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Transcript of 1 BAMS 517 Decision Analysis – II Acquiring and Using Information Martin L. Puterman UBC Sauder...
1
BAMS 517Decision Analysis – IIAcquiring and Using Information
Martin L. Puterman
UBC Sauder School of Business
Winter Term 2 2009
2
The “Monty Hall” problem
Monty Hall was the host of the once-popular game show “Let’s Make a Deal” In the show, contestants were shown three doors, behind each of which was a
prize. The contestant chose a door and received the prize behind that door This setup was behind one of the most notorious problems in probability Suppose you are the contestant, and Monty tells you that there is a car behind
one of the doors, and a goat behind each of the other doors. (Of course, Monty knows where the car is)
Suppose you choose door #1
3
The “Monty Hall” problem
Before revealing what’s behind door #1, Monty says “Now I’m going to reveal to you one of the other doors you didn’t choose” and opens door #3 to show that there is a goat behind the door.
Monty now says: “Before I open door #1, I’m going to allow you to change your choice. Would you rather that I open door #2 instead, or do you want to stick with your original choice of door #1?”
What do you do?
4
Bayes’ rule – repeated in a more general form
Bayes’ Rule can be written in general as
where A1, A2, …,An is a partition of the sample space. Note there is also a continuous version involving integrals. In the above
P(Ai) for i=1,…,n are called the prior probabilities P(B|Ai) for i=1,…,n are called the likelihoods P(Ai|B) for i=1,…,n are called the posterior probabilities
)()|()()|()()|(
)()|()|(
2211
111
nn APABPAPABPAPABP
APABPBAP
5
The “Monty Hall” problem - analysis
It does not appear at first that Monty’s showing you a goat behind door #3 is at all relevant to whether the car is behind doors 1 or 2. So why should you switch?
A careful analysis is needed here. What is important to remember is that you chose door 1 and Monte knows where the car is. Let C1, C2, C3 denote the events of the car being behind doors 1, 2, and 3 Initially, you believe P(C1) = P(C2) = P(C3) = 1/3 Since you chose door #1, Monty could have either opened door #2 or door
#3. Denote these events by M2 and M3. He could never open door 1; since that would defeat the purpose of the show.
Suppose the car is really behind door #2. Then in order to show you a goat, Monty would have had to open door #3. Thus P(M3|C2) = 1
Suppose the car is really behind door #1. Then Monty had the choice of opening either door #2 or door #3. Assume that the probability of Monty’s opening door 3 in this case is ½. So P(M3|C1) = P(M2|C1)= ½. Note this result holds for any p; 0<p<1.
To make a decision, we need to find P(C1 | M3) and P(C2 | M3). To do this we apply Bayes’ Rule.
6
The “Monty Hall” problem - solution
Hence P(C2|M3) = 1 -1/3 = 2/3. Thus your probability of getting the car is actually better if you switch doors!
It’s 2/3, rather than 1/3 if you stay with door #1 The information Monty gives you is relevant because it is more likely Monte will chose
door #3 when the car is behind door #2 (assuming you picked door #1). Ignoring the denominator in Bayes’ rule, we can state
In this problem the prior probabilities are all equal so that the posterior probability is directly proportional to the likelihood The likelihood of Monte opening door 3 is higher when the car is behind door 2
than when it is behind door 1. Therefore the posterior probability the car is behind door 2 is higher than door 1.
We can also use this formula to compute exact probabilities.
3/13/103/113/12/1
3/12/1
)3()3|3()2()2|3()1()1|3(
)1()1|3()3|1(
CPCMPCPCMPCPCMP
CPCMPMCP
posterior ≈ prior x likelihood
7
Monty Hall Problem revisited
Another way to think about it. Suppose you do not change your guess. Then you
would have won if you if the car was behind door 1. This happens independent of Monte’s action (why?) with probability 1/3.
Suppose you switch. Then you would have won if the car was not behind door 1. (Why?) This occurs with probability 2/3.
Note if the car was behind 3, Monte would have opened door 2.
8
Another and quite different Bayes’ example Suppose demand for a product is Poisson distributed but the rate parameter λ is not
known with certainty.
For simplicity assume a prior distribution for λ given by P(λ=10)=.4 and P(λ= 15)=.6 In general we can assume any discrete or continuous distribution for λ. Picking a gamma distribution gives nice formulas for the posterior.
