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The Pricing and Risk Management of Credit Default Swaps, with a Focus on the ISDA Model Screenshot: Pricing protection leg of a CDS, by OpenGamma

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:

$$\mathbb{E}\left[e^{-\int_{t_v}^{T}r(s)ds}\mathbb{I}_{\tau<T}\right] = \int_{-\infty}^{\infty} \tau e^{-\int_{t_v}^{T}r(s)ds}\mathbb{I}_{\tau<T} d\tau = \int_{0}^{T} \tau \frac{P(\tau)}{P(t_v)}d\tau$$

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

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After some more trying, I think I have it. \begin{equation} \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} \end{equation}

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  • $\begingroup$ 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). $\endgroup$ – clarkmaio Dec 14 '17 at 14:22
  • $\begingroup$ The minus appears there because derivative of the survival function is the negative of the density. Thank you for pointing out the mistake in the subscript. $\endgroup$ – Vivek Patel Jan 4 '18 at 12:57

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