IIMC Long Duration Executive EducationExecutive Programme in Business Management

Statistics for Managerial

Decisions

Prof. Saibal Chattopadhyay

IIM Calcutta

An Outline of the Course

• 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 experimentExample 1: A coin is tossed twice

A = there is at least one headEvent 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, andSum 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)=2/3Then P(HH) = 1/9, P(HT) = 2/9, P(TH) = 2/9,P(TT) = 4/9, andP(A) = P(HH) + P(HT) + P(TH) = 5/9If B = at least 3 heads, then P(B) = 0In general, P(.) is a number between 0 & 1.P(.) close to 0 : unlikely to occurP(.) close to 1: very likely to occurProbability: An attempt to quantify the degree of

uncertainty

Some Set-theoretic Operations with Events

• Union: AB is the event which occurs if at least one of A and B occurs;

• Intersection: AB 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: AB = φ

For three events A, B and C, they are disjoint if no two can occur together, i.e.,

AB = φ, AC = φ and BC = φ

• Exhaustive events: A set of events A1, A2,…,Ak is exhaustive if at least one of them is sure to occur,

A1A2…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.

AB = {GGG, DGG, GDG, GGD} = BHere event A is contained in event B: AB.

Then AB = A.Ac = {DDD, DDG, DGD, GDD, GGG} = S – A P(Ac) = 5/8 = 1 – P(A).

B C = {GGG}; P(BC) = 1/8.

BC = {GGG, DGG, GDG, GGD, DDD};

P(BC) = 5/8.

But note that this is SAME as

P(B) + P(C) – P(BC) = 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(AB) = P(A) + P(B)

2. If they are not disjoint,

P(AB) = P(A) + P(B) – P(AB)

3. P(Ac) = 1 – P(A)

4. Any event A can occur either with a second event B jointly (i.e. AB occurs) or without the occurrence of the event B (i.e., ABc occurs). Thus

P(A) = P(AB) + P(ABc)

More than two events?

P(ABC) = P(A) + P(B) + P(C), if disjoint;ButP(ABC) = P(A) + P(B) + P(C)

- P(AB) – P(AC) – P(BC) + P(ABC).

How to tackle probability of intersections?P(AB) = ? P(ABC) = ?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(AB)/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(AB) = 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 Events1. A and B are mutually independent events if (and

only if)P(AB) = P(A).P(B)

2a) A, B and C are pair-wise independent events if P(AB) = P(A).P(B); P(AC) = P(A).P(C) andP(BC) = P(B).P(C).

2b) A, B and C are mutually independent if, in addition, we also haveP(ABC) = 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 uncertaintyRandom Variable: A real-valued function on SS = {HH, HT, TH, TT}X = Number of heads obtainedIf {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 2X 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 , andP(X=2) = P(HH) = ¼ Total Probability = ¼ + ½

+ ¼ = 1Probability distribution of

X:

X=x 0 1 2 Tot

P(X=x) 1/4 1/2 1/4 1

Mean and Variance of a distribution

Mean = E(X) = Sum(value*Probability) = 0. ¼ + 1. ½ + 2. ¼ = 1

Variance = Sum 1 – Square of MeanSum1 = Sum(value-squared*probability)

= 0. ¼ + 1. ½ + 4. ¼ = 1.5Variance = 1.5 – 1 = 0.5Standard 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

P(X=2) = 1/9

Prob. Distribution of X(=no. of heads):

Mean & Variance: Similar

X=x 0 1 2 Tot

P(X=x) 4/9 4/9 1/9 1

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

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 !

No strike Strike

Job 1 10,000

¾

2,000

¼

Job 2 20,000

½

500

½

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/bondsReturn rate of stock: 2% with prob. 0.37 &

10% with prob. 0.63Return of Bond: 7% with certaintyIndividual 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 !) isE(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 from Stock S

20% 5%

5% 0.4 0.1

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)