Hull-White Swaption Formula¶
Swaption Payoff and Change of Numeraire¶
Swap Rate S_mn is a Q^mn-Martingale¶
Since \(S_{mn}(t)=\sum_{k=m+1}^n \omega_k(t)l_k(t)=\frac{P(t,T_{m})-P(t,T_n)}{A_{mn}(t)}\) and since \(P(t,T_{m})\) and \(P(t,T_{n})\) are prices of tradable assets,
Black Type Formula¶
Assume that the swap rate follows a lognormal distribution:
Then, we can use the Black–Scholes computation with interest rate 0:
where
Hull-White Swaption Formula¶
where
Crucial Expression¶
Jamshidian Trick¶
The ZCB price \(P(T_m,T_k)=e^{A(T_k-T_m)+B(T_k-T_m)r(T_m)}\) at the swaption maturity \(T_m\) is a function of the short rate \(r(T_m)\) at \(T_m\). Actually, the functions \(r(T_m)\rightarrow P(T_m,T_k)\) are strictly decreasing. So, there exists a unique \(r^*\) such that
Using this, the swaption payoff can be decomposed into a sum of ZCB put options:
Proof
Since all \(P(T_m,T_k)\) are decreasing functions of \(r(T_m)\), the sign of each term \(K_k - P(T_m,T_k)\) changes at the same \(r^*\). Therefore the max of the sum equals the sum of the maxes:
QuantPie Derivation: Hull-White Swaption Formula¶
Change of Numeraire¶
Crucial Expression¶
where
Hull-White ZCB Price¶
Jamshidian Trick¶
The ZCB price \(P(T_m,T_k)=e^{A(T_k-T_m)+B(T_k-T_m)r(T_m)}\) at the swaption maturity \(T_m\) is a function of the short rate \(r(T_m)\) at \(T_m\). Actually, the functions \(r(T_m)\rightarrow P(T_m,T_k)\) are strictly decreasing. So, there is an \(r^*\) such that
Then
Proof¶
where
Note that
is a European put option price on a zero-coupon bond with option maturity \(T_m\), bond maturity \(T_k\), and strike \(K_k\). So,
Exercises¶
Exercise 1. Explain why Jamshidian's trick works: show that since each \(P(T_m, T_k) = e^{A(T_k - T_m) + B(T_k - T_m)r(T_m)}\) is a strictly decreasing function of \(r(T_m)\), the max of the sum equals the sum of the maxes when all terms change sign at the same critical rate \(r^*\).
Solution to Exercise 1
In the Hull-White model, each ZCB price at time \(T_m\) has the form:
Since \(B(\tau) = \frac{e^{-\lambda\tau} - 1}{\lambda} < 0\) for \(\tau > 0\), the exponent \(A + B\,r\) is a decreasing linear function of \(r(T_m)\). Therefore \(P(T_m, T_k)\) is a strictly decreasing function of \(r(T_m)\) for every \(k\).
The swaption payoff is \(\max\!\left(1 - \sum_k c_k P(T_m, T_k),\, 0\right)\). Since each \(P(T_m, T_k)\) is decreasing in \(r(T_m)\), the sum \(\sum_k c_k P(T_m, T_k)\) is also strictly decreasing in \(r(T_m)\) (with \(c_k > 0\)). Therefore \(1 - \sum_k c_k P(T_m, T_k)\) is strictly increasing in \(r(T_m)\).
There exists a unique \(r^*\) where \(\sum_k c_k P(T_m, T_k; r^*) = 1\), which is the boundary between exercise (\(r < r^*\) for a payer swaption, since low rates make the payoff positive) and no exercise.
At this critical point, we define \(K_k = P(T_m, T_k; r^*) = e^{A(T_k-T_m)+B(T_k-T_m)r^*}\), and by construction \(\sum_k c_k K_k = 1\).
