# 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 $$t_0 = t $$\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 $$E^Q[\mathbb{1}_{\tau>t_i}]?$$

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*}
• 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. Oct 19 '21 at 5:33
• 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. Oct 19 '21 at 12:31
• 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 $t.$ Oct 19 '21 at 12:34
• 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. Oct 19 '21 at 12:40