# Inequality involving Co-Terminal/Co-Initial American vs Bermudan Swaptions

Let us consider a payment schedule $$\mathcal{P}:=\{t_1,\dots,t_n\}$$ which has a corresponding fixing schedule $$\mathcal{F}:=\{t_0,\dots,t_{n-1}\}$$. We have a series of co-terminal and co-initial swaps with coinciding float and fixed payment dates, accrual conventions and notionals $$-$$ the specific float rate does not matter. Hence the value of these swaps at time $$t$$ are in the case of a payer: $$V_t(t_i,t_j)=\sum_{k=i}^j\delta_k(R_t(t_{k-1},t_k)-C)P_t(t_k)$$ where $$C$$ is the swap fixed rate, $$R_t(t_{k-1},t_k)$$ the forward value of the float rate fixing at $$t_{k-1}$$ and paid at $$t_k$$, and $$P_t(t_k)$$ the discount factor, with $$\delta_k:=t_k-t_{k-1}$$ the accrual fraction (we have dispensed with notional and assumed it is 1).

Let us introduce the notation $$\mathcal{F}_{i,j}:=\{t_i,\dots,t_j\}$$. Let $$B_t(t_i,t_n)$$ be the value at $$t$$ of a vanilla Bermudan swaption which gives the right to enter on one of the fixing dates $$t_{i-1},\dots,t_{n-1}$$ into a swap starting (payments) at $$t_i,\dots,t_n$$ respectively and ending at $$t_n$$, i.e. the exercise set of the Bermudan is $$\mathcal{F}_{i-1,n-1}$$. Let $$A_t(t_i,t_n)$$ be the corresponding vanilla American swaption value which can be exercised on any date within $$[t_{i-1},t_n]$$(1). We use equivalent notation for co-initial swaptions: $$B_t(t_1,t_i)$$ which can be entered on any dates from $$\mathcal{F}_{0,i-1}$$; and $$A_t(t_1,t_i)$$ where in this case the American swaption can be exercised within $$[t_1,t_i]$$(1).

Under which circumstances would we have:

\begin{align} &1)& A_t(t_{i-1},t_n)-A_t(t_i,t_n) & \ \pmb{>}\ %\quad \begin{array}\\>\\\geq\\=\end{array} \quad B_t(t_{i-1},t_n)-B_t(t_i,t_n)\ ? \\[6pt] &2)& A_t(t_1,t_{i+1})-A_t(t_1,t_i) & \ \pmb{>}\ %\quad \begin{array}\\>\\\geq\\=\end{array} \quad B_t(t_1,t_{i+1})-B_t(t_1,t_i)\ ? \end{align} namely the change in value from including an additional swap flow is strictly greater for American than Bermudan swaptions?

(1) If the American swaption is exercised at $$t\in[t_i,t_{i+1})$$, we assume the fixing from $$t_i$$ accrues linearly from $$t$$ to $$t_{i+1}$$ and is paid at $$t_{i+1}$$. For mathematical purposes this ensures the swap value can be represented as a continuous function.
• Should you say "a Bermudan swaption on a swap allowed to start on one of $\{t_i,t_{t+1},\cdots,t_n\}$" instead of "a Bermudan swaption on a swap starting on $t_i$" which means the swaption has to start on and only on $t_i$? – Hans Jun 14 '20 at 20:34
• @Hans I have clarified notation and setting. – Daneel Olivaw Jun 15 '20 at 13:58
• The description for the American swaption is still unclear. When can the option be excercised and once exercised when can the swap start? – Hans Jun 15 '20 at 18:20
• @Hans strictly speaking, there is only one swap, given past flows do not matter for valuation: the first accrual period is $[t_0,t_1]$, paid on $t_1$; the second accrual period is $[t_1,t_2]$, paid at $t_2$; etc. The last accrual period is $[t_{n-1},t_n]$, paid at $t_n$. The Bermudans allow to enter this swap on any accrual date: $t_0$, $t_1$,..., $t_{n-1}$. – Daneel Olivaw Jun 17 '20 at 9:24
• @Hans The Americans allow to enter it on any time point: for example, for $A_t(t_i,t_n)$, you can enter at $t_{ex}\in[t_{i-1},t_i]$ in which case you receive the accrual coupon for the truncated period $[t_{ex},t_i]$ based on the fixing at $t_{i-1}$, then proceed as normal. – Daneel Olivaw Jun 17 '20 at 9:25

I have the following idea: consider the limit of low volatility. Then, exercise strategy is known simply by comparing coupon with remaining swap rate. If the addition of the additional period at the start creates an optimal exercise at the first possible date for both $$A$$ and $$B$$, then $$A=B$$ after the addition, whereas it is possible that $$A>B$$ before the addition.
For example, assume for simplicity the single period swap from $$t_1$$ to $$t_2$$, where $$t_2-t_1$$ = 1 year. Let us take $$C=2$$, and overnight forward interest rates are constant at 0% for $$[t_1, (t_1+t_2)/2]$$ and 3% for $$[ (t_1+t_2)/2,t_2]$$. Then $$B$$ has only one possible exercise date, which is out of the money (coupon = 2, whereas remaining swap rate=1.5= (0+3)/2 ). So $$B=0$$. However $$A$$ is non - zero, because the optimal exercise strategy is to wait till $$(t_1+t_2)/2$$ then exercise, delivering value equal to (3-2)/2= 1/2. Now, add the period $$[t_0, t_1]$$ to the front, and let the overnight interest rate within $$[t_0, t_1]$$ be equal to 5%. Then it is easy to show that optimal exercise strategy for both A and B is to exercise at $$t_0$$. Hence $$A=B$$ following the addition, violating the proposed inequality.