Suppose we observe one week’s demand and it equals 11. How does this change our assessment of P(λ=10) and P(λ= 15)?
The likelihoods are f(11|10) =.1137, f(11|15) =.0663 So the Posterior estimates of the probability are obtained from Bayes’ Rule in the form
on slide 6: P(λ=10|n =11) ≈ .1137• .4 = .0455 P(λ=15|n =11) ≈ .0663 • .6 = .0398
Since these two quantities have to sum to one P(λ=10|n =11) = .0455/ (.0455+.0398) = 0.534 P(λ=15|n =11) = .0398 / (.0455+.0398) = 0.466
!)|(
n
exf
n
9
More about Bayesian updating; beta prior and binomial likelihood
Suppose our event has a Binomial distribution;
Special case; when n = 1 is a Bernouilli distribution. Now suppose that our prior distribution on θ has a Beta Distribution;
Emphasis: this is a distribution on θ and has support [0,1] It is highly flexible and can represent a wide range of shapes We will call it Beta(α,β). Its mean is α/α+β and its variance is αβ/(α+β)2(α+β+1)
Suppose we have one realization of the binomial and observe x. Then the posterior distribution is Beta( α+x, n-x+β)!
This is quite amazing. The proof is quite simple and just involves some renormalization in the integration over θ.
Thumbtack experiment revisited. In last class our prior looked pretty flat. After observing one “tack up” in one toss our prior had become sharper and less spread. This is supported by this theory. We could have fit a prior distribution to our data and then done formal analysis.
nxx
nxXP xnx ,,1,0)1()(
11 )1(),(
1),|(
B
10
What is our new assessment of probabilities of observing outcomes after our observation? This is a bit more complicated
The likelihood remains as before but what we need is the marginal distribution to know is the unconditional probability of observing x.
In our example; the prior is Beta(α,β) and the likelihood is Binomial(n,x). It is simple to derive the marginal distribution of x successes
which is.
where Γ(x) = (x-1)! when x is integer. If α and β are integers the above simplifies.
dfxlxp )()|()(
)()()(
)()()()(
n
xnx
x
nxp
11
Conjugate priors
A prior distribution is said to be conjugate to a likelihood if the posterior distribution has the same form as the prior distribution.
Important examples include Beta prior and Binomial likelihood Normal prior and Normal likelihood Gamma prior and Poisson likelihood.
See the above link for more examples. Certain cases have nice marginal distributions
too.
12
Degrees of certainty
In some cases, like flipping a coin or drawing a ball from an urn at random, we feel pretty sure of the probabilities of the outcomes We really expect the probability of heads to be at or very near to 50%
In other cases, we may not be so sure Consider flipping a thumbtack into the air. It could land in one of either two
ways: ‘pin up’ or ‘pin down’:
What is the probability of the tack landing pin up on the next throw? Suppose we assess it to be 50%. However, we may not feel as though the precision of this assignment is as great as the answer we gave for the coin. We just don’t know enough about the tack to say for sure
‘pin up’ ‘pin down’
13
Degrees of certainty
How can we distinguish between such ‘vague’ and ‘precise’ probability assessments in our analyses?
