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					Uncertainty and probability

      Using probabilities
      Using decision trees
      Probability revision
           Today’s agenda
• Important terms
• Simple review (objective, subjective,
  marginal, joint, and conditional
  probabilities)
• Examples: outcomes, expected values, risk
  attitudes
• Examples: action choices, decision trees
                     Vocabulary
• A probability is a number between zero and one
  representing the likelihood of the occurrence of
  some event.
• Probability
  –   objective vs. subjective
  –   marginal vs. joint
  –   joint vs. conditional
  –   prior vs. posterior
  –   likelihood vs. posterior
       Vocabulary continued

•   Outcomes or payoffs (mutually exclusive)
•   Action choices
•   States of nature
•   Decision tree
•   Expected value
•   Risk
                    Probability

Imagine an urn containing 1500 red, pink, yellow, blue
and white marbles.
Take one ball from the urn. What is:
         P(black) = 0

         P(~black) = 1            ~ = NOT

Probabilities are all greater than or equal to zero and less
than or equal to one.
                        Same urn:
Suppose the number of balls is as follows:
     Red                400
     Pink               100
     Yellow             400 What is:
     Blue               500
     White              100 P(Red) =         400/1500 = .267
     Total             1500 P(Pink) =        100/1500 = .067
                               P(Yellow) = 400/1500 = .267
                                P(Blue) =    500/1500 = .333
                                P(White) =   100/1500 = .067
                                Total =          1
         Joint probabilities and
             independence

Define A as the event “draw a red or a pink marble.”
We know 500 marbles are either red or pink.
What are:     P(A) =                = .33


               P(~A) = (1 - P(A)) = .67
       Joint probabilities and
    independence (we’re getting
               there)

Define B as the event, “draw a pink or white marble.”

We know 200 marbles are pink or white.

What are:     P(B) =    .133
              P(~B) = .867
           Joint probabilities and
               independence
Define A as the event “draw a red or a pink marble.”
Define B as the event “draw a pink or white marble.”

What is:      P(A, B) = P(A  B)
This is the joint probability of A and B.
 What color is the marble?       Pink

        P(A, B) = P(pink) =             = .0667
         Joint probabilities and
             independence
Are A and B independent?
Note that P(A, B)  P(A) * P(B) = .33 * .13 = .0429

Are A and B mutually exclusive?
What is the probability of A or B?

P(A or B) = P(A  B) = P(A) + P(B) - P(A  B)
                       = .40
           Joint probabilities and
               independence
Suppose we draw one marble from the urn and
replace it. Then, we draw a second marble.

What is:
            P(Red, Red) =           = .071

            P(Red, Blue) = .088
Are (Red, Red) independent?
Are (Red, Blue) independent?
  Joint and marginal probabilities

                                 What are:

                             P(B) =      = .133

                                 P(~B) = .867
                      1500


P(A, B) =    P(A or B) =
       Conditional probabilities
The probability that a particular
                                        What is
event will occur, given we already
know that another event has
                                        P(A | B) =
occurred.

We have information to bring            P(~A | ~B) =
to bear on the base rate  1500
probability of the event

P(A | ~B) =               P(~B | A) =
       Definition of independence
P(B | A) =

Events A and B are independent if   P(B | A) = P(B)

                                     Here P(B | A) =

                                             P(B)


                                        P(B) =

                          1500
     Marginal and joint probability
                table:
The joint probabilities are in the box. The marginals
are outside. How do you compute conditionals from this?
                            A          ~A

 P(B | A) =         B


                   ~B
           Joint probability tables
          A         ~A
                                     What are:
B
                                   P(A, B) = .0667
~B
                                   P(~A, ~B) = .6



     P(~A) = .667   P(~B) = .867   P(~B | A) = .8
           Outcomes or payoffs
Example:    Win $1,000 if you draw a pink marble,
            win $0 otherwise.

          Outcomes: $1,000 or $0
           Events:     A pink marble or a marble of
                       another color
        Probabilities:    P(Pink) = 1/15
                          P(~Pink) = 14/15
The expected value of this gamble:

E(gamble) = (1/15)*($1,000) + (14/15)*($0) = $66.67
                  Example
We expect to sell 10,000 computers if the market is
good and to sell 1,000 computers if the market is
bad. The marketing department’s best estimate of
the likelihood of a good market is .5.

Outcomes: 10,000 computers sold or 1,000 computers
             sold.
Probabilities:     P(good market) = .5
                   P(bad market) = .5
 Are these objective or subjective probabilities?
 What are expected computer sales?          5,500
         More expected values
With discrete outcomes, an expected value is the
probability-weighted sum of the outcomes for the
decision of interest.

