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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!
Conditional Probability
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.
Conditional Probability
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.
Conditional Probability
EFP |
0 where , |
EPEP
EFPEFP
E
F
S
Conditional Probability
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.
Conditional Probability
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
|
Conditional Probability
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 |
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 |
Conditional Probability
Implications:
FPEPFEP
EPFP
FEPEPFEP
|
So, two events E and F are independent if this is true.
Conditional Probability
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
|
Conditional Probability
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
,,,,,,,,,,,
,,,,
Conditional Probability
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
Conditional Probability
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
Conditional Probability
Soln: Find .
Since A and S are independent,
SAP
SAPSPAPSAP
14.02.07.0
SPAPSAP
Conditional Probability
Soln:
There is a 76% chance of being accepted by a graduate school.
76.014.02.07.0
SAPSPAPSAP
Conditional Probability
Independence holds for complements as well.
Ex: Using previous example, find the probability of being accepted by Arizona and not by Stanford.
Conditional Probability
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
Conditional Probability
Sol: (continued)Since Arizona and Stanford are mutually exclusive (you can’t attend both universities)
(using independence)
CCCC ASPSAPASSAP
CC APSPSPAP
Conditional Probability
Independence holds across conditional probabilities as well.
If E, F, and G are three events with E and F independent, then
GFPGEPGFEP |||
Conditional Probability
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
Conditional Probability
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
|
Conditional Probability
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.
Conditional Probability
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
|
Conditional Probability
Focus on the Project: Technically, we have the following:
So,
BR
BRBRBRBR YP
YFPYFPYFP
||
5607.0| 239134 YFP
Conditional Probability
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
Conditional Probability
Focus on the Project:
This indicates that looking at only the years of experience, we should foreclose (guaranteed $2.1 million)
000,897,1$5607.0000,2504393.0000,000,4
Failure Prob. Failure Success Prob. Success
YZE
Conditional Probability
Focus on the Project: Of course, we haven’t accounted for the other two factors (education and economy)
Using similar calculations, find the following:
CFPCSPTFPTSP | and,|,|,|
Conditional Probability
Focus on the Project:
5581.0| 1154644 TFP
4419.0| 1154510 TSP
5217.0| 1547807 CSP
4783.0| 1547740 CFP
Conditional Probability
Focus on the Project: Let represent value of a loan work out for a borrower with a Bachelor’s Degree
Let represent value of a loan work out for a borrower with a loan during a Normal economy
TZ
CZ
Conditional Probability
Focus on the Project: Find and TZE CZE
000,907,1$
5581.0000,2504419.0000,000,4Failure Prob. Failure Success Prob. Success
TZE
000,206,2$4783.0000,2505217.0000,000,4
Failure Prob. Failure Success Prob. Success
CZE
Conditional Probability
Focus on the Project: So, two of the three individual expected values
indicates a foreclosure:
000,897,1$YZE
000,206,2$CZE 000,907,1$TZE
Conditional Probability Focus on the Project:
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
Conditional Probability Focus on the Project:
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 |
Conditional Probability
Focus on the Project: We can find
since Y, T, and C are independent
Also
SCPSTPSYPSCTYP |||| SCTYP |
FCPFTPFYPFCTYP ||||
Conditional Probability
Focus on the Project:
Similarly:
0714.01470105
in number and in number
||
BR
BRBR
BRBR
SSY
SYPSYP
5823.01386807
| SCP
5301.0962510
| STP
Conditional Probability
Focus on the Project:
0753.01779134
| FYP
5222.01417740
| FCP
5314.01212644
| FTP