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

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

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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*}

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  • $\begingroup$ 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 '21 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 '21 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 '21 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 '21 at 12:40

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