Conditional Default Probabilities in CVA Calculations
Summary
The note addresses how default survival indicators enter conditional expectations used in a credit valuation adjustment calculation. It explains that, under a hazard-rate construction in which the random variable used to generate default is independent of the market-information filtration, conditional survival to a given time is represented by the exponential of the negative integrated hazard rate. That survival factor can then be taken into the conditional expectation alongside an exposure value measurable at the relevant time.
The response applies this step to exposure valued at both the prior time point and the default interval endpoint, yielding expectations weighted by survival probability. The argument depends on the assumed default-time model and filtration: independence of the auxiliary random variable is essential to the stated conditional survival expression. The note does not develop the full CVA formula or discuss dependence between credit and market risk, recovery assumptions, or alternative default models, so its result should be read within those modeling assumptions.
Key ideas
- Under the stated hazard-rate construction, conditional survival is given by the exponential of the negative integrated hazard rate.
- The default time can be generated using an independent uniform random variable and the integrated hazard process.
- The survival probability can weight an exposure value inside a conditional expectation when the model assumptions apply.
- The result relies on the specified filtration and independence assumption and does not cover wrong-way risk or alternative default models.
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Full text
# conditional expectation formula of default in CVA
# conditional expectation formula of default in CVA
Here is the formula of CVA in page 74 in book `Modern Derivatives Pricing and Credit Exposure Analysis`.
Here $t_0 = t<t_1<\cdots<t_n = T;$ $\tau$ is the default; $X(t)$ is any value.
I don't much understand how we get the second equation: $$E^Q[\mathbb{1}_{\tau>t_i}X(t_{i-1})|\mathcal{F}_t] = E^Q\Big[E^Q[\mathbb{1}_{\tau>t_i}]X(t_{i-1})|\mathcal{F}_t\Big]$$ It hints that
> This is possible for the expectations containing $X(t_i)$ and $X(t_{i−1})$ since these are both $\mathcal{F}(t_i)$-measurable; apply the tower law of conditional expectations.
Does that mean $$E^Q[\mathbb{1}_{\tau>t_i}X(t_{i-1})|\mathcal{F}_t] = E^Q\Big[E^Q[\mathbb{1}_{\tau>t_i}X(t_{i-1})|\mathcal{F}_{t_{i-1}}]|\mathcal{F}_t\Big]=E^Q\Big[E^Q[\mathbb{1}_{\tau>t_i}|\mathcal{F}_{t_{i-1}}]X(t_{i-1})|\mathcal{F}_t\Big].$$
But how to convert $E^Q[\mathbb{1}_{\tau>t_i}|\mathcal{F}_{t_{i-1}}]$ to $E^Q[\mathbb{1}_{\tau>t_i}]?$
## Answer by Gordon (score 3, accepted)
https://quant.stackexchange.com/a/68428
Note that, for any $u > 0$, \begin{align*} E^Q(1_{\tau > u} \mid \mathscr{F}_u) = e^{-\int_0^u \lambda(s)ds}. \end{align*} For example, given $\lambda$, we can define the default time $\tau$ as \begin{align*} \tau = \inf\left\{t \in \mathbb{R}_+: e^{-\int_0^t \lambda_s ds} \le \xi \right\}, \end{align*} where $\xi$ is independent of $\mathscr{F}_{\infty}$ and is uniformly distributed over $(0, 1)$.
Then \begin{align*} E^Q\left(1_{\tau>t_i} X(t_{i-1}) \mid \mathscr{F}_t \right) &= E^Q\left(X(t_{i-1})E^Q(1_{\tau>t_i} \mid \mathscr{F}_{t_i}\big) \mid \mathscr{F}_t \right)\\ &=E^Q\left(X(t_{i-1}) e^{-\int_0^{t_i} \lambda(s)ds} \mid \mathscr{F}_t \right). \end{align*} Similarly, \begin{align*} E^Q\left(1_{\tau>t_i} X(t_i) \mid \mathscr{F}_t \right) &= E^Q\left(X(t_i)E^Q(1_{\tau>t_i} \mid \mathscr{F}_{t_i}\big) \mid \mathscr{F}_t \right)\\ &=E^Q\left(X(t_i) e^{-\int_0^{t_i} \lambda(s)ds} \mid \mathscr{F}_t \right). \end{align*}Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.