Potential Future Exposure (PFE)

[allenpee85]

Hi Guys,

Does anyone know what's the formula and advice available to calculate derivatives "PFE" exposures?
E.g. Monte Carlo......

Thanks!

 

[brian.field]

PFE is simply the analog to a relative VaR (if I remember correctly.6. So you need a confidence level, an interval length, and the corresponding parametric election or data and the PFE would be sigma*z(*notional)

 

[allenpee85]

Mind to share what's the meaning of "z"?

Also, is there anywhere we can find the relevant article about this formula?

Appreciate.

 

[brian.field]

Gregory's text covers this - I suggest you look there. z is a standard normal random variable (which presupposes a normality assumption.)

 

[allenpee85]

Any excel or example?

Still don't understand.......

Many thanks.

 

[QuantMan2318]

As @brian.field has pointed out, PFE is the maximum exposure at a given confidence level. It talks about the maximum amount that is under risk of counterparty default or any other credit risk. Therefore, it is a similar metric to VaR. Whereas VaR states the maximum amount of loss that would not be exceeded at a given confidence level, PFE is the Maximum amount of Exposure of recoverables that wouldn't be exceeded with a given Confidence level.

In other words, PFE = mu+z value*sigma, the same as that of VaR with a positive sign for the Mean.

I suggest you have a look at Gregorys excellent set of lessons that come with GARP syllabus on Credit Risk. If you want the spreadsheet, you can visit, http://cvacentral.com/books/credit-value-adjustment/spreadsheets/ that covers the spreadsheets associated with his excellent book on the CVA that also covers PFE. Download the spreadsheet for Chapter 7, that has calculations for the PFE

Thanks

 

[allenpee85]

As @brian.field has pointed out, PFE is the maximum exposure at a given confidence level. It talks about the maximum amount that is under risk of counterparty default or any other credit risk. Therefore, it is a similar metric to VaR. Whereas VaR states the maximum amount of loss that would not be exceeded at a given confidence level, PFE is the Maximum amount of Exposure of recoverables that wouldn't be exceeded with a given Confidence level.

In other words, PFE = mu+z value*sigma, the same as that of VaR with a positive sign for the Mean.

I suggest you have a look at Gregorys excellent set of lessons that come with GARP syllabus on Credit Risk. If you want the spreadsheet, you can visit, http://cvacentral.com/books/credit-value-adjustment/spreadsheets/ that covers the spreadsheets associated with his excellent book on the CVA that also covers PFE. Download the spreadsheet for Chapter 7, that has calculations for the PFE

Thanks

Dear QuantMan,

Thanks for the reply.

I have downloaded the chapter 7 as suggested, when looking at the <Spreadsheet 7.1>, can we know how could we derive the below:-

a) Mu (Mean)
b) Standard Deviation (sigma)
c) Alpha (confidence level)

Aside, I saw list of figure on right hand side with "Probability", these data we need for?

Or maybe, i give an example, I have a transaction as below:
Notional Amount = 6,000,000
Credit Conversion Factor = 5%
Replacement Cost (MtM) = 148,390.86

Are we able to calculate the PFE solely based on above data?

Thank you.

 

[David Harper CFA FRM]

Hi @allenpee85 I don't think it's enough information. Credit exposure is probably near to (or equal to) the replacement cost, but that is likely a current metric. Projecting the replacement cost into the future and is basically retrieving an expected exposure which is just the future's MtM (conditional on being positive). But that's a conditional mean. The PFE, like VaR, is a worst expected exposure with some confidence (probability) so it presupposes a distribution (analytical or simulated). Gregory's 7.1. makes the simplest possible assumption that that the exposure is normally distributed, so it's just solving for a normal quantile. So you are missing a distributional assumption, or the assumptions upon which to run a simulation in order to model a distribution. Thanks,

 

[Stuti]

Hi @David Harper CFA FRM , @QuantMan2318 - I am reading http://radoudoux.free.fr/last2/jGregoryCPTY Risk.pdf on CP, however i do not understand the derivation of EE which is as follows:

I am unable to understand how the integral is computed in this case.

