CVA Recovery Impact (Gregory)
- Thread starter Delo
- Start date
Hi @Delo
I didn't know at first glance what he means, either
So I retrieved the spreadsheet for this chapter (they are available at http://cvacentral.com/books/credit-value-adjustment/spreadsheets/)
Here it is at http://trtl.bz/0503-gregory-cva-2nd-chap12 (Interesting!).
The CVA (in column col G) is a product of (Def Prob)*(EE)*(Discount factor); i.e., the essential equation 12.2 but just inside the summation
Then his (Def Prob) is a direct function of the hazard rate, λ, assumption; e.g., 8.33% in his "default" model
And the hazard is given by Spread/(1-Recovery) ... so he is using the (classic) credit risk approximation λ = Spread/(1-R) where he has the counterpartry spread as a (constant) input. So, that's the tactical explanation: given constant spread, S, in λ = S/(1-R), higher recovery, R, implies higher hazard rate. For example, his default assumptions are 40% recovery and Counterparty 5Y Spread = 500, such that λ = 500/(1-0.40)*1/10,000 = 8.33%; if we increase R to 50%, then λ = 10.5%. I hope that helps!
I didn't know at first glance what he means, either
So I retrieved the spreadsheet for this chapter (they are available at http://cvacentral.com/books/credit-value-adjustment/spreadsheets/)Here it is at http://trtl.bz/0503-gregory-cva-2nd-chap12 (Interesting!).
The CVA (in column col G) is a product of (Def Prob)*(EE)*(Discount factor); i.e., the essential equation 12.2 but just inside the summation
Then his (Def Prob) is a direct function of the hazard rate, λ, assumption; e.g., 8.33% in his "default" model
And the hazard is given by Spread/(1-Recovery) ... so he is using the (classic) credit risk approximation λ = Spread/(1-R) where he has the counterpartry spread as a (constant) input. So, that's the tactical explanation: given constant spread, S, in λ = S/(1-R), higher recovery, R, implies higher hazard rate. For example, his default assumptions are 40% recovery and Counterparty 5Y Spread = 500, such that λ = 500/(1-0.40)*1/10,000 = 8.33%; if we increase R to 50%, then λ = 10.5%. I hope that helps!
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