Here are some two-outcome lotteries. Compute the
expected values.
L1 :   Win $1,000 with probability .5 or lose $500 with
       probability .5.    E(L1) = $250
L2 :   Win $2,000 with probability .5 or lose $1,000
       with probability .5.   E(L2) = $500
                  More lotteries
Here are some two-outcome lotteries. Compute the
expected values.
 L3 :    Win $1,500 with probability 1/3 or lose
         $750 with probability 2/3.      E(L3) = $0
L4 :    Win $750 with probability 2/3 or lose $750
        with probability 1/3.             E(L ) = $250
                                              4
L5 :    Win $300 with probability .5 or lose $200 with
        probability .5.                    E(L5) = $50
L6 :    Win $10,000 with probability 9/10 or lose
        $85,000 with probability 1/10.
                                           E(L6) = $500
L1 :   Win $1,000 with probability .5 or lose $500 with
       probability .5.    E(L1) = $250
L2 :   Win $2,000 with probability .5 or lose $1,000
       with probability .5.    E(L2) = $500
L3 :   Win $1,500 with probability 1/3 or lose
       $750 with probability 2/3. E(L3) = $0
L4 :   Win $750 with probability 2/3 or lose $750
       with probability 1/3.  E(L4) = $250
L5 :   Win $300 with probability .5 or lose $200 with
       probability .5.  E(L5) = $50
L6 :   Win $10,000 with probability 9/10 or lose
       $85,000 with probability 1/10. E(L6) = $500
                 Action choices

If I build a large hotel (cost = $5,000,000) and tourism
is high (P(high) = 2/3), I will make $15,000,000 in
revenue, but if it is low, I will make $2,000,000.

If I build a small hotel (cost = $2,000,000) and tourism
is high, I will make $5,000,000, but if tourism is low, I
will make $2,000,000.
I can also choose to do nothing.
                       Action choices
If I build a large hotel (cost = $5,000,000) and
tourism is high (P(high) = 2/3), I will make
$15,000,000 in revenue, but if it is low, I will
make $2,000,000.

                                  If I build a small hotel (cost = $2,000,000) and
                                  tourism is high, I will make $5,000,000, but if
                                  tourism is low, I will make $2,000,000.

 I can also choose to do nothing.

Action choices:            Do nothing, build large, build small
Outcomes: $15,000,000; $2,000,000; $5,000,000, $0
States of nature: high tourism, low tourism
 Probabilities: P(high) = 2/3; P(low) = 1/3
                  Decision trees
Suppose I need to decide whether to invest $10,000 in
the market or leave it in the bank to earn interest. If I
invest, there is a 50% chance that the market will
increase 20% over the coming year and a 50% chance
that the market will be stagnant (no change). If I
leave the money in the bank, there is an 80% chance
that interest rates will increase to 10% and a 20%
chance that interest rates will remain at 5%.

What should I do?     Use a decision tree.
           A decision by an individual is required

           Nature makes these decisions

                                                  $2,000 or
                 Nature decides                .5 EV = $1,000
                                    (Good)

                                  (Stagnant)
I decide                                       .5
             Stock                                   $0 or EV = $0
             Market          $1,000
                                                      $1,000 or
                                                .8    EV = $800
             Bank                     (Increase)

                                      (Same)
                                                   .2 $500 or
                $900
                                                      EV = $100
          Homework assignment
Problems 5-15 and 16
Consider two urns:         Urn 1           Urn 2
             Red balls         7              4
             Black balls       3              6
P(R1) = Probability of red on first draw
P(R2) = Probability of red on second draw
P(B1) = Probability of black on first draw
P(B2) = Probability of black on second draw
a(1) Take one ball from urn 1, replace it, and take a
second ball. What is the probability of two reds being
drawn?          P(R1, R2) = .7 x .7 = .49
                  Homework
a(2) What is the probability of a red on the second
draw if a red is drawn on the first draw?

  P(R2 | R1) =             =

a(3) What is the probability of a red on the second
draw if a black is drawn on the first draw?

   P(R2 | B1) =
                   Homework
b(1) Take a ball from urn 1; replace it. Take a ball
from urn 2 if the first ball was black; otherwise, draw
a ball from urn 1.

What is the probability of two reds being drawn?

        P(R1, R2) = .7 x .7 = .49
b(2) What is the probability of a red on the second
draw if a red is drawn on the first draw?
   P(R2 | R1) =
                    Homework
b(3) What is the probability of a red on the second
draw if a black is drawn on the first draw?


    P(R2 | B1) =

 What is the unconditional probability of red on the
 second draw?