Let’s consider the following outcomes: the coin has probability p of landing heads, where p can be any number in {0, .01, .02, …, .99, 1} Label these events as 0, …, 100 Of course, probabilities can be any real number between 0 and 1. To keep things
simple and discrete, we are only considering probabilities up to 2 decimals here Now consider the thumbtack: it has probability q of landing pin up, where q can be
any number in {0, .01, .02, …, .99, 1}. Label these events 0, …, 100 Suppose we assign prior probabilities to each of 0, …, 100 and 0, …, 100. The
prior probabilities for the coin’s likelihood of coming up heads should be much more concentrated around 50% than the prior probabilities for the tack’s likelihood of landing pin up Don’t confuse the event j – the event that the coin has probability j/100 of landing
heads when tossed – with the prior probability P(i) assigned to this event It may help to think of P(i) as the probability that the coin is ‘disposed’ to land heads j
percent of the time
14
0
0.002
0.004
0.006
0.008
0.01
0.012
0.014
0.016
0.01
0.07
0.13
0.19
0.25
0.31
0.37
0.43
0.49
0.55
0.61
0.67
0.73
0.79
0.85
0.91
0.97
Probability of landing pin up (phi)
Pri
or
Pro
bab
ilit
y (P
(ph
i))
0
0.010.02
0.03
0.040.05
0.06
0.070.08
0.09
0.01
0.07
0.13
0.19
0.25
0.31
0.37
0.43
0.49
0.55
0.61
0.67
0.73
0.79
0.85
0.91
0.97
Probability of landing heads (theta)
Pri
or
Pro
bab
ilit
y (P
(th
eta)
)
Degrees of certainty
The graphs display representative prior probabilities for the coin and the tack
Note our prior distribution for the probability of the tack landing pin up is more ‘spread out’ than that for the probability of the coin landing heads
Let H and U be the events of the coin landing heads and the tack landing pin up in a given toss, respectively P(H) = i P(H | i) P(i) ; P(U) = i P(U | i) P(i) For the examples given above, P(U) = P(H) = 0.5, even though P(i) P(i)
Coin Thumbtack
15
Degrees of certainty
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.01
0.07
0.13
0.19
0.25
0.31
0.37
0.43
0.49
0.55
0.61
0.67
0.73
0.79
0.85
0.91
0.97
Probability of landing heads (theta)
Pri
or
and
Po
ster
ior
Pro
bab
ilit
y (P
(th
eta)
)
0
0.01
0.02
0.03
0.04
0.05
0.06
0.01
0.07
0.13
0.19
0.25
0.31
0.37
0.43
0.49
0.55
0.61
0.67
0.73
0.79
0.85
0.91
0.97
Probability of landing pin up (phi)
Pri
or
and
Po
ster
ior
Pro
bab
ilit
y (P
(ph
i))
The more certain one is about the occurrence of an event, the more reluctant one will be to accept that this event does not occur in the light of evidence to the contrary
Suppose you flip the coin/tack 10 times and find that all 10 times it lands heads/pin up. Denote this event by E. This would suggest that the coin/tack is strongly biased in favor of landing heads/pin up. However, since we are more sure that the coin is fair than is the tack, we may be less willing to accept that the coin is strongly biased than the tack, and more inclined to write off the event as a fluke random event
The resulting posterior probabilities P(i | E) and P(i | E) are plotted in red below. The posterior probabilities for the tack have responded greatly to the new evidence, but those for the coin have shifted only slightly. Also, now P(H | E) = .55, but P(U | E) = .86
Coin Thumbtack
16
Degrees of certainty If we have no information regarding the likelihood of events 1,…,n, then we
should assign them equal (uniform) probabilities in our prior assessment. If we are more certain that some events will occur, we assign them a higher prior
probability than they would receive under the uniform assignment Suppose you attribute a probability of 0 to the event A. This indicates that
you are absolutely certain that A will not occur, and no amount of information will then sway you to believe that A might occur P(A | E) ≈ prior • likelihood = 0, since the prior P(A) = 0
This also implies that if you assign probability 1 to an event, then no evidence will make you believe that A will not occur
As a general rule, you should avoid assigning probabilities of 0 or 1 to events that might be revised in light of subsequent information Cromwell’s Rule :
0 < P(A | H) < 1, unless H logically implies A or Ac
Use small or large probabilities like .001 or .999 instead
17
Complex decisions
So far, the decisions we have studied have been quite simple, usually consisting of a single decision and a single uncertain event
We’ll now show how to solve more complex decision problems, by repeatedly using the principles we have developed for simple problems
The basic tool we will use to structure complex problems is the decision tree
The major new principle used in solving complex decisions is the idea of backward induction
18
Example: Hatton Realty
Imagine you are a real estate agent in Vancouver. One day, a new client comes to you and says:
“I currently own several properties in the West Side of Vancouver that I would like to sell. I would like to list three of my properties in Dunbar, Kitsilano, and Point Grey through your agency, but on the following conditions:
1) You sell the Dunbar property for $20,000, the Kitsilano property for $40,000, and the Point Grey property for $80,000
2) You will receive a 5% commission on each sale3) You must sell the Dunbar property first, and within a month4) If you sell the Dunbar property, then you may either sell the Kitsilano or Point Grey
property next; otherwise, you will lose the right to sell these properties5) If you sell this second property within a month, then you may list the third property;
otherwise, you lose the right to sell the third property” In considering this proposition, you assess the promotional of listing these
properties, as well as the probability of selling them to be:Property Listing Cost Prob. of SaleDunbar $800 .6Kitsilano $200 .7Point Grey $400 .5
19
Structuring complex decision problems
In the Hatton Realty problem, several events and decisions are made in sequence
The first step in analyzing a more complex decision such as this is to map the chronology of decisions and events in a timeline. Trace every possible progression of decisions and events that may occur What is the first decision that needs to be made? For each action that you might take at this point, what uncertain events that
may impact future decisions or outcomes follow? What are their probabilities of occurrence?
