Here is the formula of CVA in page 74 in book Modern Derivatives Pricing and Credit Exposure Analysis.

enter image description here

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 $E^Q[\mathbb{1}_{\tau>t_i}]?$


1 Answer 1


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*}

  • $\begingroup$ could you explain more on why \begin{align*} E^Q(1_{\tau > u} \mid \mathscr{F}_u) = e^{-\int_0^u \lambda(s)ds}. \end{align*} Since $1_{\tau > u}$ is $\mathscr{F}_u$-measurable, i think it should be $E^Q(1_{\tau > u} \mid \mathscr{F}_u) = 1_{\tau > u}.$ Namely at time $t_i,$ we already know whether the default has happened. $\endgroup$ Oct 19, 2021 at 5:33
  • $\begingroup$ Note that $\xi$ is independent of $\mathscr{F}_{\infty}$. From my definition of the default time, $1_{\tau>u}$ is not $\mathscr{F}_u$ measurable. See also this question for further discussions. $\endgroup$
    – Gordon
    Oct 19, 2021 at 12:31
  • $\begingroup$ ok, get your point, it is already the result after the filtering switch formula. Then $1_{\tau>u}$ is actually independent on $\mathcal{F}_t$ for any $t.$ $\endgroup$ Oct 19, 2021 at 12:34
  • $\begingroup$ That is true if $\lambda$ is deterministic. In general, $\lambda$ can be a process adapted to $\{\mathscr{F}_t\}$, that is, for any $t>0$, $\lambda_t$ is $\mathscr{F}_t$ measurable. $\endgroup$
    – Gordon
    Oct 19, 2021 at 12:40

Your Answer

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

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