# CVA formula for a call option

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?

• 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 $T$. Jan 21 '17 at 17:56
• When you define the $V(t, T)$ - did you actually mean $c(t, T)$? Jan 21 '17 at 21:09
• @LocalVolatility Yes, of course I have made a typo. I have corrected. Jan 22 '17 at 7:01

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.
• 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? Jan 24 '17 at 14:15
• @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. Jan 26 '17 at 10:15
• @RandomGuy it's the probability that the counterpart will default between $t$ and $t+dt$, i.e.: $dPD(0, t) = PD(0,t+dt) - PD(0,t)$ Feb 1 '17 at 17:06