2
$\begingroup$

When calculating ECLs for loans under IFRS 9, one of the requirements is that the PD estimates have to be Point-in-time ($PD_{PIT}$) rather than through-the-cycle ($PD_{TTC}$).The setting is as follows: we have a rating model with ratings $X$ ranging from 1-10 with 10 being the worse . The estimated probability of default $PD^{TTC}_i$ for each of the buckets is as follows

| X  | PD TTC |
|----|--------|
| 1  | 0.62%  |
| 2  | 0.84%  |
| 3  | 0.93%  |
| 4  | 1.23%  |
| 5  | 2.10%  |
| 6  | 2.79%  |
| 7  | 3.80%  |
| 8  | 5.04%  |
| 9  | 7.01%  |
| 10 | 31.22% |

The overall $PD_{TTC}$ for the entire portfolio is 5.74%. Lets say we estimate that in the coming year our $PD_{PIT}$ will be 8%. We now want to calibrate the probability for each rating to reflect the increase in the overall default rate of the portfolio. I was told that this can be done using the following varsion of the Bayes formula:

$$PD^{PIT}_i = \frac{(1-PD_{TTC})*PD_{PIT}*PD^{TTC}_i}{PD_{TTC}*(1-PD_{PIT})*(1-PD^{TTC}_i)+(1-PD_{TTC})*PD_{PIT}*PD^{TTC}_i}$$ where
$PD_{TTC}$: Overall portfolio TTC default rate
$PD_{PIT}$: Overall portfolio PIT default rate
$PD^{TTC}_i$: TTC default rate for rating grade $i$

For example the calibrated PD for rating 1 would be $$PD^{PIT}_1 = \frac{(1-0.0574)*0.08*0.0062}{0.0574*(1-0.08)*(1-0.0062)+(1-0.0574)*(0.08)*0.0062}$$ $$PD^{PIT}_1 = 0.0088$$

The fully calibrated rating scale would be as follows

| X  | PD TTC | PD PIT |
|----|--------|--------|
| 1  | 0.62%  | 0.88%  |
| 2  | 0.84%  | 1.20%  |
| 3  | 0.93%  | 1.32%  |
| 4  | 1.23%  | 1.75%  |
| 5  | 2.10%  | 2.97%  |
| 6  | 2.79%  | 3.94%  |
| 7  | 3.80%  | 5.34%  |
| 8  | 5.04%  | 7.04%  |
| 9  | 7.01%  | 9.72%  |
| 10 | 31.22% | 39.33% |

Can someone please explain to me the reasoning behind this particular application of Bayes' formula and if possible provide a derivation showing why it is valid in this context?

$\endgroup$
4
  • $\begingroup$ Do the PDs average out to the aggregate PDs - e.g., average of individual PIT PDs comes out to 8%? $\endgroup$ – Magic is in the chain Dec 21 '19 at 0:06
  • $\begingroup$ No, they do not $\endgroup$ – Serge Kashlik Dec 22 '19 at 6:13
  • $\begingroup$ Can you clarify how you derive the PD-PIT(i) formula? $\endgroup$ – user48040 Jul 10 '20 at 7:39
  • $\begingroup$ That is the question of this thread. What is the intuition of this formula and how is it derived? $\endgroup$ – Serge Kashlik Jul 15 '20 at 12:05
0
$\begingroup$

My view is that the equation has nothing to do with Bayes' theorem, except that its form looks like it is derived from Bayes' theorem.

As shown below, the equation can be derived using the concept of odd, which is used in the realm of default probability models.

Note: I only address how the equation, I think, was obtained. The merit of the equation or approach is open to debate, but that is another topic all together.

