Pricing a Swaption by Integrating the Swap-Rate Payoff

# Integration to calculate expected value of swap rate


# Integration to calculate expected value of swap rate












In Hagan's paper on valuing CMS swaps (Convexity Conundrums: Pricing CMS Swaps, Caps, and Floors), there is:

> So the swap rate must also be a Martingale, and $$E \big[ R_s(\tau) \big| \mathcal{F}_0 \big]=R_s(0) = R_s^0$$ To complete the pricing, one now has to invoke a mathematical model (Black’s model, Heston’s model, the SABR model, . . . ) for how $R_s(\tau)$ is distributed around its mean value $R_s^0$. In Black’s model, for example, the swap rate is distributed according to $$R_s(\tau) = R_s(0)e^{\sigma x\sqrt{\tau}-\frac{1}{2}\sigma^2\tau}$$ where x is a normal variable with mean zero and unit variance. One completes the pricing by integrating to calculate the expected value.

And he stopped here. So I am trying to "complete the pricing by integrating", but I am not sure of what he actually meant. I think he just wants to get the swap rate $R_s(0)$, but I am not sure.

This integration would lead to:

$$ R_s(0) = E \big[ R_s(\tau) \big| \mathcal{F}_0 \big] = \int_{-\infty}^{+\infty} R_s(\tau) \frac{-\frac{x^2}{2}}{\sqrt{2\pi}}dx = \int_{-\infty}^{+\infty} R_s(0)e^{\sigma\sqrt{\tau}x-\frac{1}{2}\sigma^2\tau} \frac{-\frac{x^2}{2}}{\sqrt{2\pi}}dx $$

This does not make sens, since $R_s(0)$ is on both side of the equation. Maybe I am not looking at it the right way, and he meant something else. Or maybe my integration is wrong.

I am a bit lost here. Any help would be appreciated to understand how to "complete this pricing".

## Answer by Gordon (score 2, accepted)

https://quant.stackexchange.com/a/25632

He meant to compute the swaption price given by (2.4a), that is, \begin{align*} V_{opt} = L_0 E\left((R_s(\tau)-R_{fix})^+ \mid \mathcal{F}_0 \right). \end{align*} Under the swap measure (i.e., with $L_t$ as the numeraire), the swap rate process $\{R_s(t), t \ge 0\}$ is a martingale, and is assumed to be of the form \begin{align*} R_s(\tau) = R_s^0 e^{\sigma X \sqrt{\tau} -\frac{1}{2}\sigma^2\tau }, \end{align*} where $X$ is a standard normal random variable. Then \begin{align*} V_{opt} &= L_0\int_{-\infty}^{\infty}\left(R_s^0 e^{\sigma x \sqrt{\tau} -\frac{1}{2}\sigma^2\tau }-R_{fix} \right)^+ \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx\\ &=L_0\big[R_s^0 N(d_1) - R_{fix} N(d_2) \big], \end{align*} where \begin{align*} d_1 = \frac{\ln \frac{R_s^0}{R_{fix}}+\frac{1}{2}\sigma^2 \tau}{\sigma \sqrt{\tau}}, \end{align*} and \begin{align*} d_2 = d_1 -\sigma \sqrt{\tau}. \end{align*}

## Answer by Rerito (score 1)

https://quant.stackexchange.com/a/25584

Let's recall the definition of a Martingale first: it is a stochastic process $X(t)$ that has the following property: let $0 \leq t < T$ two real numbers. Let $\mathcal{F}_t$ be a filtration for the process $X$ at time $t$. We have then:

$$ \mathbb{E}[X(T)|\mathcal{F}_t] = X(t) $$

Now, if you use Black's model, you describe your asset price using a geometric Brownian motion. This is not a martingale. Let's study it a bit more:

$$ S(t) = S_0 \exp{(\sigma \sqrt{t} W(t) - \frac{1}{2}\sigma^2 t)} $$

If you consider the logarithm of this process, you'll establish that it follows a normal law of mean $\ln S_0 - \frac{1}{2}\sigma^2t$ and variance $\sigma^2 t$. By going a bit more into calculus you can establish it has the Markov property, but this is not a martingale.