P(R2) = P(B1, R2) + P(R1, R2) = (.3 x .4) + (.7 x .7) = .61
                   Homework
5-16. Draw a tree diagram for Problem 5-15a

                           Red      P(R1, R2) = .49
                  Draw 2
                             .7
            Red                .3
 Draw 1
            .7             Black    P(R1, B2) = .21

             .3            Red      P(B1, R2) = .21
           Black             .7

                              .3
                           Black    P(B1, B2) = .09
                  Homework
5-17. Draw a tree diagram for Problem 5-15b

                           Red     P(R1, R2) = .49
                 Draw 2
                             .7
          Red
Draw 1                        .3
            .7            Black    P(R1, B2) = .21

            .3             Red     P(B1, R2) = .12
                             .4
         Black
                             .6
                          Black    P(B1, B2) = .18
               Homework
5-29. This is the survey problem involving home-
ownership and income levels. The results can be
summarized by the table below
                                         Survey




A. Suppose a reader of this magazine is selected at
random and you are told that the person is a home-
owner. What is the probability that the person has
income in excess of $25,000?
       P(>$25,000 | homeowner) =
                                        Survey




b. Are home ownership and income (measured only as
above or below $25,000) independent factors for this
group?       They are NOT independent.
If yes, then P(>$25 | home) = P(>$25)
But, P(>$25) = .7 and P(>$25 | home) = .75
                  Homework

5-38. The president of a large electric utility has to
decide whether to purchase one large generator (Big
Jim) or four smaller generators (Little Arnies) to
attain a given amount of electric generating capacity.
On any given summer day, the probability of a
generator being in service is 0.95 (the generators are
equally reliable). Equivalently, there is a 0.05
probability of a failure.
                   Homework
5-38.

a. What is the probability of Big Jim’s being out of
service on a given day?

 Let P(out) = the probability of any generator being
              out of service = .05
 If P(BJout) = the probability of Big Jim’s being out
               of service.

 Then P(out) = P(BJout) = .05
5-38.               Homework
b. What is the probability of either zero or one of the
four Arnies being out? (At least three will be running.)
Since the probability of a failure (f) for one Arnie is .05,

  We want P(f  1|n=4, p=.05)
  P(f = 0 | n=4, p=.05) =

  P(f = 1 | n=4, p=.05) =

  P(f  1|n=4, p=.05) = .8145 + .1715 = .9860
                 Homework
5-38.

c. If five Little Arnies are purchased, what is the
probability of at least four operating?

P(f  1 | n=5, p=.05) =
5-38.               Homework
d. If six Little Arnies are purchased, what is the
probability of at least four operating?

 P(f  2 | n=6, p=.05) =
                   Homework
5-40. Newspaper articles frequently cite the fact that in
any one year, a small percentage (say, 10%) of all
drivers are responsible for all automobile accidents.
The conclusion is often reached that if only we could
single out these accident-prone drivers and either
retrain them or remove them from the roads, we could
drastically reduce auto accidents. You are told that of
100,000 drivers who were involved in one or more
accidents in one year, 11,000 of them were involved in
one or more accidents in the next year.

A. Given the above information, complete the entries
in the joint probability table in Table 5-21.
        Accidents: Joint probability table
  A1 = accident in year 1, A2 = accident in year 2
  Given: P(A2 | A1) = 11,000/100,000 = .11

 P(A2 | A1)=


                            P(A1, A2) =
                             .011
                                           .089
 Therefore,                          .11
 P(A1, A2) =
 P(A2 | A1)x                    .089       .811
 P(A1) =
.11 x .10 = .011
                   Accidents
B. Do you think searching for accident-prone
drivers is an effective way to reduce auto accidents?
Why?


If the information in the problem is representative,
then searching for accident-prone drivers will not
be very helpful, since having had an accident in
Year 1 has only a minor effect on the probability of
an accident in Year 2.
Uncertainty continued . . .

     Probability revisions

    Continue decision trees
            Today’s agenda
• Finish the homework problems
• Work through a decision tree example that
  – Uses no information
  – Uses perfect information
  – Uses imperfect information
• Briefly discuss Freemark Abbey Winery
• Group problem solving
                  Homework
5-42. A safety commissioner for a certain city
performed a study of the pedestrian fatalities at
intersections. He noted that only 6 of the 19 fatalities
were pedestrians who were crossing the intersection
against the light (i.e., in disregard of the proper
signal), whereas the remaining 13 were crossing with
the light. He was puzzled because the figures seemed
to show that it was roughly twice as safe for a
pedestrian to cross against the light as with it. Can
you explain this apparent contradiction to the
commissioner?
5-42.
                      Homework
 The commissioner is looking at the wrong conditional
 frequencies (probabilities).
        P(?) = 6/19    It’s a conditional probability

P(crossing against the light | killed at intersection) = 6/19

 It is not the probability of being killed if you cross
 against the light. Further, 13/19 is not the probability
 of being killed if you cross with the light.
 5-42.              Homework
The relevant probabilities are:
      P(killed | crossed with light) and
      P(killed | crossed against light)

The deaths must be considered relative to the number
of pedestrians who cross with and against the light.