Since all terms \(K_k - P(T_m, T_k)\) change sign simultaneously at \(r^* = r(T_m)\) (because each \(P(T_m,T_k)\) is monotonically decreasing), the region where \(1 - \sum c_k P > 0\) is exactly the same as the region where every \(K_k - P(T_m,T_k) > 0\). Therefore:
This decomposition converts the max of a sum into a sum of maxes, each of which is a standard ZCB put option.
Exercise 2. For a payer swaption with \(T_m = 5\), \(T_n = 10\), annual payments, \(K = 0.04\), and \(N = 1{,}000{,}000\), write out the coefficients \(c_k\) for \(k = 6, 7, 8, 9, 10\). Verify that \(c_{10} = K\tau_{10} + 1 = 1.04\).
Solution to Exercise 2
For a payer swaption with \(K = 0.04\), \(\tau_k = 1\) (annual payments), and \(n = 10\):
Verification of \(c_{10}\): The coefficient \(c_{10} = 1.04\) accounts for both the final fixed coupon payment (\(K\tau_{10} = 0.04\)) and the principal repayment (1) at the last payment date. This arises because the swap's floating leg at the last date pays \(l_{10}\tau_{10}\) plus returns the notional, while the fixed leg pays \(K\tau_{10}\) plus the notional. The "crucial expression" absorbs the notional exchange into \(c_n\).
Summary:
| \(k\) | \(c_k\) |
|---|---|
| 6 | 0.04 |
| 7 | 0.04 |
| 8 | 0.04 |
| 9 | 0.04 |
| 10 | 1.04 |
Exercise 3. The critical rate \(r^*\) satisfies \(\sum_{k=m+1}^n c_k e^{A(T_k - T_m) + B(T_k - T_m)r^*} = 1\). Describe a numerical method (e.g., Newton-Raphson) for solving this equation. Why is the solution unique?
Solution to Exercise 3
The equation \(\sum_{k=m+1}^n c_k e^{A(T_k - T_m) + B(T_k - T_m)r^*} = 1\) must be solved for \(r^*\).
Define \(f(r) = \sum_{k=m+1}^n c_k e^{A(T_k - T_m) + B(T_k - T_m)r} - 1\).
Newton-Raphson iteration:
where
A good initial guess is the par swap rate implied by the forward curve at \(T_m\).
Uniqueness: The function \(f(r)\) is strictly monotone. Since \(B(\tau) < 0\) for \(\tau > 0\) and \(c_k > 0\), each term \(c_k e^{A+Br}\) is strictly decreasing in \(r\), so \(f(r)\) is strictly decreasing. Furthermore, \(f(r) \to +\infty\) as \(r \to -\infty\) and \(f(r) \to -1\) as \(r \to +\infty\). By the intermediate value theorem and strict monotonicity, there is exactly one root.
Newton-Raphson converges rapidly (typically 3-5 iterations to machine precision) due to the smoothness of \(f\).
Exercise 4. Compare the Black swaption formula (assuming lognormal swap rates) with the Hull-White swaption formula (sum of ZCB puts). Under what market conditions might the two formulas produce significantly different prices?
Solution to Exercise 4
Black's swaption formula assumes the swap rate \(S_{mn}(t)\) is lognormal under the annuity measure \(\mathbb{Q}^{mn}\), giving the standard Black-Scholes-type expression. This formula treats the swaption as a single option on the swap rate.
Hull-White swaption formula decomposes the swaption into a portfolio of ZCB puts via Jamshidian's trick. The short rate is Gaussian, and each ZCB put has a closed-form solution.
The two formulas produce different prices when:
-
Far from ATM strikes: The lognormal and Gaussian-exponential distributions differ in their tails. Black's formula implies a flat smile (constant implied vol across strikes), while the Hull-White model generates a skew.
-
High vol-of-vol or long maturities: For long-dated swaptions, the cumulative effect of the different distributional assumptions becomes more pronounced. The mean reversion in Hull-White reduces effective volatility for long maturities, while Black's formula applies a constant percentage volatility.
