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.
– 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
• 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
Demand
0
1 2 3 4
Probability 0.2 0.1 0.4 0.1 0.2
• Conclusion made by Poindexter:
Oops Demand
Demand
0 1 2 3 4
Probability 0.1 0.2 0.4 0.2 0.1
– 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
• To improve Variance's profitability, Poindexter proposes a new, twotier 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
Oops
Demand
0
1
2
3
4
0
0.02
0.04
0.08
0.04
0.02
0.2
1
0.01
0.02
0.04
0.02
0.01
0.1
2
0.04
0.08
0.16
0.08
0.04
0.4
3
0.01
0.02
0.04
0.02
0.01
0.1
4
0.02
0.04
0.08
0.04
0.02
0.2
0.1
0.2
0.4
0.2
0.1
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?
• 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
• 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
– 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: ...
0
Oops
Demand
0
1
2
3
4
FearUs Demand
1
2
3
4
0.08
0.07
0.03
0.02
0.05
0.02
0
0.07
0.21
0
0
0.03
0
0.01
0.11
0.1
0.2
0.4
0.01
0.01
0.2
0.01
0
0.1
0.10
0.02
0.4
0.06
0.01
0.1
0.02
0.06
0.2
0.2
0.1
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
• 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:
• 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
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
• 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
•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
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.
6. If A and B are independent events, then
P(A and B) = P(A  B) = P(A)  P(B). (Markov??)
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.
• 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
• 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
•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)
• 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 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
•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
• 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
•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
•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] =Sx xPX(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)] =Sx x2PX(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 X has a finite number of possible values.
• The probability distribution of X lists the values and their
probabilities.
Value of X
Probability
x1
p1
x2
p2
x3
p3
…
…
xk
pk
–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
• Possible values of X and
corresponding probability.
• A relative frequency
histogram for a very large
number of trials.
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
• 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
• 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.
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?
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 x P(X=x)
where the summation is over all x that have P(X = x) > 0. It
is sometimes denoted mX or m.
•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 s2(X) or .