# 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.

• Hi, what do you mean with wrong results? I guess your Delta is ok, as the Black76 call is linear in the forward swap rate as seen at time t – Gabriele Pompa Jul 30 '15 at 18:42
• When you say Bringo, do you mean D. Brigo, the author and his famous textbook? – Alex C Aug 2 '15 at 12:46
• @GabrielePompa Hi, I added an example. I hope it is clear now. – math Aug 2 '15 at 13:49
• @AlexC Yes, I mean D.Brigo and his famous text book – math Aug 2 '15 at 13:50
• Sorry, but I still do not understand what's the problem with that Delta. Is it because it's greater than one that disturbs you? – Gabriele Pompa Aug 2 '15 at 22:00

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.
• thanks a lot for your explanation and the link. What would be interesting to know why exactly $A_{\alpha,\beta}$ is considered to measure the change. Sure it is a tradable asset, but the connection to the swaption / swap is not fully clear yet (to me). For example, a swap can also be decomposed into FRA's contracts which are tradable asset. I'm new to these things and still lacking of the intuition. Thanks for your help and the pdf. – math Aug 4 '15 at 17:55
• @user8: I made some changes, where I changed $A_{\alpha, \beta}$ to the value of the underlying swap. This is now comparable to equity options. That is, the delta hedge ratio is the derivative with respect to the underlying asset. – Gordon Aug 4 '15 at 18:24
• thanks gordon. I really like the paper! Maybe one last question. In the equity case, i.e. call on a future $F$: $C=\beta(t)E_Q[\frac{1}{\beta(T)}(F-K)^+|\mathcal{F}_t]$. Using Black 76 we see that the forward delta, $\frac{dC}{dF} = \frac{1}{\beta(t)}N(d_1)$. $F$ is the underlying and can be traded. Why do we have there a discounting factor? It seems to me, one can't really make a linke between equity and the FI case. – math Aug 5 '15 at 11:45