Swaption Delta: Hedge the Underlying Swap Value, Not the Rate Alone

# Greeks of a swaption using Brigo


# Greeks of a swaption using Brigo












I struggeling with calculating the delta of a swaption. In the interest rate case I usually mess around with the multiple cash flows over time so that the discounting is more complex than in the equity case.

Let me first introduce some notation. We denote with $D(0,T)$ the discounting factor with maturity $T$, $P(0,T)$ the price of a zero coupon bond with maturity $T$ and let $Q$ denote the risk neutral measure.

By simple risk neutral valuation we know:

$$D(0,0)V_0 = V_0 = E_Q[V_TD(0,T)|\mathcal{F}_t]$$

No we are interested in a swaption, where we expiry of the option is at $T_\alpha$ and the underlying swap has a tenor $T_\beta$. The discounted value of the swpation can be writen as

$$D(t,T_\alpha)(S_{\alpha,\beta}(T_\alpha)-K)^+\sum_{i=\alpha + 1}^\beta\tau_iP(T_\alpha,T_i)$$

where $\tau_i$ is the daycount convention between $T_{i-1}$ and $T_i$.

Now regarding valution using the above two equations:

$$ V_0 = E_Q[D(0,T_\alpha)(S_{\alpha,\beta}(T_\alpha)-K)^+\sum_{i=\alpha + 1}^\beta\tau_iP(T_\alpha,T_i)|\mathcal{F}_0]$$

using a smart change of numeraire, the swap measuer $S$, i.e. the numeraire introduced by $\sum_{i=\alpha + 1}^\beta\tau_iP(t,T_i)$ yield

$$ V_0 = E_Q[D(0,T_\alpha)(S_{\alpha,\beta}(T_\alpha)-K)^+\sum_{i=\alpha + 1}^\beta\tau_iP(T_\alpha,T_i)|\mathcal{F}_0]=\sum_{i=\alpha + 1}^\beta\tau_iP(0,T_i)E_S[(S_{\alpha,\beta}(T_{\alpha})-K)^+|\mathcal{F}_0]$$

We know that under the measure $S$, the forward swap rate $S_{\alpha,\beta}(t)$ is a martingale. For the price we could now simple apply Black formula, if we assume that the forward swap rate is normally distributed.

Now my question, if I would apply the normal calculation for the delta I would get $\sum_{i=\alpha + 1}^\beta\tau_iP(0,T_i) N(d_1)$, where $d_1$ is the expression from Black 76 formula. However this term $\sum_{i=\alpha + 1}^\beta\tau_iP(0,T_i)$ annoys me. I get completely wrong results. If I used just $N(d_1)$ I would get reasonable result. So my question, is the delta given by $N(d_1)$ for a swaption as well? If so, where is my mistake?

For simplicity I add an example with concrete numbers.

example We take a swaption with expiry $5$ years and underlying tenor of $5$ years. $S_{\alpha,\beta}(0) = 0.0271$, $\sigma = 0.34$, $r = 0.011$, $T=5$, $K = 0.028$ and annuity $A=4.92$. Using Black 76 we should get for $\Delta$:

$$\Delta = A\cdot N(d_1),$$ where

$$d_1 = \frac{\log{\frac{S_{\alpha,\beta}(0)}{K}}+\frac{\sigma^2\cdot T}{2}}{\sigma\cdot\sqrt{T}}$$

Here I get the values $N(d_1) = 0.332296$ and $\Delta = 1.634896$, which doesn't make sense.

## Answer by Gordon (score 7, accepted)

https://quant.stackexchange.com/a/19172

As the swap rate is not tradable, the delta hedge ratio with respect to the spot swap rate is not really useful. However, note that \begin{align*} V_0 &= \sum_{i=\alpha+1}^{\beta}\tau_i P(0, T_i)\big[S_{\alpha, \beta}(0)N(d_1) - k N(d_2) \big]\\ &= \sum_{i=\alpha+1}^{\beta}\tau_i P(0, T_i) S_{\alpha, \beta}(0)N(d_1) - N(d_2) k \sum_{i=\alpha+1}^{\beta}\tau_i P(0, T_i)\\ &= \Big[P(0, T_{\alpha}) - P(0, T_{\beta})\Big]N(d_1) - N(d_2) k \sum_{i=\alpha+1}^{\beta}\tau_i P(0, T_i)\\ &= \bigg[P(0, T_{\alpha}) - P(0, T_{\beta})- k \sum_{i=\alpha+1}^{\beta}\tau_i P(0, T_i)\bigg]N(d_1) \\ & \qquad\qquad\qquad\qquad + \Big[N(d_1)- N(d_2)\Big] k \sum_{i=\alpha+1}^{\beta}\tau_i P(0, T_i). \end{align*} Here, \begin{align*} A_{\alpha, \beta} &\triangleq P(0, T_{\alpha}) - P(0, T_{\beta}) - k \sum_{i=\alpha+1}^{\beta}\tau_i P(0, T_i)\\ &= \sum_{i=\alpha+1}^{\beta}\tau_i P(0, T_i)\big[S_{\alpha, \beta}(0) -k \big] \end{align*} is the value of the underlying swap, and \begin{align*} k \sum_{i=\alpha+1}^{\beta}\tau_i P(0, T_i) \end{align*} is the value of a portfolio of zero-coupon bonds. We define the delta hedge ratio of the swaption to the derivative of the swaption value with respect to the swap value $A_{\alpha, \beta}$. Note that \begin{align*} S_{\alpha, \beta}(0) = \frac{A_{\alpha, \beta}}{\sum_{i=\alpha+1}^{\beta}\tau_i P(0, T_i)} + k. \end{align*} Then \begin{align*} \frac{\partial V_0}{\partial A_{\alpha, \beta}} &= \frac{\partial V_0}{\partial S_{\alpha, \beta}(0)} \frac{\partial S_{\alpha, \beta}(0)} {\partial A_{\alpha, \beta}}\\ &= N(d_1), \end{align*} which is, for hedging purpose, the quantity of the underlying swap to buy.

See alos the discussion in http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.28.7064&rep=rep1&type=pdf.