For every possible event realization, what subsequent decisions or events can take place? Etc.
It is usually helpful to think of there being a sequence of decision points, between each of which may occur one or more uncertain events
Once you come to the end of each possible chain of decisions and events, you need to enter the total value of reaching that point.
20
Decision trees
A useful way to represent complex decisions is through a decision tree Every decision point is represented by a square node
in the tree. Every “branch” leading out of such a node represents a possible decision
Each uncertain event is represented by a round node, and every “branch” leading out of such a node represents a possible realization of the event. These branches are labeled with the probabilities of each of these realizations
The “root” of the tree corresponds to the first decision that must be taken
The “leaves” of the tree represent final outcomes
21
The Hatton Realty Decision Problem
Refuse to List Dunbar
$0
List Dunbar
-$800
Don’t Sell Dunbar
Sell Dunbar
.6
.4
$200
Refuse next property
List Kits
List Pt. Grey
Sell PG
Don't Sell
-$200
$3,800
Refuse Kits
Accept Kits
.5
.5
.7
.3
Sell
Don't Sell
$3,600
$5,600
$0Don't Sell
.5
.5
Sell
Don't Sell
$1,600
$5,600
Refuse Pt Grey
Sell Kits
Accept Pt Grey
.7
.3 $2,000
22
Backward induction
Once we have structured the decision as a decision problem, how do we determine the correct actions to take at each decision node?
We use the following node replacement method to reduce the size and complexity of the tree: Replace any terminal event node
with the leaf
Replace any terminal decision node
with the leaf
Replacing these nodes, we work backwards from the end of the tree, computing expected utilities and decision values as we go, until we replace the entire tree with a single leaf
In replacing the decision nodes, we need to record which decisions achieve the maximum – these are the decisions we want to take if we ever reach this node
p
1-p
a
bpa+(1-p)b
a
bmax(a,b)
23
Solving Hatton Realty
Refuse to List Dunbar
$0
List Dunbar
-$800
Don’t Sell Dunbar
Sell Dunbar
.6
.4
$200
Refuse next property
List Kits
List Pt. Grey
Sell PG
Don't Sell
-$200
$3,800
Refuse Kits
Accept Kits
.5
.5
.7
.3
Sell
Don't Sell
$3,600
$5,600
$0Don't Sell
.5
.5
Sell
Don't Sell
$1,600
$5,600
Refuse Pt Grey
Sell Kits
Accept Pt Grey
.7
.3 $2,000
24
Solving Hatton Realty
Refuse to List Dunbar
$0
List Dunbar
-$800
Don’t Sell Dunbar
Sell Dunbar
.6
.4
$200
Refuse next property
List Kits
List Pt. Grey
Sell PG
Don't Sell
-$200
$3,800
Refuse Kits
Accept Kits
.5
.5
$5,000
$0Don't Sell
Refuse Pt Grey
Sell Kits
Accept Pt Grey
.7
.3 $2,000
$3,600
25
Solving Hatton Realty
Refuse to List Dunbar
$0
List Dunbar
-$800
Don’t Sell Dunbar
Sell Dunbar
.6
.4
$200
Refuse next property
List Kits
List Pt. Grey Sell PG
Don't Sell
-$200
.5
.5
$0Don't Sell
Sell Kits
.7
.3
$3,600
$5,000
26
Solving Hatton Realty
Refuse to List Dunbar
$0
List Dunbar
-$800
Don’t Sell Dunbar
Sell Dunbar
.6
.4
$200
Refuse next property
List Kits
List Pt. Grey
$2,400
$2,500
27
Solving Hatton Realty
Refuse to List Dunbar
$0
List Dunbar
-$800
Don’t Sell Dunbar
Sell Dunbar
.6
.4
$2,520
28
Solving Hatton Realty
Refuse to List Dunbar
$0
List Dunbar
$1,192
29
The value of the proposition is $1192 You will choose the following policy
List Dunbar If Dunbar Sells, List Kits; If Kits sells list Pt Grey
Note that we used the expected monetary values as outcomes In this problem, none of the probabilities assigned to each event
depended on prior decisions We assumed that the probabilities of each sale were independent of the
order in which the properties went on the market This will not always be the case – sometimes the probabilities of events
will depend on previous decisions and previous events The probability assigned to any event realization should be the
probability of that event conditioned on all decisions and events that preceded it up to that point
Solving the realty problem
30
Payoff distribution under the optimal policy
Amount Probability
-800 .4
0 .18
1600 .21
5600 .21
Mean Payoff ? Standard deviation of payoff?
