Converting Annual Default Probability to Payment-Period Risk

# Marginal Probability of Default for Credit Risk


# Marginal Probability of Default for Credit Risk












I am working on a model to predict credit defaults. We have worked out PD's of clients using logistic regression.

When calculating the default amount, we have to convert PDs to marginal PDs. The formula I am using to convert PD into marginal PD is:

$$mPD = (1+PD)^{\frac{1}n} -1,$$

where $n$ is the number of payments made by the client in one year.

So for every repayment schedule, we see how many payments are being made in an year and we use that number as n. but mostly in the last year, payments may not be for the full year but rather for a shorter period. For example for two full years client made 6 payments in an year but in last year only 3 payments would be made.

So I have two questions;

- Am I right in how I am calculating $n$?

- For the last year $n$ would be 3 or we keep it 6

## Answer by Cettt (score 1, accepted)

https://quant.stackexchange.com/a/51365

I think what you are calling marginal PD is simply the intra year PD. PD usually refers to the 1-year default probability, so if the default time is denoted by $\tau$ then $$ PD = \Bbb P (\tau \leq 1 \ \text{year}). $$

What you are refering to as marginal PD is the probability that you default within a shorter period of time, e.g. one month ($n = 12$) or one quarter ($n = 4$). It makes sense to align $n$ with the payments the client has to make, but you can compute $PD_n$ for any $n$. However you formula is slightly wrong: if we make the general assumption that the intra-year PD is constant than $$ 1 - PD = (1 - PD_n)^n, $$ which is equivalent to $$ PD_n = 1 - (1-PD)^{\frac 1n}. $$

In your case I would recommend to first compute $PD_6$. Given $PD_6$ you can easily compute $PD_3$:

$$ 1 - PD_3 = (1 - PD_6)^2. $$ In the first year you would rather work with $PD_6$ and in the second year you can work with $PD_3$ instead.