Definitions:

Let $\displaystyle \alpha $ be the odd based on the overall portfolio TTC PD and $\displaystyle \beta $ be the odd based on the overall forecasted portfolio PIT PD: \begin{equation*} \alpha \ =\ \frac{( 1\ -\ PD_{TTC})}{PD_{TTC}} \end{equation*} \begin{equation*} \beta \ =\ \frac{( 1\ -\ PD_{PIT})}{PD_{PIT}} \end{equation*} In addition, let the TTC PD of rating $\displaystyle i$ be defined as \begin{equation*} PD^{i}_{TTC} \ =\frac{B_{i}}{G_{i} +B_{i}} \ \end{equation*} where $\displaystyle B_{i}$ and $\displaystyle G_{i}$ are the bad count and the good count in raitng $\displaystyle i$, respectively.

Concept of Calibratiion:

The concept of calibration is to adjust the probability of default estimates so that, on average, they will not deviate from the long-run central tendency of the observable default rates.

Derivations:

After obtaining the forecasted $\displaystyle PD_{PIT}$ by whatever means, one can calculate $\displaystyle \beta $. However, $\displaystyle PD_{PIT}$ and $\displaystyle PD_{TTC}$ must be related in some way. That is what calibration is all about. The relationship could be specified as \begin{equation*} \beta \times \alpha ^{-1} \end{equation*} But note that the relationship can be written as

\begin{equation*} \left[\frac{( G_{PIT} /( G_{PIT} +B_{PIT})}{( B_{PIT} /( G_{PIT} +B_{PIT})}\right] \times \left[\frac{( B_{TTC} /( G_{TTC} +B_{TTC})}{( G_{TTC} /( G_{TTC} +B_{TTC})}\right] \end{equation*} \begin{equation*} =\ \left[\frac{G_{PIT}}{B_{PIT}}\right] \times \left[\frac{B_{TTC}}{G_{TTC}}\right] \end{equation*} where $\displaystyle G_{k}$ and $\displaystyle B_{k}$ are the good count and the bad count for $\displaystyle TTC$ and $\displaystyle PIT$, respectively. Over the long run, $\displaystyle \beta \ \times \alpha ^{-1}$ is expected to be 1 because $\displaystyle G_{PIT}$, $\displaystyle G_{TTC} ,$ $\displaystyle B_{PIT}$ and $\displaystyle B_{TTC}$ will cancel out each other.

The relationship is then brought to the granular rating level by scaling the good count within a particular rating with $\displaystyle \beta \ \times \alpha ^{-1}$. The reason of choosing good over bad is that in some ratings, bad count can be zero. Assuming the portfolio is well diversified, all ratings will have non-zero good counts.

With the above, the calibrated PIT PD for rating $\displaystyle i$ is:

\begin{equation*} Calibrated\ PD^{i}_{PIT} \ =\frac{B_{i}}{\beta \alpha ^{-1} G_{i} +B_{i}} \end{equation*} \begin{equation*} =\ \frac{Bi}{\left(\frac{1-PD_{PIT}}{PD_{PIT}}\right)\left(\frac{PD_{TTC}}{1-PD_{TTC}}\right) G_{i} +B_{i}} \end{equation*}

\begin{equation*} =\frac{( PD_{PIT}) \ (1-PD_{TTC} )\ Bi}{( 1-PD_{PIT}) \ ( PD_{TTC}) \ G_{i} +( PD_{PIT}) \ (1-PD_{TTC} )\ B_{i}} \end{equation*}

\begin{equation*} =\left[\frac{( PD_{PIT}) \ (1-PD_{TTC} )\ B_{i}}{( 1-PD_{PIT}) \ ( PD_{TTC}) \ G_{i} +( PD_{PIT}) \ (1-PD_{TTC} )\ B_{i}}\right]\left[\frac{1/( Gi+B_{i})}{1/( Gi+B_{i})}\right] \end{equation*}

\begin{equation*} =\left[\frac{( PD_{PIT}) \ (1-PD_{TTC} )\ PD^{i}_{TTC}}{( 1-PD_{PIT}) \ ( PD_{TTC}) \ \left( 1-PD^{i}_{TTC}\right) +( PD_{PIT}) \ (1-PD_{TTC} )\ PD^{i}_{TTC}}\right] \end{equation*}

End Note:

Is the calibration approach sound? My view is that we can only tell by validating it with actual data.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.