BAYES THEOREM (Dated 26th March 2013)
In business , many things can change very fast. Stocks that are predicted to sell well in the market, sometimes don’t because of few reasons. Political reasons when riots happened everywhere not only in the our asian streets but every where. ‘He’ has to revise his chances of making profits and staying in business. Without changing his strategy, stock will pile up and he will incur loss and more loss.
Similar is in football games , Mr Alex Ferguson (Manchester United) thought he can assemble the best players to win the F.A. Cup but because of injuries to important players like Rooneyand Giggs , that CUP seems to be drifting away from their hands. Thus he has to think of revising his chances, his probabilities etc etc.
In both cases, certain probabilities were altered after the people involved got additional information. The new probabilities are known as revised , or posterior, probability. Because probabilities can be revised as more information is gained, probability theory is of great value in managerial decision making like I.T industries and railway operations.
Thomas Bayes (1702-1761) came with this basic formula for conditional probability under dependence
.. this is called Bayes Theorem. He used mathematics to study God actually. He didn’t meet the God as I was told (perggg!).
Bayes Theorem offers powerful statistical method of evaluating new information and revising our prior estimates (based on limited information only) of the probability that things are in 1 state or another. If correctly used, it makes it unnecessary to gather masses of data over long periods of time in order to make decision based upon probabilities.
Calculating posterior probabilities
Elementary event / Probability of elementary event / P( getting Ace |elementary event) / P(Ace, elementary event)Type 1 / 0.5 / 0.4 / 0.4*0.5=0.20
Type 2 / 0.5 / 0.7 / 0.7*0.5= 0.35
1.0 / P(ace)= 0.55
Idea?
The sum of the probabilities of the elementary events (drawing either a type 1 or a type 2 dice) is 1.0 because there are only two types of dice. The probability of each type is 0.5. the two types constitute mutually exclusive list.
The sum of probability(ace|elementary event) does not equal to 1.0 the figures 0.4 and 0.7 simply represent the conditional probability of getting an ace, given type 1 and type 2 respectively.
The fourth column shows the joint probability of ace and type 1 occurring together (0.4 *0.5 = 0.20), and the joint probability of ace and type 2 occurring together (0.7*0.5=0.35) . the sum of these joint probabilities (0.55) is the marginal probability of getting an ace. Notice that in each case , the joint probability was obtained by using the formulae
P(AB) = p(A|B) * p(B)
To find the probability that the die we have drawn is type 1, we use the formula for conditional probability under statistical dependence
P(B|A) =
Converting to our problem , we have: p(type 1|ace) =
Or p( type 1|ace) =
Thus , the probability that we have drawn a type 1 die is 0.364
How about the probability that the die is type 2 ?
P(type 2|ace} =
Conclusion after 1 roll
What we got here is.. with 1 additional information made available to us?
What inferences that have we been able to draw from 1 roll of the die?
Before roll of the dice , the best we can say there is 0.5 chance it is a type 1 dice. And type 2 dice is also 0.5. however after rolling , we have been able to alter or revise our prior probability estimate.
Our new posterior estimate is that there is a higher probability (0.636) that the dice we have in our hand is a type 2 than that it is a type 1 (only 0.364).
How about more and more information?
Answer: life getting a bit difficult/ complicated/ time consuming. But we should see. No worry?
By rolling the second dice
See the table below
Elementary event / Probability of elementary event / P(Ace| elementary event) / P( 2 aces |elementary event) / P( 2 aces, elementary event)Type 1 / 0.5 / 0.4 / 0.16 / 0.16*0.5=0.080
Type 2 / 0.5 / 0.7 / 0.49 / 0.49*0.5= 0.245
1.0 / P(2 aces) = 0.325
Assume we rolled a die a second time and again comes up the ace. What is the further revised probability that the dice is type 1?
Now we have a new column p(2 aces |elementary event). This column gives the joint probability of 2 aces on 2 successive rolls if the die is type 1 and if it type 2 : p( 2 aces|type1) = -.4*0.4=0.16
And p(2 aces|type 2 ) = 0.7 *0.7 = 0.49
In the last coloumn , we see the joint probabilities of 2 aces on 2 successive rolls and the elementary events (type 1 and type 2). That is, p( 2 aces, type 1) equals p(2 aces|type 1) times the probability of type 1, or 0.16*0.5 = 0.080.
And p(2 aces, type 2) equals p(2 aces|type 2) times the probability of type 2 or 0.49*0.5 = 0.245.
The sum of these (0.325) is the marginal probability of 2 aces on 2 successive rolls of a dice.
Now we are ready to compute the probability of 2 aces on 2 successive rolls.
To compute the probability that the die we have drawn is type 1, given an ace on each of two successive rolls. Using the same general formula as before, we convert this to
P( type 1| 2 aces) =
Similarly, p( type 2|2 aces) =
What we have got here?
Two rolls?
When we first drew the die, all we knew was there was 0.5 chance to get type 1 or type 2 dice.
In other words 50 50 chance right?
After rolling the dice once and getting an ace, we revised these original probabilities to the following
Prob( type 1) = 0.364
Prob( type 2) = 0.636
After the second roll(another ace), we revised the probability again
P( type 1) = 0.246
P(type 2) = 0.754
What is the Conclusion now?
We have changed the original probabilities from 0.5 for each type to 0.246 for type 1 and 0.754 for type 2. This means that we can now assign a probability of 0.754 that if a die turns up ace on 2 successive rolls, it is type 2 .
What We did was- roll the die twice; observe its behavior, and draw inferences from the behavior without any monetary cost.
Reference:
Levin R.I. and Rubin, D.S. (1990). Statistics for management. 5th edition. USA: Prentice-Hall ISBN 0-13-847203-3
Exercise:
Q1. A doctor has decided to prescribe 2 new drugs to 200 heart patients as follows: 50 get drug A , 50 get drug B, and 100 get both. Drug A reduces the probability of a heart attack by 35 percent, drug B reduces the probability by 20 percent. And two drugs taken together work independently.
The 200 patients were chosen so that each has an 80 percent chance of having a heart attack. If a randomly selected patient has a heart attack, what is the probability that the patient was given both drugs?
Answer: 0.4178