# Bayes Rule - need help

Hi I have problem complete with notations of elements in Bayes rule I am not sure if I properly distinguish prior probabilities Practice problem: We have mortgage company and they did research saying that: 1% of mortgagees default and lose property 90% mortgages are late 2mos and default as compared to 45% of those who do not default. What is the probability that a mortgagee with two or more late monthly payments will default on the mortgage and lose the property? Is my understanding correct: a) first look for the mutually exclusive and exaustive events which are Probability of not being default = .45 => probability of default .55 ? notations: p(not default) = .45 vs P(default) = .55 are the above prior probabilities? then rest notations: are they properly noted? P(loose property/default)=.01 or P(default/lose property) P(not loose/default) =.9 or (default / not lose) please help me it is eating me up

Bayes rule remained to be the only formula I never understood in my undergrad studies and in CFA level 1 as well … so I have a long but easy to remember way for doing Bayes rule - I came up with it myself in order to solve the problem lol When I’m home tonight I will write it in a word document and send it to you and hopefully it will help …

Can you restate this part? This is confusing: (90% mortgages are late 2mos and default) as compared to (45% of those who do not default). Do you mean 90% of the 2 month late mortgages default? Would be easier if you just type the exact words from the question.

Omar, Could you please email me this document also - tommyrogulj@gmail.com Ive studied up to Equity in the CFA curriculum and Bayes was the only thing I didn’t understand and skipped. Thanks mate

schweser has a great trick method to solve these, not sure what page

I really struggled with this too. If you run a search for Bayes, you will see my post a while back and a few helpful suggestions from others. I think I have finally got to grips with it, here’s how I understand it. (This might be obvious to some but it wasn’t to me). P(AB) = P(BA) This is self-explanatory, no justification needed, I think. Therefore we can substitute in the following formula (ie if you were given a problem to solve for P(AB) or P(BA) these are what you would use): P(A|B)*P(B) = P(B|A)*P(B) Bayes formula is just the equation above, rearranged! – for me, I find it easier to start with the one above, and do the rearranging myself. I don’t know why this helped me understand it/remember it, but it did. Cheers

full text is somewhat unclear A mortgage holding company has found that 1% of its mortgage holders default on their mortgage and lose the property. Furthermore, 90% of those who default are late on at least two monthly payments over the life of their mortgage as compared to 45% of those who do not default. What is the probability that a mortgagee with two or more late monthly payments will default on the mortgage and lose the property?

Kiakaha Wrote: ------------------------------------------------------- > I really struggled with this too. If you run a > search for Bayes, you will see my post a while > back and a few helpful suggestions from others. I > think I have finally got to grips with it, here’s > how I understand it. (This might be obvious to > some but it wasn’t to me). > > P(AB) = P(BA) This is self-explanatory, no > justification needed, I think. Therefore we can > substitute in the following formula (ie if you > were given a problem to solve for P(AB) or P(BA) > these are what you would use): > > P(A|B)*P(B) = P(B|A)*P(B) > should not be: P(A|B)*P(B) = P(B|A)*P(A) ?

That actually clears it up. They are saying: P(Default) = 0.01 P(Late | Default) = 0.9 P(Late | !Default) = 0.45 They want you to find: P(Default | Late) Bayes’ theorem says: P(Default | Late) = P(Late | Default) * P(Default) / P(Late) P(Late) = P(Default)*P(Late|Default) + P(!Default)*P(Late|!Default) = 0.01*0.9 + 0.99*0.45 = 0.4545 So: P(Default | Late) = 0.9*0.01/0.4545 = 0.019802…

yes rafal – you are right – thanks for catching my typo!

Ohai thanks alot I had a problem with notations here i guess Wonder how I can improve proper notations of Bayes problem Maybe decision tree?

Trees are much easier in my opinion. When I was doing this stuff, I never used Bayes’ formula (though, I did here for illustrative purposes).

I will prepare and send the file tomorrow night - hopefully - as I got busy today … Sorry for that

Notations is mof essence in these examples thanks again Ohai tree for this example: >>>>>>>>> -------L-----.9 (we look for this path from the end) so 0.01 x .9 /( .01x.9+.99x.45) -------D----.01 >>>>>>>>> -------!L----.1 >>>>>>>>> -------L-----.45 -------!D—.99 >>>>>>>>> -------!L----.55 @Omar I will appreciate that thanks;) Rafal

I posted an example of Baye’s rule as well as an explanation behind the mathematics that goes into it. hopefully you will find that useful. http://www.analystforum.com/phorums/read.php?11,1227764,1228039#msg-1228039

Attached is the file (link will expire in a week): https://fileshare.aus.edu/pnp7/files/EylciNKYo/Bayes_Law.pdf I have added the first 3 pages (though not related to Bayes law) to introduce you to the way I use in solving Bayes questions. The file was prepared by me for introductory math students in undergraduate studies and this is why it is illustrated in details (you can skip the irrelevant stuff). The last 2 pages have 2 examples that show how to solve any Bayes question using a systematic way. I have just prepared it after a long day so I’m sorry if there are mistakes with calculations Good luck Omar

Hey Omar, I checked out the examples. Thanks for providing those. Great practice. One question, the second example asks for P(Higher R | FA) Can’t we calculate it using total probability only? P(Higher R) = P(Higher R | TA) *P(TA) + P(Higher R | FA) *P(FA) .5729 = .47 * 5/7 + P(Higher R | FA) * 2/7 We get the value of P(Higher R | FA) as 0.83. Is there something wrong here?

anish Wrote: ------------------------------------------------------- > Hey Omar, > > I checked out the examples. Thanks for providing > those. Great practice. > > One question, the second example asks for P(Higher > R | FA) > > Can’t we calculate it using total probability > only? > P(Higher R) = P(Higher R | TA) *P(TA) + P(Higher R > | FA) *P(FA) > .5729 = .47 * 5/7 + P(Higher R | FA) * 2/7 > > We get the value of P(Higher R | FA) as 0.83. > > Is there something wrong here? That’s exactly what I did in the solution except that I filled in the entire table in order to be able to solve any question asked. For example, now if you are asked about the probability of stock being selected using FA given that return is not higher than R, you look at the table and: 0.0508/0.4271 No need for complex calculations once you have the table up there …

Omar Adnan Wrote: ------------------------------------------------------- > > > That’s exactly what I did in the solution except > that I filled in the entire table in order to be > able to solve any question asked. For example, now > if you are asked about the probability of stock > being selected using FA given that return is not > higher than R, you look at the table and: > > 0.0508/0.4271 > > No need for complex calculations once you have the > table up there … Oh Ok. So we dont really need Bayes’ formula here. I was wondering if I have missed something. Thanks.