As an extreme possibility, it may be that the only six
persons who crossed against the light were killed, a
fatality rate of 100%; whereas, 1 million crossed with
the light, a fatality rate of .000013. It isn’t likely that
this is the case, but the commissioner’s data do not
rule it out.
 5-43.             Homework
Probability revision

Suppose a new test is available to test for drug
addiction. The test is 95 percent accurate “each way”;
that is, if the person is an addict, there is a 95 percent
chance the test will indicate “yes”; if the person is not
an addict, then 95 percent of the time the test will
indicate “no.”

Suppose it is known that the incidence of drug addiction
in urban populations is about 1 out of 1,000. Given a
positive (yes) test result, what are the chances that the
person being tested is addicted?
   Homework: Probability revision
 5-43.
We were given P(“yes” | addicted) = .95
and P(“no” | not addicted) = .95        and
 the prior P(addicted) = .001

We want P(addicted | “yes”) and P(not addicted | “no”)

 This is a different conditional probability called a
 “revised” or “posterior” probability.
       Test for drug addiction

We know:

   P(addicted | yes) =


and

   P(yes | addicted) =                 = .95

   P(addicted) = .001
                         We can solve for the joint.
         Test for drug addiction
 P(addicted, yes)


What is P(yes)?     It consists of two joint probabilities.
The test can say “yes” and the subject is addicted
OR
The test can say “yes” and the subject is not addicted

P(addicted, yes) + P(not addicted, yes) = P(yes)
           Test for drug addiction
P(not addicted, yes)


Therefore, P(yes) = .00095 + .04995 = .0509

Given a positive test result, the probability that a
person chosen at random from an urban population
is a drug addict is
        P(addict | yes) = .00095/.0509 = .0187
This is a posterior probability.
Drug test: joint probability table



              .95


                      .95
       Bayes Theorem and the
        Multiplication Rule

      P(Ai | B) =


and
      Another revision example
Priors:      P(disease) = .01
             P(~disease) = .99
Test accuracy: P(positive | disease) = .97
               P(positive | ~disease) = .05
               P(negative | disease) = .03
               P(negative | ~disease) = .95

Note that:        false positives > false negatives
          Disease detection continued
Priors:      P(disease) = .01
             P(~disease) = .99
Test accuracy:      P(positive | disease) = .97
                    P(positive | ~disease) = .05
                    P(negative | disease) = .03
                    P(negative | ~disease) = .95
 What is the probability that an individual chosen at
 random who tests positive has the disease?
   P(positive, disease) = .97 * .01 = .0097
   P(positive) = (.97 * .01) + (.05 * .99) = .0592
   P(disease | positive) = .0097/.0592 = .1639
    Disease detection continued
Suppose the tested individual was not chosen
from the population at random, but instead
was selected from a subset of the population
with a greater chance of getting the disease?
Prior: Suppose P(disease) = .2
Then,
     P(disease | positive) =
5-45 Revision
                   Homework
This is a classical probability problem. Try
out your intuition before solving it systematically.

Assume there are three boxes and each box has two
drawers. There is either a gold or silver coin in each
drawer. One box has two gold, one box two silver,
and one box one gold and one silver coin. A box is
chosen at random and one of the two drawers is opened.
A gold coin is observed. What is the probability of
opening the second drawer in the same box and
observing a gold coin?
         Coin and box problem
Here is a helpful visualization:




We know we chose a box with a gold coin.

We want P(gold2 | gold1) =
            Buying information
As manager of a post office, you are trying to decide
whether to rearrange a production line and facilities
in order to save labor and related costs. Assume that
the only alternatives are to “do nothing” or “rearrange.”
Assume also that the choice criterion is that the expected
savings from rearrangement must equal or exceed
$11,000.
Operating costs if you do nothing will be $200,000
If you rearrange successfully, operating costs will be
$100,000.
If you rearrange unsuccessfully, operating costs will
be $260,000.
Buying information
            Post Office Example
Operating costs if you do nothing will be $200,000
If you rearrange successfully (P(success) = .6),
operating costs will be $100,000.
If you rearrange unsuccessfully (P(fail) = .4),
operating costs will be $260,000.

