PQ-external Part 2 practice question on expected loss

[janicekg]

How should this problem be approached? Is the joint PD relevant in answering this question?

Suppose Bank Z lends EUR 1 million to X and EUR 5 million to Y. Over the next year, the PD for X is 0.2 and for Y is 0.3. The PD of joint default is 0.1. The loss given default is 40% for X and 60% for Y. What is the expected loss of default in one year for the bank?

 

[Nicole Seaman]

How should this problem be approached? Is the joint PD relevant in answering this question?

Suppose Bank Z lends EUR 1 million to X and EUR 5 million to Y. Over the next year, the PD for X is 0.2 and for Y is 0.3. The PD of joint default is 0.1. The loss given default is 40% for X and 60% for Y. What is the expected loss of default in one year for the bank?

Hello @janicekg

Is this a question from one of the GARP practice exams? I just want to make sure it is posted in the correct section, as we have a full section for GARP practice questions. Also, I want to make sure you are aware of our search function in the forum. There is a great deal of discussion in the forum regarding the concepts that you are asking about.

Thank you,

Nicole

 

[janicekg]

@Nicole Seaman My source is not GARP.

 

[David Harper CFA FRM]

Hi @janicekg I think the easiest approach is to sketch the probability matrix; see below, yellow are the given assumptions
(XLS is here https://www.dropbox.com/s/g3izk6tceezr25q/090518-bank-xyz.xlsx?dl=0). I hope that's helpful!

 

[janicekg]

Thanks David!

 

[Gareth]

Hi David

Stupid question, but how did you get the other probabilities in the matrix apart from the 0.1?

Was it using Bayes' Theorem? If you would you mind doing the calc for just one of the entries?

Thanks!

 

[David Harper CFA FRM]

Hi @Gareth I was done for the day (only to move to other work!) but this is quick indeed so .... no, you don't need Bayes to populate the probability matrix, although I've come to see that, if you understand the prob matrix, Bayes is trivial to understand.

You just need three (two, really) features of the matrix:

1. The cells inside must sum to 100%: they are mutually exclusive joint probabilities, if they don't sum to 100%, you don't have a probability distribution, although I'm not sure we even need this feature here
2. The sum of unconditional probabilities, for each variable, must be 100% (ie, each random variable on its own must be characterized by an unconditional probability distribution). Visually, that simply means the right-most column outside the matrix must sum to 100%; in this case 30% + 70% = 100% because unconditional Prob (Y = D) + unconditional Prob (Y = D') need to equal 100%. Same for the bottom row where unconditional Prob (X = default) + unconditional Prob (X = not default) = 100%; in this case, 20+80% = 100%. So far, we can see that in order to have valid probabilities, not only do the joint probabilities (inner cells) need to sum to 100%, but also the sum of each of the unconditional probabilities,
3. Finally, a row or column of joint probabilities (inside the matrix) needs to sum to its associated unconditional probability (outside the matrix); e.g., at the top row, the joint 10% + joint 20% must equal the unconditional 30%; because these are each of the possible outcomes that contribute to the unconditional outcome. If unconditional Prob(Y = D) = 30%, there are two outcomes that can achieve this.

So given we know the yellow highlighted squares, we can deduce:

• inside the matrix, lower-left cell must be 10% because it is the X such that, in terms of the column, 10% + X = 20%
• inside the matrix, upper-right cell must be 20% because it is the X such that, in terms of the row, 10% + X = 30%
• That means the lower right must be 60% because inside the matrix, we know that 10% + 20% + 10% + X = 100%; but alternatively we could deduced (bottom row, outside the matrix) that 80% = 100 - 20%; and (right most column) that 70% = 100% - 30%. This would allow us to retrieve the 60% either as (column-wise) 20% + X = 80% or (row-wise) 10% + X = 70%. Okay phew i can't seem to write briefly, thanks,

 

[Gareth]

Ah amazing. So obvious now. And as always thank you for the quick and clear response David.