Question about two bond's probability of default from transition matrix

[dadalee1102]

Hi, David,
I was doing a typical question of calculating probability of default of a bond based on the transition matrix from the Handbook. I just though about the following possible scenario:
If I am given a transition matrix, and I am asked to calculate two bonds' probability of default at the same time given their prob of default are independent, do I just multiply their individual prob of default ?
For example, if from the transition matrix, I calculate prob of default of B-rated bond is 4.2%, and prob of default of C-rated bond is 5%, so given their prob of default are independent, I simply multiply 4.2% by 5% ??

Is the above concept right?

Thank you so much for your help!!!

 

[David Harper CFA FRM]

Hi hsintalee,

Oh, that's a great idea for a 6th question i can add to http://forum.bionicturtle.com/threads/l1-t2-202-variance-of-sum-of-random-variables.4968/

If they are independent, you are correct! The reason it will make for a nice practice question is that it generalizes:
Let B = PD(bond b) = 4.2% and C = PD(bond c) = 5% are two Bernouilli random variables (only two outcomes), so our general formula is:
Prob[both bonds default] = E(BC) = E(B)*E(C) + covariance(B,C); and your independence --> cov(B,C) = 0.
So, if you added (eg.) assumption that default covariance(B,C) is 0.1, then E(BC) =4.2%*5% + 0.1

Hope that helps, Thanks, David

 

[David Harper CFA FRM]

FYI, I added 202.6 as a variation on this http://forum.bionicturtle.com/threads/l1-t2-202-variance-of-sum-of-random-variables.4968/

 

[dadalee1102]

David, Thanks for the reply.
One more question, I can't recognize where did the following formula come from: Prob[both bonds default] = E(BC) = E(B)*E(C) + covariance(B,C) ? E here is denoted as expectation ? I am a bit confused here. Did you simply derive this formula from the independent of two RV : P(AB) = P(A) * P(B) ?

 

[David Harper CFA FRM]

Hi hsintalee,

See Stock & Watson 2.34 (page 35), or better, it's really just a covariance property: http://en.wikipedia.org/wiki/Covariance
covariance (X,Y) = E[XY] - E[X]E[Y], such that E[XY] = E[X]E[Y]+ covariance (X,Y)

It's worth memorizing as, notice if X = Y, we have the useful variance formula:
Cov(X,X) = variance(X) = E[X^2] - E[X]^2

Thanks, David

 

[dadalee1102]

okay, I see. Thank you so much, David!
sorry that I got another question to bug you. I know that under independent, P(AB) = P(A) * P(B), but what if they are not under indpendent? what's P(AB) ?
Thank you!!!

 

[David Harper CFA FRM]

sure thing.
P(AB) depends on the marginal distributions and dependence structure; i.e., need more information.
P(AB) = P(A)P(B) if and only if (iif) independent.

Thanks, David