# Question on volatility equation for CMS pricing

In "Full implications of CMS convexity (Cedervall and Piterbarg, 2012)", a.k.a. "CMS: covering all bases (idem)", the authors develop a CMS model equipped with an annuity mapping function which captures normal volatilities $\sigma_i$ and correlations $\rho_{i,j}$ for a set of swap rates $S_1,\dots,S_n$ of tenors $T_1,\dots,T_n$ coming from a swaption smile model. We're interested in a derivative written on the swap rate $S_n$.

Letting $S_i(T)$ be the swap rate of tenor $T_i$ at time $T$, the authors work with a one-factor lognormal model:

$$1+\delta_iS_i(T)=(1+\delta_iS_i(0))e^{\mu_i+\nu_i\sqrt{T}X}$$

where $X\sim\mathcal{N}(0,1)$ and $\delta_i$ is the float leg accrual factor of the swap of tenor $T_i$. Then the article states that:

The parameters $\nu_i$ in [the equation above] can be linked to normal volatilities and correlations by projecting the rates [i.e. computing the expectation of one rate conditional on the other rate]:"

$$\nu_i\triangleq\frac{\color{blue}{\delta_i}}{1+\delta_iS_i(0)}\sigma_i\rho_{n,i}$$

Now, I understand if we assume a two-factor Gaussian model such that:

\begin{align} \text{d}S_n(t)&=\sigma_n\text{d}W_t \\[3pt] \text{d}S_i(t)&=\sigma_i\left(\rho_{n,i}\text{d}W_t+\sqrt{1-\rho_{n,i}^2}\text{d}\tilde{W}_t\right) \end{align}

where $W_t$ and $\tilde{W}_t$ are two independent Brownian Motions, we get:

$$E\left(S_i(T)|W_T\right)=S_i(0)+\sigma_i\rho_{n,i}W_T$$

If we now assume a single risk factor, then the normal volatility of $S_i(T)$ is $\sigma_i\rho_{n,i}\sqrt{T}$ $-$ this is basically a trick to capture correlation information from the swaption smile model $-$ and using the fact that:

$$\sigma_{\text{Normal}}\approx\sigma_{\text{Log Normal}}\times \text{Spot}$$

We set:

$$\nu_i\triangleq\frac{1}{1+\delta_iS_i(0)}\sigma_i\rho_{n,i}$$

But I don't see where the accrual factor comes from. Does anybody understand how it ends up there?

• If the accrual factor tends to zero then the volatility of (1+accrual factor *rate) must tend to zero- this wouldn't be the case if it weren't there I think? – dm63 Jun 14 '18 at 1:50

## 1 Answer

Let $X_i(t) = 1+\delta_i S_i(t)$. Then $\nu_i$ is the log normal volatility for $X_i(t)$, and because $dX_i(t) = \color{blue}{\delta_i} dS_i(t)$ we get $\nu_i X_i(t) dW_t = \color{blue}{\delta_i} \sigma_i \rho_{n,i} dW_t$ and $\nu_i \approx (\color{blue}{\delta_i} \sigma_i \rho_{n,i} ) / X_i(0)$.

• Exactly, I was messing up $1+\delta_iS_i(t)$ and $S_i(t)$. Cheers. – Daneel Olivaw Jun 14 '18 at 8:41