31
Newsvendor Problem
Items cost c, you sell them for p and if they can’t be sold you receive a scrap value s
For every unit sell you make a profit of G = p-c For every unit you order and do not sell you incur a loss
L=c-s Demand D is unknown and either discrete with
distribution P(D=d) or continuous with density f(d) How many should you order to maximize expected
profit? Handout to be distributed.
32
Using Information
33
Value of Information
Suppose in the Hatton Realty case you could do some market research before accepting the offer.
What would it be worth to know in advance whether or not the Dunbar property would sell? Or how much would you pay a clairvoyant to get this
information?
Let’s simplify the Hatton Realty decision tree to see how to take the availability of this information into account.
Assume for now the only option available is to list the Dunbar property by itself.
34
Revised realty problem – Dunbar only
Refuse to List Dunbar
$0
List Dunbar
-$800
Don’t Sell Dunbar
Sell Dunbar
.6
.4
$200
35
Solving revised problem
Do not List Dunbar
$0
List Dunbar
-$200
$0
36
Revised realty problem – Dunbar only; but with perfect information
Dunbarwon’t sell
Dunbar will sell
$200List Dunbar
Don’t list Dunbar
.4
.6 $0
$-800List Dunbar
Don’t list
Dunbar$0
37
Revised realty problem – Dunbar only; but with perfect information
Dunbarwon’t sell
Dunbar will sell
$200List Dunbar
Don’t list Dunbar
.4
.6 $0
-$800List Dunbar
Don’t list
Dunbar$0
$0
$200
$120
38
Selling Dunbar and the Value of Perfect Information
If we knew in advance that Dunbar would sell we would list it and make $200
If we knew in advance that Dunbar would not sell, we would not list it and earn $0. (saving $ 800).
Thus the expected value under this information would be $120. If we didn’t know in the outcome of this chance event we would not list
Dunbar and have an expected value of $0. Thus knowing in advance whether or not Dunbar will sell is worth $120. This is called the expected value of perfect information (EVPI).
In general the EVPI is the difference in expected values between the situation when the uncertain event is resolved and the expected value without this information (the base case).
Usually our base case is the no information case. The EVPI is the most we would pay for any information about the event be it a survey,
market research or …. Exercise; What would is the EVPI for the event Point Grey Sells? Kitsilano
Sells? They both sell? … What is the EVPI for the newsvendor problem?
39
Expected value of perfect information
The difference in the expected utility value between a decision made under uncertainty and the same decision made with the outcomes of the uncertainties known prior to the decision is the expected value of perfect information (EVPI)
It is not difficult to see that by removing the uncertainties from a decision problem, the value of that decision problem is increased, so that the EVPI is always positive: Under normal conditions when the outcomes of the uncertain
event are not known, the value of the decision is DU = maxi j Cij P(j)
When the uncertainties are removed prior to choosing di, the value of the decision becomes
DPI = j [maxi Cij] P(j) Since the terms in the second sum always exceed the first sum,
we can see that DPI ≥ DU
40
Suppose someone has available some information X which is relevant to a decision problem you are facing. Suppose X could be take on one of several values X1, X2, …, XK
You have a prior P(X1), …, P(XK) for each of these possible messages
The expected payoff that results from having this partial information is
k maxi j Cij P(j | Xk) P(Xk) Exercise; Show that acquiring any additional (even
imperfect) information can never reduce the value of a decision problem
Partial information
41
Example: Balls and UrnsRaiffa - 1970 A room contains 1000 urns; 800 Type 1 and 200 Type 2 Type 1 urns contain 4 red and 6 white balls Type 2 urns contain 9 red and 1 white ball Decisions;
A1 - Pick and urn and say type 1, A2 - Pick and urn and say type 2 A3 - Do not play.
