conditional expectation formula of default in CVA

Question

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 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*}