Download Decision Making Under Uncertainty

IIMC Long Duration Executive Education
Executive Programme in Business Management
Statistics for Managerial
Decisions
Prof. Saibal Chattopadhyay
IIM Calcutta
An Outline of the Course
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Probability Theory: Basic Concepts
Distribution Theory:Random Variables
Utility Theory: Decisions under Uncertainty
Different Probability Distributions and their
applications: Modeling real data
Bi-variate Data Analysis: Correlation and
regression
Multivariate Data Analysis: Multiple and partial
correlations; multiple regression
Sampling Theory: Different Methods
Statistical Inference
Uncertainty and Randomness
– Theory of Probability
– Random variables & Probability Distributions
– Mean & Variance of a distribution
Decisions under uncertainty
- Utility Theory
- Decision Making using expected utility
Theory of Probability
Preliminaries
• Random Experiment – outcomes bound by chance
• Events – outcomes of a random experiment
Example 1: A coin is tossed twice
A = there is at least one head
Event A has the following decompositions:
A1: Head in both tosses (HH)
A2: Head in 1st toss, Tail in 2nd (HT)
A3: Tail in 1st toss, Head in 2nd (TH)
No further decomposition possible – these are simple
events
• Simple events – can’t be decomposed further
• Sample Space – Collection of all simple
events (Sure event : S)
• Impossible event: events impossible to occur
(φ )
B = at least 3 heads : B = φ
Probability Space : (S, P)
S = Sample Space
P = Probability function on simple events in S
P(.)  0 for any simple event, and
Sum of all probabilities for simple events = 1.
Then P(A) = Sum of the probabilities for those
simple events which constitute event A.
Example 1: S = {HH, HT, TH, TT}
P(HH) = P(HT) = P(TH) = P(TT) = ¼
(equally likely outcomes – set-up of classical
definition)
A = at least one head = {HH, HT, TH}
Therefore, P(A) = P(HH) + P(HT) + P(TH) = ¾
If outcomes not equally likely (general case):
Suppose the coin is biased, with P(H) = 1/3 &
P(T)=2/3
Then P(HH) = 1/9, P(HT) = 2/9, P(TH) = 2/9,
P(TT) = 4/9, and
P(A) = P(HH) + P(HT) + P(TH) = 5/9
If B = at least 3 heads, then P(B) = 0
In general, P(.) is a number between 0 & 1.
P(.) close to 0 : unlikely to occur
P(.) close to 1: very likely to occur
Probability: An attempt to quantify the degree of
uncertainty
Some Set-theoretic Operations with Events
• Union: AB is the event which occurs if at
least one of A and B occurs;
• Intersection: AB is the event which
occurs if A and B occur together;
• Difference: A – B is the event which occurs
if A occurs, but not B;
• Complement: Ac is the event which occurs
if event A does not occur
Some special types of events
• Mutually Exclusive Events: A and B are
mutually exclusive (disjoint) if they cannot
occur together
Notation: AB = φ
For three events A, B and C, they are
disjoint if no two can occur together, i.e.,
AB = φ, AC = φ and BC = φ
• Exhaustive events: A set of events A1,
A2,…,Ak is exhaustive if at least one of
them is sure to occur,
A1A2…Ak = Sample space = S
• Partition of Sample Space: A1, A2, …, Ak
form a partition of S if they are mutually
exclusive as well as exhaustive.
Example: A and Ac are disjoint (by definition)
and are exhaustive; so they form a partition
An Example
Experiment consists of selecting three items
from a manufacturer’s output and observing
whether or not each item is defective.
Call Defective = D and non-defective = G.
S = Sample space
= {DDD, DDG, DGD, GDD, DGG, GDG,
GGD, GGG}
Total no. of elements = 23 = 8
Suppose all are equally probable; then
probability of each simple event = 1/8
A = exactly one defective item
= {DGG, GDG, GGD}; P(A) = 3/8.
B = at most one defective
= {GGG, DGG, GDG, GGD}; P(B) = 4/8=1/2
C = items drawn are all of the same type
= {GGG, DDD}; P(C) = 2/8.
AB = {GGG, DGG, GDG, GGD} = B
Here event A is contained in event B: AB.
Then AB = A.
Ac = {DDD, DDG, DGD, GDD, GGG} = S – A
P(Ac) = 5/8 = 1 – P(A).
B C = {GGG}; P(BC) = 1/8.
