Conditional Probability. So far for the loan project, we know how to: Compute probabilities for the...

Conditional Probability

Conditional Probability So far for the loan project, we know how to:

Compute probabilities for the events in the sample space: S = {success, failure}.

Use the loan values to compute the expected value of a workout

Use database filtering to get the information you need from the loan records

What we haven’t learned yet is how to use the characteristics of the borrower— education, experience, and economic conditions!


Many of the records are irrelevant for us, because they represent borrowers who are very different from John Sanders.

We want to target our computations to the ”right kinds” of borrowers.

This kind of targeting is called conditioning: we place conditions on the records we consider.

Conditional Probability The basic principle of conditioning is this:

Conditioning permits us to adjust probabilities based on new or more specific information, which we then take for granted.

Business can be fast-moving, and new information is always coming in — we need a way to adapt and adjust our expectations based on it.

Once new information is assimilated, any historical data that doesn’t fit its pattern may be discarded as irrelevant, so our predictions can be more accurate to the current situation.


Think of conditioning as pulling weeds in the sample space of a probability experiment.

When we condition on an event E having happened, we eliminate any outcomes outside of E, and consider E itself to be the new sample space!

E

F

S

Notation Means the probability of F happening given that E

has already occurred

Definition

In words, this is saying what proportion does F represent out of E.


EFP |

0 where , |

EPEP

EFPEFP

E

F

S


The formula implies:

|EP

EFPEFP

EFPEPEFP |

FEPFPEFP |

Notice the reversal of the events E and F

Note: EFPFEP || Very Important!These are two

different things. They aren’t always equal.


Ex: In a classroom of 360 students, 120 students play the flute and 120 students are male. There are 10 flute-playing males. Let E be the event that a randomly-selected

student is male Let F be the event that a randomly-selected

student plays flute. What percentage of male students play the

flute?

Conditional Probability Sol: The proportion of F that makes up the sample

space, P(F) = . The proportion of F that makes up E, however, is P(F | E) = .

E

F

S

360120

12010

E

EF

EPEFP

EFP within outcomes Total within outcomes

12010

36012036010

|


Ex: Suppose 22% of Math 115A students plan to major in accounting (A) and 67% on Math 115A students are male (M). The probability of being a male or an accounting major in Math 115A is 75%. Find and . MAP | AMP |


Sol:

First find

MP

MAPMAP

|

MAP

14.075.067.022.0

MAPMPAPMAP


Sol:

2090.067.014.0

|

MPMAP

MAP


Sol:

6364.022.014.0

|

APAMP

AMP

Conditional Probability Sometimes one event has no effect on another

Example: flipping a coin twice

Such events are called independent events

Definition: Two events E and F are independent if or EPFEP | FPEFP |


Implications:

FPEPFEP

EPFP

FEPEPFEP

|

So, two events E and F are independent if this is true.


The property of independence can be extended to more than two events:

assuming that are all independent.

nn EPEPEPEEEP 2121

nEEE ,,, 21

Conditional Probabilities

INDEPENDENT EVENTS AND MUTUALLY EXCLUSIVE EVENTS ARE NOT THE SAME

Mutually exclusive:

Independence:

0FEP

FPEPFEP

EPFEP

|


Ex: Suppose we roll toss a fair coin 4 times. Let A be the event that the first toss is heads and let B be the event that there are exactly three heads. Are events A and B independent?

TTTTTTTHTTHTTTHHTHTTTHTHTHHTTHHHHTTTHTTHHTHTHTHHHHTTHHTHHHHTHHHH

S

,,,,,,,,,,,

,,,,


Soln:For A and B to be independent,

and

Different, sodependent

TTTTTTTHTTHTTTHHTHTTTHTHTHHTTHHHHTTTHTTHHTHTHTHHHHTTHHTHHHHTHHHH

S

,,,,,,,,,,,

,,,,

21

168 AP 4

1164 BP

1875.0163 BAP

BPAPBAP

125.081

41

21 BPAP


Ex: Suppose you apply to two graduate schools: University of Arizona and Stanford University. Let A be the event that you are accepted at Arizona and S be the event of being accepted at Stanford. If and , and your acceptance at the schools is independent, find the probability of being accepted at either school.

7.0AP 2.0SP


Soln: Find .

Since A and S are independent,

SAP

SAPSPAPSAP

14.02.07.0

SPAPSAP


Soln:

There is a 76% chance of being accepted by a graduate school.

