5
$\begingroup$

I have a very quick question. Suppose that I buy a European call option from party S with expiry $T$. I want to determine the general formula for the CVA of the option, at time $t$. If I let $T_1\leq T$ be the default time of $S$ and I denote by $\xi=S$ the default event of $S$ (I am following here the notation of Quantitative Risk Management by McNeil, Frey, Embrechts, Chapter 17), and I denote by $c(t,T)$ the risk-neutral, default-free (Black-Scholes) price of the option at time $t$, is the following formula correct: $$ CVA(t) = LGD\cdot\mathbb E^Q[\mathbb 1_{\{T_1\leq T\}}\cdot\mathbb 1_{\{\xi=S\}}D(t,T_1)c(T_1,T) |\mathcal F_t] ? $$

My point here is that the evaluation of the default-free expected cash flow of the option is just the price of the option at time $t$, and since this cannot be negative, the term $c(T_1,T)$ is always positive, so there's no need to take its positive part.

So is my formula above correct?

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ I'm not familiar with that notation, but it seems OK. Indeed with european options the CVA Formula simplifies than further to $CVA = LGD \cdot PD \cdot Optionprice$. This is due to the fact, that the price is always positive and that that an european option has only one cashflow at $T$. $\endgroup$
    – Ami44
    Jan 21, 2017 at 17:56
  • 1
    $\begingroup$ When you define the $V(t, T)$ - did you actually mean $c(t, T)$? $\endgroup$
    – LocalVolatility
    Jan 21, 2017 at 21:09
  • 1
    $\begingroup$ @LocalVolatility Yes, of course I have made a typo. I have corrected. $\endgroup$
    – RandomGuy
    Jan 22, 2017 at 7:01

1 Answer 1

Reset to default
3
$\begingroup$

No need to overcomplexify things. The CVA gives you simply the amount that you expect to lose if and when your counterparty defaults, discounted to today.

In your case, you are buying an option. So, you already paid a premium but still expect to receive the payoff at expiry. In a sense, your counterparty owes you the payoff and you can lose this amount in case of a default.

Assuming the LGD is constant, and that the counteparty default probabilities and the call value are independant, the CVA can be expressed as follows: $$ \mathrm{CVA} = LGD \int_0^T \mathbb{E} [ D(0,t) c(t,T) ] dPD(0,t)$$ Which can be discretized as follows ($t_0 = 0,\dots \ , t_n = T$): $$ \mathrm{CVA} \approx LGD \sum_{i=0}^{n-1} \mathbb{E} [ D(0,t_i) c(t_i,T) ] PD(t_i,t_{i+1}) \\ $$ So it is equal to the sum of the expected call value at each time-step (until its expiry), weighted by the default probability of the counterparty during this time-step.

Improve this answer
$\endgroup$
5
  • 2
    $\begingroup$ I am rather unfamiliar with CVA calculations. But I would have expected to see sth. like $\mathbb{E} \left[ c(t, T) \right]$ in the integrand. Or is that what you notation wants to express? $\endgroup$
    – LocalVolatility
    Jan 24, 2017 at 14:15
  • 1
    $\begingroup$ You are right, it wasn't clear. I edited to correct in both formulae. $\endgroup$
    – byouness
    Jan 24, 2017 at 14:25
  • $\begingroup$ @uness I actually find the expression of your integral more complex than mine.... What exactly is the differential $dPD(0,t)$ in the integral? What does it mean, are you integrating over a family of measures indexed by $t$?? Thanks for the clarification. $\endgroup$
    – RandomGuy
    Jan 26, 2017 at 10:15
  • 1
    $\begingroup$ @RandomGuy it's the probability that the counterpart will default between $t$ and $t+dt$, i.e.: $dPD(0, t) = PD(0,t+dt) - PD(0,t)$ $\endgroup$
    – byouness
    Feb 1, 2017 at 17:06
  • $\begingroup$ I am trying to understand why you made the (good) point about already having paid the premium and only worrying about the payoff? How does this reflect in your formulation of CVA? $\endgroup$
    – gmarais
    Nov 23 at 8:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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