# Vanilla swap payoff

The payoff of a plain vanilla swap with respect to measure $Q$ is : $$V_{\mathrm{swap}}(t) = \beta(t) \sum_{n=0}^{N-1} \tau_n E_t^Q \left( \dfrac{1}{\beta(T_{n+1})} ( L(T_n, T_n, T_{n+1}) - k) \right)$$

How do we get : $$V_{\mathrm{swap}}(t) = \beta(t) \sum_{n=0}^{N-1} \tau_n E_t^Q \left( \dfrac{1}{\beta(T_{n})} ( L(T_n, T_n, T_{n+1}) - k) \, P(T_n, T_{n+1})\right)$$

given

$$P(T_n, T_{n+1}) = E^Q_t \left(\dfrac{1}{\beta(T_n, T_{n+1}}\right)$$

Note that the money-market numeraire is defined in terms of the short-term interest rate as

$$\beta(t) = \exp\left(\int_0^tr(s)ds\right),$$

and the forward price of a zero-coupon bond issued at the future date $T_n$ and maturing at $T_{n+1}$ is the risk-neutral conditional expectation of the discounted payoff:

$$P(T_n, T_{n+1}) = E^Q \left(\left.\exp\left(-\int_{T_n}^{T_{n+1}}r(s)ds\right)\right|\mathcal{F_n}\right)=E^Q \left(\left.\frac{\beta(T_n)}{\beta(T_{n+1})}\right|\mathcal{F_n}\right).$$

Using the tower property of conditional expectation $$E_t^Q \left( \dfrac{1}{\beta(T_{n+1})} ( L(T_n, T_n, T_{n+1}) - k) \right)\\=E_t^Q \left[E^Q\left( \left.\dfrac{1}{\beta(T_{n+1})} ( L(T_n, T_n, T_{n+1}) - k)\right|\mathcal{F}_n\right) \right]\\=E_t^Q \left[E^Q\left( \left.\frac1{\beta(T_n)}\dfrac{\beta(T_n)}{\beta(T_{n+1})} ( L(T_n, T_n, T_{n+1}) - k)\right|\mathcal{F}_n\right) \right]$$

Taking what is known at time $T_n$ out of the inner conditional expectation we get

$$E_t^Q \left( \dfrac{1}{\beta(T_{n+1})} ( L(T_n, T_n, T_{n+1}) - k) \right)\\=E_t^Q \left[\frac1{\beta(T_n)}( L(T_n, T_n, T_{n+1}) - k)E^Q\left( \left.\dfrac{\beta(T_n)}{\beta(T_{n+1})} \right|\mathcal{F}_n\right) \right]\\=E_t^Q \left( \dfrac{1}{\beta(T_{n})} ( L(T_n, T_n, T_{n+1}) - k) \, P(T_n, T_{n+1})\right)$$

• Awesome! Thanks a lot. – ZeroCool Aug 15 '14 at 14:22
• @ZeroCool: You're welcome. I edited the answer with a correction for the money market as the exponential of the integral of the short-rate process. – RRL Aug 15 '14 at 14:30

In case any one is interested I figured out a short answer :

Let $M(T_{n+1}) = L(T_n, T_n, T_{n+1})-k$

Given $M(T{n+1}$ is $F_{T_{n+1}}-measurable$ the payoff at the end of each period becomes :

$$E \left( E \left( \dfrac{M(T_{n+1})}{\beta(T_n) \beta(T_n, T_{n+1})} \; | \; \mathcal{F}_{T_{n+1}}\right) | \; \mathcal{F_t}\right)$$

$$= E \left( \dfrac{M(T_{n+1})}{\beta(T_n)} E \left( \dfrac{1}{\beta(T_n, T_{n+1})} \; | \; \mathcal{F}_{T_{n+1}}\right) | \; \mathcal{F_t}\right)$$

$$= E \left( \dfrac{M(T_{n+1})}{\beta(T_n)} P(T_n, T_{n+1}) | \; \mathcal{F_t}\right)$$

CQD