Explore the concepts of expectation standard deviation variance and covariance
This presentation is the property of its rightful owner.
Sponsored Links
1 / 33

Explore the concepts of expectation, standard deviation, variance, and covariance PowerPoint PPT Presentation


  • 155 Views
  • Uploaded on
  • Presentation posted in: General

Explore the concepts of expectation, standard deviation, variance, and covariance. It is based on a lecture given by Professor Costis Maglaras at Columbia University. Inventory management problem. Variance Associates (VA) is a commercial aircraft leasing firm.

Download Presentation

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

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


Explore the concepts of expectation standard deviation variance and covariance

Explore the concepts of expectation, standard deviation, variance, andcovariance

It is based on a lecture given by Professor Costis Maglaras at Columbia University.


Inventory management problem

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.


Analysis

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?


A new strategy

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


Analysis1

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?


Is there a better strategy

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?


Risk in cash flow

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


Cash flow

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


    Demand pattern

    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?


    Optimum proposal

    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?


    Summary questions

    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?


    Probability the study of randomness

    Probability: the study of randomness


    Randomness

    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.


    Probability

    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...


    Probability terms

    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.


    Basic probability rules

    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??)


    Probability models

    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.


    Explore the concepts of expectation standard deviation variance and covariance

    • 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.


    Independent dependent

    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}


    Conditional probability

    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.


    P a b p a b p b p b a p a

    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


    Random v ariables

    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, …


    Random variable

    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.


    Distribution discrete

    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.


    Distribution continuous

    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.


    Expectation

    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).


    Discrete random variable

    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.


    Probability histogram

    Probability histogram

    • Possible values of X and corresponding probability.

    • A relative frequency histogram for a very large number of trials.


    Commonly used continuous distribution

    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.


    Bernoulli processes and the binomial distribution

    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?


    Bernoulli process

    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.


    Analysis2

    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.


    How many reservation

    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 .


  • Login