BC = {GGG, DGG, GDG, GGD, DDD};
P(BC) = 5/8.
But note that this is SAME as
P(B) + P(C) – P(BC) = 4/8 + 2/8 – 1/8.
Earlier we also noted that P(Ac) = 1 – P(A).
Are these mere coincidences, or are
generally true?
True, depending on specific conditions.
Some Probability Results
1. (For disjoint events): If A and B are disjoint,
P(AB) = P(A) + P(B)
2. If they are not disjoint,
P(AB) = P(A) + P(B) – P(AB)
3. P(Ac) = 1 – P(A)
4. Any event A can occur either with a second
event B jointly (i.e. AB occurs) or without
the occurrence of the event B (i.e., ABc
occurs). Thus
P(A) = P(AB) + P(ABc)
More than two events?
P(ABC) = P(A) + P(B) + P(C), if disjoint;
But
P(ABC) = P(A) + P(B) + P(C)
- P(AB) – P(AC) – P(BC)
+ P(ABC).
How to tackle probability of intersections?
P(AB) = ? P(ABC) = ?
Need some further concepts !!
Conditional Probability – Updating prior belief
Knowing some event A to have occurred already,
what is the chance that another event B will also
occur ?
P(B | A) = conditional prob. of B, given A
= P(AB)/P(A), if P(A) > 0.
What happens if P(A) = 0? Then A = φ.
P(B | A) is not defined in this case.
In general, P(A) = P(A | S)
When information given about event A is trivial (A is
a sure event), conditional probability of B, given A
is same as unconditional probability of B, since
no extra information is provided.
Some useful Results
Result 1: P(AB) = P(A).P(B | A)
= P(B).P(A | B)
Result 2: P(A) = P(B).P(A | B) + P(Bc).P(A|
Bc)
Use: Helps in updating our belief about
chance of occurrence of random events,
given additional information.
An Application in Medical Science
Following information are given:
1. P(a doctor diagnose disease X correctly) = 0.60
2. P(a patient will die by his treatment after correct
diagnosis) = 0.40;
3. P(patient will die by his treatment after wrong
diagnosis) = 0.70;
Question 1: What is P(patient will die) ?
Call B = doctor diagnose disease X correctly
Then, Bc=doctor diagnose wrongly
Given that P(B) = 0.60; we have
P(Bc) = 1 – P(B) = 0.40.
Call A = patient will die. To find P(A).
Given: P(A | B) = 0.40, and
P(A | Bc) = 0.70.
Then P(A) = P(B).P(A|B) + P(Bc).P(A| Bc)
= (0.60)(0.40) + (0.40).(0.70)
= 0.24 + 0.28 = 0.52
How do we update our prior belief?
Suppose we now know that the person (with
disease X) died;
Question 2: What is the probability that his
disease was diagnosed correctly?
Without knowing anything extra, this is
P(doctor diagnosed correctly) = P(B) = 0.60
But now we know that the person had died
(i.e., event A has already occurred).
Given this extra information, what is P(B)?
[should we expect it to be less now?]
 Conditional probability of B, given A
= P(B | A) = P(A and B) / P(A)
=P(B).P(A | B) / P(A) = (0.60)(0.40)/(0.52)
= 0.46 ( Bayes’ Theorem)
Bayes’ Theorem
Suppose B1, B2, …, Bk form a partition of S.
Then for any event A which is known to have
occurred, we have, for any i=1,2,...,k,
P(Bi | A) = P(Bi) P(A|Bi) / P(A), where
P(A) = P(B1).P(A | B1) + P(B2).P(A | B2) +
…. + P(Bk).P(A | Bk).
Note: A can occur only if one of B1, B2, …, Bk
occurs; knowing A had occurred we now take a
fresh look at the initial events B1, B2, …, Bk and
examine if we have to update our prior belief
about their occurrences.
 Posterior probabilities of B1, B2, …, Bk
Statistical Independence of Events
1. A and B are mutually independent events if (and
only if)
P(AB) = P(A).P(B)
2a) A, B and C are pair-wise independent events if
P(AB) = P(A).P(B);
P(AC) = P(A).P(C) and
P(BC) = P(B).P(C).
2b) A, B and C are mutually independent if, in
addition, we also have
P(ABC) = P(A).P(B).P(C)
Note: If A and B independent, the so are Ac and Bc.
