Joint Default Probabilities from Marginals and Correlation

# Joint probability of default


# Joint probability of default












Had a couple of questions from Jorion's FRM book (5th edition, page 438, Table 18.2 shown below). The book has a very stylized example as shown in the table below. The example shows how to calculate the probability of joint default. Once that is calculated, all other probabilities can be calculated using the individual marginal probabilities (e.g. P (A defaults, but B does not) = marginal probability of A defaulting less the joint probability of default.

Questions:

- Do the marginal distributions have to be identical? When I made the marginal default probabilities unequal, I get a negative probability of default (Prob A defaults, but B does not). So what kind of constraints do we need on the joint PDF to make this viable? Alternatively, if I specify one set of marginal probabilities (say for event A defaulting), and a correlation, how would I calculate the rest of the marginal distribution for B - is this possible?

- Is it possible to calculate P(A defaults, but B does not) directly? I did attempt....but the answer does not tie out to the calculations in the table.

Would appreciate some guidance on where to look for material related to this....a google search prints out stuff that is way more advanced than what I'm looking for. Thanks!

Thanks

!Jorion - FRM 5th edition, pg 438]1

## Answer by ir7 (score 2, accepted)

https://quant.stackexchange.com/a/63495

(I didn't quite understand where exactly you are going with your questions, but I inserted a few statements below that might be useful.)

Jorion's table shows: $$ \begin{bmatrix} P(A\cap B) & P(A\cap B^c) & : & P(A)\\ P(A^c\cap B) & P(A^c\cap B^c) & : & P(A^c)\\ .. & .. & & \\ P(B) & P(B^c) & & \end{bmatrix} $$

The four probabilities of event intersections sum up to $1$.

(Q2)

Given $P(A)$ and $P(A\cap B)$,

$$ P(A\cap B^c) = P(A) - P(A\cap B). $$

Given $P(B)$ and $P(A|B^c)$, via Bayes,

$$ P(A\cap B^c) = P(A|B^c)(1-P(B)) $$

Similar connections: $$ P(A|B^c) = \frac{P(A\cap B^c)}{P(B^c)} = \frac{P(A)- P(A\cap B)}{1-P(B)} $$ $$ \stackrel{Bayes}{=} \frac{P(A)- P(A| B)P(B)}{1-P(B)}$$ $$ \stackrel{(alt)Bayes}{=} \frac{P(A)- P(B| A)P(A)}{1-P(B)} =P(A)\frac{1- P(B|A)}{1-P(B)} $$

(Q1)

Given

$$\rho = \frac{P(A\cap B) - P(A)P(B)}{\sqrt{P(A)(1-P(A))P(B)(1-P(B))}} $$

and $P(A)$ and $P(A\cap B)$, we can calculate $P(B)$. Also, we note:

$$ P(A\cap B) = P(A)P(B) + \rho \sqrt{P(A)(1-P(A))P(B)(1-P(B))}, $$

$$P(A|B) = P(A) +\rho \sqrt{\frac{P(A)}{P(B)}(1-P(A))(1-P(B))} $$