Payoffs A1 - if correct win $40, if wrong lose $20 A2 – if correct; win $100, if wrong lose $5 A3 - $0
For $8 you can draw and observe a ball from the urn before guessing What should you do?
42
Balls and urns
Draw a decision tree for problem without sampling.
The expected value of A1= $28, A2 = 16 and A3 =0
So best to guess Type I. Expected value under PI = .8•40+.2•100 = $52 So EVPI = $52- $28 = $24 which is the most we
would pay for any information
43
Draw a ball sub tree (without probabilities)
White
Red
Guess Type 1
92
Guess Type 2
Do not playType 1
Type 1
Type 2
Type 2
Guess Type 1
Guess Type 2
Type 1
Type 1
Type 2
Type 2
Do not play
-8
-8
32
-28
-13
92
32
-28
-13
44
Probabilities for sub tree
Knowns; Prior; P(1) = .8; P(2) = .2 Likelihood; P(red|1) = .4; P(red|2) =.9
Unknowns Posteriors; P(1|red); P(1|white) Marginals; P(red); P(white)
Use Bayes to find them P(red)=P(white) = .5 P(1|red) = .64; P(1|white) = .96
45
Draw a ball sub tree (with probabilities)
White
Red
Guess Type 1
92
Guess Type 2
Do not playType 1
Type 1
Type 2
Type 2
Guess Type 1
Guess Type 2
Type 1
Type 1
Type 2
Type 2
Do not play
-8
-8
32
-28
-13
92
32
-28
-13
.5
.5
.64
.64
.36
.36
.96
.96
.04
.04
46
Draw a ball sub tree (with decisions)
White
Red
Guess Type 1
92
Guess Type 2
Do not playType 1
Type 1
Type 2
Type 2
Guess Type 1
Guess Type 2
Type 1
Type 1
Type 2
Type 2
Do not play
-8
-8
32
-28
-13
92
32
-28
-13
.5
.5
.64
.64
.36
.36
.96
.96
.04
.04
10.40
24.80
29.60
-8.80
24.80
29.60
27.20
47
Analysis
The optimal strategy is draw a ball; if red, guess type 2, if white, guess type 1. The expected value of this strategy is $27.20 (or $0.80 less than no information case.
So the optimal strategy for the combined problem is to guess urn 1. What is the most you would pay for drawing a ball?
Suppose instead you could draw 2 balls for $12? Suppose for $9 you could draw one ball, look at it and
decide whether or not to draw a second ball (with your without replacement) for $4.50. These last two problems will be part of HW #1.
48
Example: determining a sample size
Suppose you are testing the effectiveness of a new drug. You will administer the drug to a group of patients to test their reaction to it
If the company decides to market the drug and it is better than existing treatments, then your company will make profits of $1M. If it is not better, then the company will lose $1M. If the company decides not to market the drug, it will make $0
The company is risk-neutral Existing treatments have a cure rate of 60%. You
think that either the new drug will be as effective as the existing treatment, or it will cure 80% of patients
Your prior probability for the drug being better than existing treatments is 0.5
It costs $5000 per experimental patient. How many patients should you test?
49
Example: determining a sample size
The more people you test, the more sure you will be of the effectiveness of the drug
In selecting a sample size, you are effectively choosing the amount of information you will receive
The more information you have, the greater the value of the end decision: do you market the drug or not?
If you choose a sample size of N, there is a gain in the expected value of the information in the sample, which is increasing in N, as well as a cost of $5000 N The goal in choosing a sample size is to maximize the “profit” of
experimenting, i.e., the expected value of partial information (EVpI), minus the cost of experimenting
While the cost grows linearly, the expected value of partial information increases up to a finite limit – the expected value of perfect information, which equals $500,000 in this case
Therefore, there will be a sample size that will be too large in the sense that the profit of experimenting will be negative
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Sample Size
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EVpICost
Example: determining a sample size
The computations for this problem are straightforward but tedious, and can be easily done on a computer
The graph at right shows the EVpI and the cost as a function of the sample size
EVpI – Cost is maximized at a sample size of 22
expected value of perfect information
max profit