Use: Helps in calculation of probabilities
Another way to quantify uncertainty
Random Variable: A real-valued function on S
S = {HH, HT, TH, TT}
X = Number of heads obtained
If {HH} occurs, X = 2;
If {HT} occurs, X = 1;
If {TH} occurs, X = 1:
If {TT} occurs, X = 0.
X takes 3 values: 0, 1, and 2
X is a random variable.
Are all values of X equally probable?
May be not, even if simple events are !
Consider equally likely
simple events:
P(HH) = P(HT) = P(TH) =
P(TT) = ¼. Then
P(X=0) = P(TT) = ¼ ,
P(X=1) = P(HT) + P(TH)
= ½ , and
P(X=2) = P(HH) = ¼
Total Probability = ¼ + ½
+¼=1
X=x
0
1 2
Tot
Probability distribution of
P(X=x) 1/4 1/2 1/4 1
X:
Mean and Variance of a distribution
Mean = E(X) = Sum(value*Probability)
= 0. ¼ + 1. ½ + 2. ¼ = 1
Variance = Sum 1 – Square of Mean
Sum1 = Sum(value-squared*probability)
= 0. ¼ + 1. ½ + 4. ¼ = 1.5
Variance = 1.5 – 1 = 0.5
Standard deviation = SQRT(Variance)
= SQRT(0.5) = 0.707
What if simple events are not equally likely?
With P(H) = 1/3 and P(T) =
2/3, we get
P(HH)=1/9, P(HT) = P(TH)
= 2/9, P(TT) = 4/9
Now P(X=0) = P(TT) = 4/9
P(X=1) = P(HT) + P(TH) =
4/9
X=x
0 1
2
Tot
P(X=2) = 1/9
Prob. Distribution of X(=no. P(X=x) 4/9 4/9 1/9 1
of heads):
Mean & Variance: Similar
Expected Utility Theory
Some Math preliminaries:
1. Function: y=f(x) is a mapping between two sets of
elements ( or numbers)
Example: Y=a + bx : linear function
Or, Y = a + bx + cx2 : second degree etc.
2. Optimization (maxima & minima) of a function:
Differentiable function:
f ’(x) = 0: solve for x (say x = x0)
f ”(x) at x=x0 is negative: f(x) is maximum at x = x0
f ”(x) at x=x0 is positive: f(x) is minimum at x = x0
In decision making under uncertainty, a
decision d may lead to several levels of wealth:
w1, w2, …, wk, with corresponding probs.
p1, p2, …, pk, total prob.=sum(pi) = 1.
Wealth is usually transformed into consumption,
and hence utility (for example, in a business
decision, it may be profit of the company)
Utility function over wealth: u(w)
Different levels of utility: u(w1), u(w2),.., u(wk)
These are random quantities, with respective
probalilities p1, p2, …, pk.
Expected Utility of a decision d
E(u(d)) = average of these utilities
= u(w1).p1 + u(w2).p2 +… + u(wk).pk
An optimal decision d depends on:
a) Optimization of expected utility
b) Choice of the utility function u(w)
Some Applications
Example 1(Dilemma of a Contractor)
A contractor has to choose one of the two
contracting jobs:both having chances of labour
problems (say, strike).
Profit possibilities:
Job1: 10K if no strike, 2K if strike
Job 2: 20K if no strike, 0.5K if strike
Chance of strike:
P(strike in Job 1) = ¼ ; P(no strike) = ¾
P(strike in Job 2) = ½ ; P(no strike) = ½
Job 1
No strike
Strike
10,000
2,000
¾
Job 2
20,000
¼
500
½
½
Expected Profit from Job 1 = (10000)(3/4) +
(2000)(1/4) = 8,000
Expected Profit from Job 2 = (20000)(1/2) +
(500)(1/2) = 10,250
Want to maximize your profit ? Choose Job 2 !
Any other consideration for his choice?
What if he is a born pessimist?
Expects the worst: there will be a strike!
Choose Job 1: it maximizes his minimum profit.
What if he is an optimist?
Expects no strike or neglects the chance of it
Choose Job 2: it may give him 20,000
Anything else?
Chance of strike not known !
How does he choose?
Go for a randomized decision rule:
Choose Job 1 with prob. p and Job 2 with prob.
(1-p) such that his profit must be same whether
he chooses Job 1 or Job 2 and whether there is
a strike or not.
His profit is:
A = 10000 p + 20000 (1-p) if no strike;
B = 2000 p + 500 (1-p) if strike;
Find p such that A = B; p > 1(check); Can’t be !
