Partition Theorem Homework: Finding Probability of Lying

  • Thread starter Thread starter Philip Wong
  • Start date Start date
  • Tags Tags
    Partition Theorem
Click For Summary
SUMMARY

The discussion centers on applying Bayesian Theorem to calculate the probability of an insurance claimant lying based on lie detector test results. The calculations show that the probability of a claimant being lying given they failed the test (IP(L|F)) is 62%, while the probability of lying given they passed the test (IP(L|P)) is 36%. The correct formula for overall probability of lying (IP(L)) is confirmed as IP(L) = IP(L|F)IP(F) + IP(L|P)IP(P), which accounts for both outcomes of the lie detector test.

PREREQUISITES
  • Understanding of Bayesian Theorem
  • Familiarity with conditional probability
  • Knowledge of probability notation (e.g., IP, P(A|B))
  • Basic statistics concepts related to probability distributions
NEXT STEPS
  • Review Bayesian Theorem applications in real-world scenarios
  • Explore advanced conditional probability techniques
  • Learn about probability distributions and their implications in insurance contexts
  • Study the impact of lie detector test accuracy on insurance claims
USEFUL FOR

Statisticians, data analysts, insurance professionals, and students studying probability and statistics who want to deepen their understanding of Bayesian analysis in practical applications.

Philip Wong
Messages
95
Reaction score
0

Homework Statement


Assume that it is appropriate to transfer the probabilities IP(F|L) and IP(F|T) from the police context to the insurance context.
Define the following new events for the insurance context:
L = “insurance claimant is lying”;
T = “insurance claimant is truthful”;
F = “insurance claimant failed lie-detector test on phone”;
P = “insurance claimant passed lie-detector test on phone”.
An insurance company finds that a massive 52.5% of claimants fail the liedetector
test on the phone. What is the probability that a claimant is actually
lying?

IP(F) = 0.525 IP(P) = 1-0.525=0.475
IP(F|L)=0.38 IP(F|T)=0.23
IP(P|L)=0.14 IP(P|T)=0.25

Homework Equations


Bayesian TheoremP(B |A) = P(A|B)P(B)/ P(A)
IP(L) = IP((L|F) \cap (L|P))


The Attempt at a Solution


IP(L|F)=(0.38*0.525) / (0.38*0.525+0.23*0.525)=0.1995/0.32025=0.62
IP(L|P)=(0.14*0.475) / (0.14*0.475+0.25*0.475)=0.0665/0.18525 = 0.36

IP(L) = IP((L|F) \cap (L|P))
= IP(L|F) * IP(L|P)
= 0.62 * 0.36 = 0.2232

is my workings right? I'm kind of worried that I used the wrong formula to work out IP(L), so it would be nice if someone could double check that part too
thanks
 
Physics news on Phys.org
Though I haven't checked numbers, your use of Bayes to get IP(L|F) and IP(L|P) is on the right track, and given the results show someone who fails the test is likely to be lying (62%) or someone who passes the test is unlikely to be lying (36%) is encouraging

I don't understand what you've done in the last step however. I would notice (or assume) that the events P&F span the entire probability universe, and use the following:

IP(L) = IP(L|F)IP(F) + IP(L|P)IP(P)
 
Last edited:
lanedance said:
Though I haven't checked numbers, your use of Bayes to get IP(L|F) and IP(L|P) is on the right track, and given the results show someone who fails the test is likely to be lying (62%) or someone who passes the test is unlikely to be lying (36%) is encouraging

I don;t understand what you've done in the last step however. I would notice (or assume) that the events P&F, span the entire probability universe and use the following:

IP(L) = IP(L|F)IP(F) + IP(L|P)IP(P)

you are right, I just figured it out about half an hour ago. I knew I had implied some of the formula wrongly.

Thanks for confirming that with me!