 

[Dr. Jayanthi Sankaran]

Hi @Stuti,

Please refer to the following thread: P2.T6.411. Expected (EE) and potential future exposure (PFE). Please refer to David's answer to 411.3.

I did go through the formula in Gregory's appendix. It does makes a lot of intuitive sense to me. However, the derivation in the last step through me off! I don't think the FRM Exam would ever want you to know this derivation - my math is not that great enough!

Thanks!

 

[QuantMan2318]

Hi there @Stuti , the earlier discussion pointed out by @Dr. Jayanthi Sankaran is highly instructive of what the formula really means and what would be tested in the exam. Your derivation would not be tested; However, if you are really interested in knowing what it really means, consider the following:

1. You see the EE is equivalent in design to the ES, in just the same way that the PFE is equivalent to VaR, therefore, the EE is the Average of the values of the PFE, just like the ES is the average of the VaR. For summation of all the values, we use the integral for a continuous function.

2. Consider a coherent risk measure as ∫phi(x)*q dx, in the same vein, we have the EE as the ∫ norm(x) *(mu+sigma*x) dx as the addition of the values comprising the PFE at various ranges and intervals weighted by the norm.dis. value

3. When you break that integral, we know that the integral of the pdf, becomes the cdf, and hence the cumulative normal function for the first term. The second term occurs as a result of breaking the integral.

This is what probably occurs:

mu∫norm(x)dx + sigma∫x*norm(x)dx
= mu*cum.norm(x) + sigma*norm(x) as a result of integration by parts @David Harper CFA FRM ?
and hence your answer for the EE

Again, I am no Maths guy either, as you can see, I was only able to go so far

 

[Dr. Jayanthi Sankaran]

Hi @QuantMan2318,

Totally agree with the above derivation. Just one more thought:

From your equation:

mu∫norm(x)dx + sigma∫x*norm(x)dx
= mu*cum.norm(x) + sigma*norm(x) as a result of integration by parts.............(1)

mu∫norm(x)dx + sigma∫x*norm(x)dx
= mu*cum.norm(x) + sigma*norm(x)......................(2)

shouldn't the RHS in equation (2) be:
= mu*cum.norm(x) + (sigma*x^2)/2?

or am I mistaken?

Thanks!

 

[QuantMan2318]

Well there are two terms there @Dr. Jayanthi Sankaran , x and norm(x), I think it involves an integration by parts over there,
your answer would hold good when
sigma∫x dx = sigma*x^2/2

Anyway, thats why I have asked @David Harper CFA FRM to verify the veracity of my derivation

 

[Dr. Jayanthi Sankaran]

Yes you are right - in which case it would be:


mu∫norm(x)dx + sigma∫x*norm(x)dx
= mu*cum.norm(x) + sigma*[x*cum.norm(x) + norm(x)*(x^2)/2].......(3)

Have to look up integration by parts.....Thanks!

 

[Stuti]

i tried recalling the integration by parts and gave up after that. That is why i was confused with how the final result came. Thank you so much for answering my questions.

 

[brian.field]

You would never need to compute the integral by hand on the exam, in my opinion - if a similar question appeared on an exam, it would probably require something much simpler since it is not an exam of calculus skill. In the real world, you would simply use some approximation (quadrature, etc.) carried out via a computer program like Maple or Matlab.

Regarding integration by parts, recall that:



also, since PFE is analogous to VaR, the exam could have a question, I suppose, that assumes a specific distribution for thee exposure, say normal, and then asks for the PFE at a specified confidence and interval, which would be the same process as calculating a VaR.

 

[QuantMan2318]

Yes Brian, what you are saying is the general form of integration by parts, while I didn't try that form for the above, I found out that the following version would be more relevant for the above calculation on the RHS
∫u(x)v'(x) dx = u(x)*v(x) - ∫u'(x)*v(x) dx
This would perfectly accommodate x and norm(x)

But I couldn't proceed with it beyond a point, I get integral of a cum.norm variable and I couldn't simplify the equation beyond that for the second part of the equation ( I wish I had taken more core Maths courses )

 

[Dr. Jayanthi Sankaran]

Hi @QuantMan2318,

I did something similar to yours and also got integral of a cum.normal variable

Hi Brian,

The classic example of integration by parts - tried that....