Hi, can you help to think of an efficient way to tackle this problem with Bayes’ formula? Thx. A sccoring system shows if a bond’s score < 20, P(default)=85% in five years. If its score >= 20, P(default)=40% in five years. Given a randomly chosen bond: P(score<20)=25%, what’s P(default | score >=20) ?
p(default/score>=20)= p(score>=20/default)/p(score>=20)*p(default). p(score>=20/default)=40% p(score>=20)=1-p(score<20)=1-.25=.75 p(default)= p(default/<20) + p(default/>=20)= .85+.4=1.25 therefore: p(default/score>=20)= 40%*1.25/.75=.67 I hope it is right. Not sure.
I don’t think you can have 1.25 as a probability of a default. Can we remove the joint prob of a default —> 1.25 - (.4)(.85) = 0.91 = (.4*.91) / .75 = 48.5%
i dont really get the wording when the q says: “If its score >= 20, P(default)=40% in five years.” i first interpreted that as the conditional probability of default given score >=20, i.e. P(default | score>=20), but then the question is actually asking for that conditional probability. are we to interpret that phrase as joint probability, i.e. P(default * score>=20)?
tree
approaching_c Wrote: ------------------------------------------------------- > i dont really get the wording when the q says: > > “If its score >= 20, P(default)=40% in five > years.” > > i first interpreted that as the conditional > probability of default given score >=20, i.e. > P(default | score>=20), but then the question is > actually asking for that conditional probability. > are we to interpret that phrase as joint > probability, i.e. P(default * score>=20)? I think you misunderstood the question. The given known info. of the scoring system is: In general, P(default | score < 20)=85%; P(default | score >= 20)=40% in five years. Now let’s say we randomly choose a bond. This bond has 25% probability of a score less than 20. What’s the probability that this bond defaults in the next five years had a score of 20 or higher? Yes, tree diagram is a good way to go. And I’m sort of understanding the tree way. Yet I have trouble to employ the Bayes’ Fomula to verify it - I believe it is a Bayes’ Formula question. Any advice? How can we get this done accurately in 2 min. ? Schweser shows the correctly answer is 58.5%.
.585 comes if you read the problem as What is the probability the Bond defaults given the score is >= 20 Check this out Score < 20 Default NoDefault .25 .85 .15 Score >=20 Default NoDefault .75 .4 .6 PDefault given Score < 20 = .25 * .85 = .2125 PDefault given Score >= 20 = .75 * .4 = .3 So Total Probability bond defaults = .3 + .2125 = .5125 Cases where Score >= 20 = .3 So your reqd prob is .3 / .5125 = .585
In your original post you stated: “what’s P(default | score >=20) ?” But this question is really asking you to solve: P(score>=20 | default), i.e. what is the probability of score >=20 given the bond defaults (the way CP solved it)
Excellent, CP! Quite clean table you drew, thx. I tried to apply Bayes’ Formula again. I think the question should actually be interpreted as: for the given bond whose P(score<20)=25%, if it defaults in the next 5 years, what’s the probability of its score equal or greater than 20? P (score >=20 | default) = P (default | score >=20) x P (score >= 20) / P (default) P (score >= 20) = 1 - 25% = 75% P (default | score >=20) = 40% Unconditional probability: P (default) should be the total probability of default under both score ranges. So P (default) = 25%x85% + 75%x40% = 51.25% Thus, P (score >=20 | default) = 75%x40% / 51.25% = 58.5%
approaching_c Wrote: ------------------------------------------------------- > In your original post you stated: > > “what’s P(default | score >=20) ?” > > > But this question is really asking you to solve: > P(score>=20 | default), i.e. what is the > probability of score >=20 given the bond defaults > (the way CP solved it) You are right, I was wrong! Thx!!!
Hyang, how was the question actually worded? Thanks
d=default a=above 20 b=below 20 given: p(d/b)=0.85 p(d/a)=0.4 p(b)=0.25 P(a)=0.75 using total probabily rule P(d)=0.85*0.25 + 0.4*0.75=0.5125 question is asking for P(a/d) P(a/d) = P(a&d)/P(d) = P(d/a)*p(a)/p(d) = 0.58
I love the simple notation!