# CDS protection/contingent leg pricing, taking expectation of interest and hazard rates

In the screenshot above, I am having trouble understanding the maths between equation 13 and equation 14.

Notation:

• $$N$$ = notational payment, e.g., £100
• $$RR$$ = recovery rate, the percentage of the $$N$$ recovered upon default, e.g., you get back 40%
• $$\tau$$ = time of default
• $$t_v$$ = valuation date
• $$T$$ = maturity date
• $$\mathbb{I}_A$$ = indicator function for event $$A$$
• $$r(s)$$ = instantaneous short rate at time $$s$$
• $$P(t)$$ = discount factor from time $$t > 0 =$$start date
• $$Q(t)$$ = survival probability at time $$t$$

What I have tried:

$$\mathbb{E}\left[e^{-\int_{t_v}^{T}r(s)ds}\mathbb{I}_{\tau

From here, I cannot see how equation 14 is derived.

After some more trying, I think I have it. $$\label{eq1} \begin{split} N(1-RR)\ \mathbb{E}\left[ e^{-\int_{t_v}^{\tau}r(s)ds} \mathbb{I}_{\tau<T} \right] & = N(1-RR)\ \mathbb{E}\left[ \frac{P(\tau)}{P(t_v)} \mathbb{I}_{\tau<T} \right] \\ & = \frac{N(1-RR)}{P(t_v)}\ \mathbb{E}\left[ P(\tau) \mathbb{I}_{\tau<T} \right] \\ & = \frac{N(1-RR)}{P(t_v)}\ \int_{-\infty}^{\infty} -\frac{dQ(s)}{ds}P(s) \mathbb{I}_{s<T} ds \\ & = -\frac{N(1-RR)}{P(t_v)}\ \int_{0}^{T} P(s) \frac{dQ(s)}{ds} ds \\ & = -\frac{N(1-RR)}{P(t_v)}\ \int_{0}^{T} P(s)\ dQ(s) \end{split}$$
• Ciao! why a minus appears in the third line? (in the same line I think you should put $s$ instad of $\tau$ in the characteristic function). Dec 14, 2017 at 14:22