one of these outcomes must occur. The set of basic outcomes is mutually exclusive and collectively exhaustive. • Sample Space: The totality of basic outcomes of an experiment.

16. Basic Probability Rules 1. For any event A, 0 P(A)  1. 2. If A and B can never both occur (they are mutually exclusive), then P(A and B) = P(A B) = 0. 3. P(A or B) = P(A B) = P(A) + P(B) - P(A B). 4. If A and B are mutually exclusive events, then P(A or B) = P(A B) = P(A) + P(B). 5. P(Ac) = 1 - P(A). Independent Events • Two events A and B are said to be independent if the fact that A has occurred or not does not affect your assessment of the probability of B occurring. Conversely, the fact that B has occurred or not does not affect your assessment of the probability of A occurring. • If A and B are independent events, then P(A and B) = P(A B) = P(A) P(B). (Markov??)

17. Probability models • Two parts in coin tossing. • A list of possible outcomes. • A probability for each outcome. • The Sample space S of a random phenomenon is the set of all possible outcomes. • Examples. S={heads, tails}={H,T} • General analysis is possible.

18. What is the probability of “exactly 2 heads in four tosses of a coin”? • What kind of rules that any assignment of probabilities must satisfy? • An event is an outcome or a set of outcomes. (= it is a subset of the sample space) • A={HHTT,HTHT,HTTH,THHT,THTH,TTHH} • In a probability model, events have probabilities that satisfy ... • Two events A and B are independent if knowing that one occurs does not change the probability that the other occurs. • If A and B are independent,P(A and B) = P(A)P(B)the multiplication rule for independent events.

19. Independent/Dependent • The heads of successive coin tosses are {independent, not independent}. • The colors of successive cards dealt from the same deck are {independent, not independent}. • Two successive blood pressure measurements are {independent, not independent}. • Two successive IQ test results are {independent, not independent}

20. Conditional Probability • Example: One of the businesses that have grown out of the public's increased use of the internet has been providing internet service to individual customers; those who provide this service are called Internet Service Providers (ISPs). • More recently, a number of ISPs have developed business models whereby they do not need to charge customers for internet service at all, by collecting fees from advertisers, and forcing the non-paying customers to view these advertisements. • Jupiter Communications estimates that by the end of 2003 20% of web users will have a free ISP. 6% of all web users, it is estimated, will have both a free ISP and a paid ISP account. • In 2003, what proportion of internet users is expected to do the following? a) subscribes to both a free ISP and a paid ISP. b) subscribes only to a paid ISP. c) subscribes only to a free ISP.

21. P(AB)= P(A|B)P(B)= P(B|A)P(A) • In these simple calculations, we are making use of the conditional probability formula: P(A|B) = P(A holds given that B holds) = P(A∩B)/P(B) • This relationship is known as Bayes' Law, after the English clergyman Thomas Bayes (1702-1761), who first derived it. Bayes' Law was later generalized by the French mathematician Pierre-Simon LaPlace (1749-1827). Bayes Laplace

22. Random Variables • A random variable is a variable whose value is a numerical outcome of a random phenomenon. • Sample spaces need not consist of numbers. • Examples: number of heads in 4 coin tossing, …

23. Random Variable • A random variable is called discrete if it has countably many possible values; otherwise, it is called continuous. • The following quantities would typically be modeled as discrete random variables: • The number of defects in a batch of 20 items. • The number of people preferring one brand over another in a market research study. • The credit rating of a debt issue at some date in the future. • The following would typically be modeled as continuous random variables: • The yield on a 10-year Treasury bond three years from today. • The proportion of defects in a batch of 10,000 items. • The time between breakdowns of a machine. • Sometimes, we approximate a discrete random variable with a continuous one if the possible values are very close together; e.g., stock prices are often treated as continuous random variables.

24. Distribution: discrete • If X is a discrete random variable then we denote its pmf by PX. • The rule that assigns specific probabilities to specific values for a discrete random variable is called its probability mass function or pmf. • For any value x, PX(x) is the probability of the event that X = x; i.e., PX(x) = P(X = x) = probability that the value of X is x. • We always use capital letters for random variables. Lower-case letters like x and y stand for possible values (i.e., numbers) and are not random. • A pmf is graphed by drawing a vertical line of height PX(x) at each possible value x. • It is similar to a histogram, except that the height of the line (or bar) gives the theoretical probability rather than the observed frequency. • Are a histogram close to its corresponding pmf? • The pmf gives us one way to describe the distribution of a random variable. Another way is provided by the cumulative probability function, denoted by FX and defined by FX(x) = P(X≦ x) • It is the probability that X is less than or equal to x. • The the pmf gives the probability that the random variable lands on a particular value, the cpf gives the probability that it lands on or below a particular value. In particular, FX is always an increasing function.

25. Distribution: continuous • The distribution of a continuous random variable cannot be specified through a probability mass function because if X is continuous, then P(X = x) = 0 for all x; i.e., the probability of any particular value is zero. Instead, we must look at probabilities of ranges of values. • The probabilities of ranges of values of a continuous random variable are determined by a density function. It is denoted by fX. The area under a density is always 1. • The probability that X falls between two points a and b is the area under fX between the points a and b. The familiar bell-shaped normal curve is an example of a density. • The cumulative distribution function or cdf of a continuous random variable is obtained from the density in much the same way a cpf is obtained from the pmf of a discrete distribution. • The cdf of X, denoted by FX, is given by FX(x) = P(X≦ x). • FX(x) is the area under the density fX to the left of x.