76.014.02.07.0

SAPSPAPSAP


Independence holds for complements as well.

Ex: Using previous example, find the probability of being accepted by Arizona and not by Stanford.


Soln: Find .

56.08.07.0

2.017.0

CC SPAPSAP

CSAP


Ex: Using previous example, find the probability of being accepted by exactly one school.

Sol: Find probability of Arizona and not Stanford or Stanford and not Arizona.

CC ASSAP


Sol: (continued)Since Arizona and Stanford are mutually exclusive (you can’t attend both universities)

(using independence)

CCCC ASPSAPASSAP

CC APSPSPAP


Soln: (continued)

62.006.056.0

3.02.08.07.0

CC

CCCC

APSPSPAP

ASPSAPASSAP


Independence holds across conditional probabilities as well.

If E, F, and G are three events with E and F independent, then

GFPGEPGFEP |||


Focus on the Project: Recall: and

However, this is for a general borrower

Want to find probability of success for our borrower

464.0SP 536.0FP


Focus on the Project: Start by finding and

We can find expected value of a loan work out for a borrower with 7 years of experience.

YSP | YFP |

Conditional Probability Focus on the Project:

To find we use the info from the DCOUNT function

This can be approximated by counting the number of successful 7 year records divided by total number of 7 year records

YSP |

YP

YSPYSP

|


Focus on the Project: Technically, we have the following:

So,

BR

BRBRBRBR YP

YSPYSPYSP

||

4393.0| 239105 YSP

Why “technically”? Because we’re assuming that the loan workouts BR bank made were made for similar types of borrowers for the other three. So we’re extrapolating a probability from one bank and using it for all the banks.


Focus on the Project: Similarly,

This can be approximated by counting the number of failed 7 year records divided by total number of 7 year records

YP

YFPYFP

|


Focus on the Project: Technically, we have the following:

So,

BR

BRBRBRBR YP

YFPYFPYFP

||

5607.0| 239134 YFP


Focus on the Project: Let be the variable giving the value of a loan work out for a borrower with 7 years experience

Find

YZ

YZE


Focus on the Project:

This indicates that looking at only the years of experience, we should foreclose (guaranteed $2.1 million)

000,897,1$5607.0000,2504393.0000,000,4

Failure Prob. Failure Success Prob. Success

YZE


Focus on the Project: Of course, we haven’t accounted for the other two factors (education and economy)

Using similar calculations, find the following:

CFPCSPTFPTSP | and,|,|,|



5581.0| 1154644 TFP

4419.0| 1154510 TSP

5217.0| 1547807 CSP

4783.0| 1547740 CFP


Focus on the Project: Let represent value of a loan work out for a borrower with a Bachelor’s Degree

Let represent value of a loan work out for a borrower with a loan during a Normal economy

TZ

CZ


Focus on the Project: Find and TZE CZE

000,907,1$

5581.0000,2504419.0000,000,4Failure Prob. Failure Success Prob. Success

TZE

000,206,2$4783.0000,2505217.0000,000,4

Failure Prob. Failure Success Prob. Success

CZE


Focus on the Project: So, two of the three individual expected values

indicates a foreclosure:

000,897,1$YZE

000,206,2$CZE 000,907,1$TZE


Can’t use these expected values for the final decision

None has all 3 characteristics combined: for example has all education levels and

all economic conditions included YZE


Now perform some calculations to be used later

We will use the given bank data:That is is reallyand so on…

SCPSTPSYP | and,|,|

SYP | BRBR SYP |


Focus on the Project: We can find

since Y, T, and C are independent

Also

SCPSTPSYPSCTYP |||| SCTYP |

FCPFTPFYPFCTYP ||||



Similarly:

0714.01470105

in number and in number

||

BR

BRBR

BRBR

SSY

SYPSYP

5823.01386807

| SCP

5301.0962510

| STP



0220.05823.05301.00714.0

||||

SCPSTPSYPSCTYP



0753.01779134

| FYP

5222.01417740

| FCP

5314.01212644

| FTP



0209.05222.05314.00753.0

||||

FCPFTPFYPFCTYP



Now that we have found and we will use these values to find and

SCTYP | FCTYP | CTYSP | CTYFP |

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Conditional Probability. So far for the loan project, we know how to: Compute probabilities for the...

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