# Conditional expectation proof

I hope this is the right place to ask my question - my question comes from some reading I'm doing in mathematical finance, but my question is really a question in probability theory, and is about how to prove two conditional expectations are equal.

I'm trying to follow the derivation of the conditional expectation result appearing about half way down page 9 of the following paper:

http://www.damianobrigo.it/CDS_Counterparty_fitchsolutions.pdf

I'm having trouble coming up with a rigorous proof of going from the 3rd to the 4th line, i.e. in showing that, for $u > T_j$ \begin{align} 1_{T_{j-1} < \tau_2 \leq T_j} &\mathbb{E}[ 1_{\tau_1 > u} | \mathcal{G}_{T_j}, \tau_1 > T_j, T_{j-1} < \tau_2 \leq T_j ] \\ &=1_{T_{j-1} < \tau_2 \leq T_j} \mathbb{E}[ 1_{\tau_1 > u} | \mathcal{F}_{T_j}, \tau_1 > T_j, T_{j-1} < \tau_2 \leq T_j ]. \end{align}

Notation is explained in the linked paper linked, but I'll try and explain briefly the setup in hope of getting more response. We work on a filtered probability space $(\Omega, \mathcal{F}_t, \mathbb{Q})$. There are $\mathcal{F}_t$ adapted processes, $\lambda_1(t)$ and $\lambda_2(t)$. We define $\Gamma_i(t) = \int_0^t \lambda_i(s) ds$. There are also Uniform random variables $U_1$ and $U_2$, both assumed independent of the filtration $\mathcal{F}_{\infty} = \sigma( \cup_t \mathcal{F}_t )$.

$\tau_i$ is defined as $$\tau_i := \inf \{t: \exp(-\Gamma_i(t)) <= U_i \}$$ and the joint cumulative distribution function of $(U_1, U_2)$ is given by a copula $C(u_1,u_2)$, from which we get dependence of the default times.

The filtration $\mathcal{G}_t$ is defined as $$\mathcal{G}_t : = \sigma ( \mathcal{F}_t, 1_{\tau_1 \leq s_1}, 1_{\tau_2 \leq s_2} : s_1, s_2 \leq t ).$$

My original plan had been the following. The left hand side of the above equation is just a conditional expectation on $\mathcal{G}_{T_j}$. Clearly also the right hand side of the above equation is $\mathcal{G}_{T_j}$ measurable (since we are conditioning on a sub sigma algebra of it). Consequently, if I call the the right hand side random variable of the above equation, R, I just need to show that for any $G \in \mathcal{G}_{T_j}$

$$E(1_G 1_{T_{j-1} < \tau_2 \leq T_j} 1_{\tau_1 > u} ) = E( 1_G R).$$

To this end, my plan was to show that for any $$A \in \Pi := \{ X\cap 1_{\tau_1 > s_1} \cap 1_{\tau_2 > s_2} : X \in \mathcal{F_{T_j}}, s_1, s_2 <= T_j \},$$

$$\mu_1(A) := \mathbb{E}[ 1_A \mathbb{E}[1_{T_{j-1} < \tau_2 \leq T_j} 1_{\tau_1 > u} | \mathcal{F}_{T_j}, \tau_1 > T_j, T_{j-1} < \tau_2 \leq T_j ] ]$$ and $$\mu_2(A) := \mathbb{E}[ 1_A 1_{T_{j-1} < \tau_2 \leq T_j} 1_{\tau_1 > u} ]$$ are equal. I think this would do it, since then $\mu_1$ and $\mu_2$ are measures agreeing on the $\pi$-system $\Pi$. They therefore must be equal on the sigma algebra generated by $\Pi$, which contains the sigma algebra $\mathcal{G}_{T_j}$, which proves the required conditional expectation property.

Unfortunately, I'm not able to prove $\mu_1(A) = \mu_2(A)$...essentially, I seem to need to move the $1_{\tau_2 > s_2}$ term in/out of the conditional expectation conditioned on $\sigma( \mathcal{F}_{T_j}, 1_{\tau_1 > u}, 1_{T_{j-1} < \tau_2 \leq T_j} )$ which I obviously can't do.

Any help on this problem would be greatly appreciated. Also I would be interested to hear an intuitive explanation of why the equality is true if anyone can offer one!