What is the expected value of each action choice?
Rearrange: .6 x $100,000 + .4 x $260,000 = $164,000
Do nothing: $200,000      What would you choose?
       Post Office Decision Tree
                   Do nothing
                                               $200,000
You decide

                                               $100,000
                                .6
        Rearrange                    Succeed

                                     Fail
        $164,000                .4
                                               $260,000
   You can hire a consultant, Joan Zenoff, to study the
   situation. She would then render a flawless prediction
   of whether the rearrangement would succeed or fail.
   Compute the maximum amount you would be willing
    to pay for the errorless prediction.
                                     $164,000
                Don’t buy info.               rearrange
                                                   $100,000
$140,000           $100,000
                       Positive               do nothing
                              .6                   $200,000
         Buy              .4               rearrange
                                                 $260,000
                     Negative
         $140,000
                                           do nothing
                    $200,000                     $200,000
   You can hire a consultant, Joan Zenoff, to study the
   situation. She would then render a flawless prediction
   of whether the rearrangement would succeed or fail.
   Compute the maximum amount you would be willing
    to pay for the errorless prediction.
                                   $164,000
                 Don’t buy info.              rearrange
$140,000                                           $100,000
                    $100,000
                       Positive                do nothing
                              .6                    $200,000
           Buy              .4                rearrange
                                                    $260,000
                       Negative
           $140,000
                                              do nothing
                      $200,000                      $200,000
How much would you pay for Joan’s report?

Compute the expected value of perfect information
= EVPI

EVPI = The expected value of the decision with the
report ($140,000) - The expected value of the decision
without the report ($164,000)

EVPI = $140,000 - $164,000 = -$24,000

The report saves us $24,000 in expected value
We would pay up to $24,000
Suppose now that Joan’s reports are not flawless.
Suppose you have been provided the following
posterior probabilities:




 This means that:

       P(success | optimistic) = .818, not 1
       P(failure | optimistic) = .182, not 0
  What would you now be willing to pay for Joan’s
  report?




   EVII = E(decision with imperfect information)
        - E(decision with no information)

Recall that E(decision with no information) = $164,000
 Required:

 1. Compute the expected cost assuming an optimistic
    report.                                success
                                        .818 $100,000
              We        rearrange
             decide                     .182
                                               failure
optimistic                ?
                       $129,120                 $260,000


 $129,120                do nothing
                                       $200,000
 2. Compute the expected costs assuming a pessimistic
 report.                                   success
                                       .333 $100,000
                       rearrange
             We
                                       .667
            decide                          failure
                      $206,720
                          ?                    $260,000
pessimistic

$200,000               do nothing
                                      $200,000
  3. We were given the probability of an optimistic
  report (P(optimistic) = .55) and the probability of
  a pessimistic report (P(pessimistic) = .45).
  Compute the expected value of imperfect information.
  First we the first finish thewe must show on the tree?
  What is need to decision decision tree.
                                              optimistic

                 Buy information             .55 $129,120
     We
    decide                                     .45
                                                pessimistic
                        $161,016                     $200,000

$161,016         Don’t buy info     $164,000
E(decision with imperfect information) = $161,016
E(decision with no information) = $164,000

EVII = $164,000 - $161,016 = $2,984
We were not given likelihoods. We do not know the
probability that Joan will render an optimistic report
given the rearrangement is a success.
What is that probability?
  P(optimistic | success) =
  P(success) = .6
  P(success | optimistic) =                        = .818
  P(optimistic) = .55
  P(optimistic, success) = .55 x .818 = .45
  P(optimistic | success) = .45/.6 = .75
  Also: P(pessimistic | failure) = (.667 x .45)/.4 = .75
Normally we would be given likelihoods and priors
and we would expect to compute:

1. The posterior probability of the outcome given
a particular kind of information, and

2. The marginal probability of receiving that
particular kind of information

Therefore, given the following information about
the accuracy of Joan Zenoff’s forecasts, complete
a joint probability table and compute the necessary
posterior (revised) probabilities.
                                 P(optimistic | success) =
P(optimistic | success) = .75
P(pessimistic | failure) = .75

                                          P(opt, success)
                                          = .75 x .6
                                          = .45

            .45         .10      .55
                                           P(pess, failure)
                                           = .75 x .4
                                           = .30
            .15         .30      .45
P(success | opt) = .45/.55 P(failure | opt) = .1/.55
P(failure | pess) = .3/.45   P(success | pess) = .15/.45




                    .45        .10         .55


                    .15        .30         .45




          Which ones go on the decision tree?

				
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