26. Expectation • The expected value of a random variable is denoted by E[X]. • It can be thought of as the “average” value attained by the random variable. • The expected value of a random variable is also called its mean, in which case we use the notation mX. • The formula for the expected value of a discrete random variable is this: E[X] =SxxPX(x). • The expected value is the sum, over all possible values x, of x times its probability PX(x). • The expected value of a continuous random variable cannot be expressed as a sum; instead it is an integral involving the density. • If g is a function (for example, g(x) = x2), then the expected value of g(X) is E[g(X)] =Sxx2PX(x). • The variance of a random variable X is denoted by either Var[X] or sX2. • The variance is defined by sX2 = E[(X- mX)2]= E[X2] - (E[X])2. • For a discrete distribution, we can write the variance as Sx (x- mX)2PX(x).

27. Discrete random variable • Discrete random variableX has a finite number of possible values. • The probability distribution of X lists the values and their probabilities. • The probabilities pk must satisfy ... • Every probability pi is a number between 0 and 1. • p1+ p2+... +pk=1. • The probability of any event is found by adding the probabilities pi of the particular values xi that make up the event.

28. Probability histogram • Possible values of X and corresponding probability. • A relative frequency histogram for a very large number of trials.

29. Commonly Used Continuous Distribution The Normal Distribution • History: • Abraham de Moivre (1667-1754) first described the normal distribution in 1733. • Adolphe Quetelet (1796-1874) used the normal distribution to describe the concept of l'homme moyen (the average man), thus popularizing the notion of the bell-shaped curve. • Carl Friedrich Gauss (1777-1855) used the normal distribution to describe measurement errors in geography and astronomy.

30. Bernoulli Processes and the Binomial Distribution • An airline reservations switchboard receives calls for reservations, and it is found that • When a reservation is made, there is a good chance that the caller will actually show up for the flight. In other words, there is some probability p (say for now p = 0.9) that the caller will show up and buy the ticket the day of departure. • Consider a single person making a reservation. This particular reservation can either result in the person on the flight (a success) or a “no show” (a failure). Let X (a random variable) represent the result of a particular reservation. That is, we could assign a value of 1 to X if the person shows up for the flight (X = 1), and let X = 0 if the person does not. Then, P(X = 0) = 1 - p and P(X = 1) = p. • The airline is not particularly interested in the decision made by any one individual, but is more concerned with the behavior of the total number of people with reservations. • Suppose each passenger carried on the plane provides a revenue of $100 for the airline and each bumped passenger (passengers that do not find a seat due to overbooking) results in a loss of $200 for the airline. • If a plane holds 16 people, not including pilots and crew, how many reservations should be taken?

31. Bernoulli process • This is an example of a Bernoulli process, named for the Swiss mathematician James Bernoulli (1654-1705). • A Bernoulli process is a sequence of n identical trials of a random experiment such that each trial: • (1)produces one of two possible complimentary outcomes that are conventionally called success and failure and • (2)is independent of any other trial so that the probability of success or failure is constant from trial to trial. • Note that the success and failure probabilities are assumed to be constant from trial to trial, but they are not necessarily equal to each other. • In our example, the probability of a success is 0.9 and the probability of a failure is 0.1. • The number of successes in a Bernoulli process is a binomial random variable. • Random Variable: A numerical value determined by the outcome of an experiment.

32. Analysis • If the airline takes 16 reservations, what is the probability that there will be at least one empty seat? P(at least one empty seat) = = 1 - (0.9)16 = 0.815. An 81.5% chance of having at least one empty seat! So the airline would be foolish not to overbook. • Suppose we take 20 reservations for a particular flight, let Y be the number of people who show up. • Y is a binomial random variable that takes on an integer value between 0 and 20. • What is the probability function or distribution of Y? • What is the probability of getting exactly 16 passengers? A = 0.08978 • P(Y 16) = 0.133, P(Y = 17)= 0.190, P(Y = 18)=0.285, P(Y = 19)= 0.270, P(Y = 20) = 0.122 • Consider B = number of people bumped. The load L is Y - B. • The airline's total expected revenue (call this R, then R = 100L - 200B) • E(R) = E(100L - 200B) = 100E(L) - 200E(B) = 1,182.81.

33. How many reservation? Reservation 20 19 18 17 16 E(Load) 15.943 15.839 15.599 15.132 14.396 E(Bumps) 2.057 1.261 0.600 0.167 0.000 E(Revenue) $1,183 $1,332 $1,440 $1,480 $1,440 • In this case, the best strategy is to take 17 reservations. • Expected Value: The expected value (or mean or expectation) of a random variable X with probability function P(X = x) is E(X) = S xP(X=x) where the summation is over all x that have P(X = x) > 0. It is sometimes denoted X or . • Variance: The variance of a random variable X with probability function P(X = x) is Var(X) = S (x- E(X))2P(X=x) , where the summation is over all x such that P(X = x) > 0. It is sometimes denoted 2(X) or .