-
Strongly mean-reverting environments: When \(\lambda\) is large, the Hull-White model dampens rate movements substantially, producing lower prices than Black's formula calibrated to short-term ATM vols.
At ATM, the two formulas typically agree closely because the distributional differences are minimized near the center of the distribution.
Exercise 5. Show that the swap rate \(S_{mn}(t)\) is a martingale under the annuity measure \(\mathbb{Q}^{mn}\). What is the numeraire for this measure, and why is this martingale property useful for Black's swaption formula?
Solution to Exercise 5
The annuity (or present value of a basis point, PVBP) is:
The swap rate is defined as:
The numeraire for the annuity measure \(\mathbb{Q}^{mn}\) is \(A_{mn}(t)\). To show \(S_{mn}\) is a martingale, note that the value of the payer swap at time \(t\) is:
Since the swap value divided by the numeraire \(A_{mn}(t)\) must be a martingale under \(\mathbb{Q}^{mn}\) (by the general change-of-numeraire theorem):
is a \(\mathbb{Q}^{mn}\)-martingale. Since \(K\) is a constant, \(S_{mn}(t)\) itself must be a \(\mathbb{Q}^{mn}\)-martingale:
This martingale property is essential for Black's swaption formula because it allows us to model \(dS_{mn} = \sigma_{mn} S_{mn}\,dW^{mn}\) under the annuity measure, apply Black-Scholes pricing with zero interest rate (the numeraire absorbs the discounting), and obtain the closed-form swaption price.
Exercise 6. The "crucial expression" rewrites the swaption payoff as \(1 - \sum_{k=m+1}^n c_k P(T_m, T_k)\). Derive this step by step from \(\sum_{k=m+1}^n \tau_k P(T_m, T_k)(l_k(T_m) - K)\), using the definition of the forward rate \(l_k\).
Solution to Exercise 6
Starting from the swaption payoff:
Using \(l_k(T_m) = \frac{1}{\tau_k}\!\left(\frac{P(T_m, T_{k-1})}{P(T_m, T_k)} - 1\right)\):
The first two terms telescope:
So the expression becomes:
Now define \(c_k = K\tau_k\) for \(k \neq n\) and \(c_n = K\tau_n + 1\). Then:
Therefore:
This is the "crucial expression" used in Jamshidian's trick.
Exercise 7. In Jamshidian's trick, each ZCB put has its own strike \(K_k = A(T_m, T_k)e^{-B(T_m, T_k)r^*}\). Explain why these strikes depend on \(r^*\) and how an error in the root-finding for \(r^*\) propagates to the swaption price.
Solution to Exercise 7
The strikes \(K_k = e^{A(T_k - T_m) + B(T_k - T_m)r^*}\) depend on \(r^*\) because they represent the ZCB prices evaluated at the critical short rate. Each \(K_k\) is the bond price \(P(T_m, T_k)\) when \(r(T_m) = r^*\), which is the rate at which the holder is indifferent between exercising and not exercising the swaption.
Why they depend on \(r^*\): The decomposition into ZCB puts requires that all put options switch from in-the-money to out-of-the-money at the same short rate \(r^*\). The strikes are chosen precisely to make this happen: \(K_k = P(T_m, T_k; r^*)\) ensures that \(K_k - P(T_m, T_k) > 0\) if and only if \(r(T_m) > r^*\).
Error propagation: If \(r^*\) is computed with error \(\delta r\), then:
Since each \(K_k\) enters the ZCB put pricing formula, an error in \(r^*\) shifts all strikes simultaneously. The swaption price error is approximately:
The sensitivity \(\frac{\partial V_p^{\text{ZCB}}}{\partial K}\) is largest for near-ATM puts. Because \(B(\tau) < 0\) and grows in magnitude with \(\tau\), longer-dated ZCB puts are more affected. In practice, Newton-Raphson converges to machine precision in a few iterations, so the error from root-finding is negligible compared to model error.