So p = 1. Choose Job 1 !!
Back to Utility Function and expected utility
E(u(d)) = expected utility for decision d and
utility function u(w)
= u(w1).p1 + u(w2).p2 +… + u(wk).pk
Different Choices of u(w):
1. u(w) = √w : risk averse
2. u(w) = w2 : risk seeker
3. u(w) = w : risk neutral
For a given u(w), choose a decision that
maximizes the expected utility
Example 2: An investment decision problem
(Ex. 1.22; Aliprantis-Chakrabarti)
To invest $10,000 in stocks/bonds
Return rate of stock: 2% with prob. 0.37 &
10% with prob. 0.63
Return of Bond: 7% with certainty
Individual is Risk-averse:
utility function is u(w) = √w
How much to invest in stocks?
Say a fraction ‘s’ of the total investment.
Investor has a chance 0.37 of getting an amount
(10,000 s)(1.02) + 10,000(1-s)(1.07)
=10,000(1.07 – 0.05 s)
And a chance 0.63 of getting the amount
(10,000 s)1.10 + 10,000(1-s)(1.07)
=10,000(1.07 + 0.03 s)
Investor’s expected utility (risk-averse !) is
E(s) = 0.37 √10000(1.07 – 0.05 s)
+ 0.63 √10000(1.07 + 0.03 s)
Choose s such that E(s) is maximum; s=56.9%
Invest $5690 in stocks &$4310 in Bonds.
Example 3:Choice between two stocks
(Ex.1.23 (Aliprantis & Chakrabarti):
$10,000 to invest between two stocks S & M
Probability Table for Returns from S & M
Return from Stock M
Return
20%
5%
from
5%
0.4
0.1
Stock S
20%
0.1
0.4
Invest proportion ‘s’ in stock S and proportion (1-s) in
stock M
With 5% return from stock S and 20% from M:
Wealth = w1 = 10000 s.1.05 + 10000(1-s).1.20
= 10000(1.20 – 0.15 s)
{ this event has probability = 0.40}
With 5% from S and 5% from M:
Wealth = w2 = 10000.1.05
{ probability = 0.1}
With 20% from S and 20% from M:
Wealth = w3 = 10000.1.20
{ probability = 0.1}
With 20% from S and 5% from M:
Wealth = w4 = 10000(1.05 + 0.15 s) { probability= 0.4}
For risk averse, utility function u(w) = √w
Expected utility = E(s) = Sum{(√w1)(0.40) +
(√w2)(0.10) + (√w3)(0.10) + (√w4)(0.40)}
= 40 √(1.2 – 0.15 s) + 40 √(1.05 +0.15 s)
+ 10 √1.05 + 10 √1.20
Choose ‘s’ so that E(s) is maximum
S = 0.5 = 50%
Decision for Risk averse: Invest 50% ($5000) in
Stock S and 50% ($5000) in Stock M.
A Question for you: What happens if he is risk
seeker or risk neutral?
Some remarks
Are decisions always based on expected utility?
Possibly not; consider the following lotteries:
L1: Receive $2 million with certainty
L2: Receive $10 million with prob. 0.15, $2 m
with prob. 0.75 & $ 0 with prob. 0.10
L3: Receive $2 million with prob. 0.25 and $0
with prob. 0.75
L4: Receive $10 million with prob. 0.15 and $0
with prob. 0.85
Choose one between L1 and L2 and one
between L3 and L4
If L1 is chosen over L2, then
u(2) > 0.15 u(10) + 0.75 u(2) + 0.10 u(0)
Add 0.75 u(0) to both side and get
0.25 u(2) + 0.75 u(0) > 0.15 u(10) + 0.85 u(0)
i.e., expected utility for L3 > expected utility for L4
So we should choose L3 over L4
Do you agree? I don’t !
Violations of the expected utility theory
Lotteries and Gambling
If by paying a small amount one has a
chance of winning a large amount,
individuals often ignore the negative
expected payoff, as the loss is small.
BUT
If potential loss is larger, the same
individual may choose very differently
 preference reversal in decision making
Suggested Reading
• Statistical Methods in Business and Social
Sciences: Shenoy, G.V. & Pant, M.
(Macmillan India Limited)
• Games and Decision Making: Aliprantis,
C.D. & Chakrabarti, S.K. (Oxford
University Press)
• •Complete Business Statistics: Aczel, A.D.
& Sounderpandian, J. – Fifth Edition (Tata
McGraw-Hill)