Consider a single-name total return swap (TRS) on some reference asset $S$. For concreteness, suppose the length of the contract is one year with quarterly resets, and the performance of $S$ is exchanged for LIBOR.

Then the TRS value resets at $0$ at each reset date, so for some $t$ in some period that ends at time $T$, the value of the agreement assuming there is no cancellation feature is simply $$V_{t}=\mp(S_{t}-P_{t}^{T})\pm P_{t}^{T}S_{0}L_{0}^{T}T$$ where $P_{t}^{T}$ is the discount factor observed at time $t$ for the period $[t,T]$ and $L_{0}^{T}$ is the period LIBOR (simple) rate set at time $0$ (beginning of the period) for the period $[0,T]$.

However, if at any time in the performance period of the contract either party can terminate the agreement (not settle at the market value $V_{t}$), I have seen it claimed that an accrual valuation method is appropriate and that the contract can be valued for any $t$ as $$V_{t}=\mp(S_{t}-S_{0})\pm S_{0}L_{0}^{T}(T-t).$$

I don't understand the justification for this formula. Since the bi-cancellable feature can be modeled as a long (short) call and short (long) put position (depending on the side of the contract you are on), some sort of put-call parity should be applicable, which leads to the accrual formula above, but I could not get this to quite work out (perhaps someone with more practice with these types of derivations would more easily be able to do it).

Another argument is that the hedging strategy employed by a desk selling this TRS would be to overnight repo $S$, and thus the cumulative borrowing cost upto time $t$ is in some sense $S_{0}L_{0}^{T}(T-t)$, and so if the counterparty wanted to cancel, the desk could just not roll over the repo and sell the asset to cover the position and repo loan. But besides the assumption of constant interest rates, if the term of the contract is set to $T$ at the beginning of the performance period, then ostensibly the desk takes out a $T$-period loan of $S_{0}$, charges the client the interest on this loan, and then liquidates the position at time $T$ to cover the loan and the performance gain or loss on $S$ (i.e., $S_{T}-S_{0}$). This hedging argument is how the first valuation formula is obtained (although it can also be obtained in a straight forward manner using risk-neutral pricing methods).

  • $\begingroup$ The second formula looks wrong, the interest component should be $S_0L^T_0t$ ie. accrued interest since trade inception (instead of $T-t$). $\endgroup$ Commented Nov 29, 2023 at 9:17

3 Answers 3


Any time that a contract is cancellable by either party, it will be cancelled. That's because it is always to one party's advantage to cancel rather than carry on. The exception is that the contract is worth exactly zero, which has effectively zero probability. Therefore, the value of the contract is whatever will get paid out on cancellation. I.e. The accrual value.

  • $\begingroup$ I don't follow this line, "it is always to one party's advantage to cancel rather than carry on". Which party are we referring to, the one ITM or OTM? $\endgroup$ Commented Nov 29, 2023 at 9:51
  • $\begingroup$ Theoretically, the OTM party should cancel , thereby picking up value. If desired, they could reenter a new swap at zero MTM. $\endgroup$
    – dm63
    Commented Nov 29, 2023 at 11:28
  • $\begingroup$ With the demise of Libor , the issue raised by this question has gone away, because the floating rate side of the TRS would now be daily compounded SOFR so that the two valuation methods should match. $\endgroup$
    – dm63
    Commented Nov 29, 2023 at 11:31
  • $\begingroup$ "Theoretically, the OTM party should cancel , thereby picking up value" wouldn't that also depend on the sign of rates? The OTM party has a liability to fund, with positive rates this liability would increase with time and therefore it is optimal to cancel, but with negative rates wouldn't the size of the liability reduce with time (hence waiting would be more optimal)? Although I guess the other side of the coin is that the other party's asset losses present value. I have come up with an alternative backward induction argument that seems to back up the immediate termination behavior. $\endgroup$ Commented Nov 29, 2023 at 11:39

I've been thinking about this too and for me, the answer is different from the accrual formula, but may not answer your question ^^'.

First little point, maybe your pricing formula is, for the libor part, L_0 * t, and not L_0 * (T-t) as when you break, you receive the accrued.

Analysis of the formula : The formula has no forecast term, it's like the call is exercised immediately at the valuation date.

Practical knowledge : From what I know, very little breakable swaps are cancelled the day they start. So this valuation method seems to be wrong, moreover it's not because it's possible to break that the break will occur, if we applied the same method for american option, they would have no time value and just intrinsic value ... quite gross.

Now we know the formula is wrong.

Call criteria proposition : I want to tell a little more by defining a reason to call. There are many reasons, I've chosen a marked to market criteria : if the forward value of the deal gets big for a counterparty, and too small for the other one, then the call occurs and the payoff is (by definition) the wrong formula, plus a 0 marked to market payoff. So it's like you have american down out/up out deal on an underlying that is the forward part of the marked to market. Moreover, that forward part is evanescent : it converges to 0 as the time passes.

Now you just have to diffuse your marked to market, you can use a backward method as you have an american criteria. You may specify the dynamic of your marked to market and the boundaries of each counterparty (unknown parameters ... !). You'll then get your new value.

Special examples :

  1. if the market to market is constant and equal to 0, there'll be no call, the swap is non breakable then valued as a ... VANILLA one !

  2. if 1) and moreover if you have no specified maturity, it'll last forever !

  3. In general cases, the value depends on the expected time it lasts, so from the volatility, drift of the diffusion parameters and the up and down boundaries.

To resume : it's very difficult to get an accurate model as we have to make behaviour hypothesis for the call. But for sure, the accrued style value is wrong. For short maturities the swap may be more vanilla, for long ones he may be model dependent. Thank you very much for your interest and remarks on my reflection.

P.S. : an analogy is when you can break a loan without paying interests, you don't break it as soon as you can earn one dollar you need more, maybe 1000$, 3000$, this has a value.

  • $\begingroup$ Agree with your comment on the interest payment accrual, it should be $S_0L^T_0t$ instead ie. the accrued interest since inception. However the question is about explaining why we use accrual valuation for breakable swaps; the approach should be correct because that is what the market actually uses. $\endgroup$ Commented Nov 29, 2023 at 9:12
  • $\begingroup$ +1 for considering real-world exercise criteria. This sort of thing comes up even for American-exercise options on US equities. Gray-haired practitioners know well that not everyone who "should" exercise does so, leaving money on the table for anyone with a short position who does not get fully assigned. The value of that phenomenon is highly situation-dependent, but can be nontrivial at certain times. $\endgroup$
    – Brian B
    Commented Nov 29, 2023 at 18:06

I have come up with the following approach based on methods used for American options. For that kind of problems, it is useful to assume a discrete termination schedule, apply a backward induction argument and interpret the continuous case as the limit from the discrete case.

The backward induction argument

For simplicity let us assume a single flow between parties $A$ and $B$ settled at $T>0$ and whose funding leg pays a fixed rate $r$. Suppose $A$ is long the stock return and that the deal has a bilateral breakable clause such that both parties can terminate at any date $t_k\in\{t_0,t_1,\dots,t_n\}$ where $t_0=0$ and $t_n=T$ (we write $S_k:=S_{t_k}$). Let $B_t$ be the bank account and $P_t(T)$ the zero-coupon bond with expiry $T$.

At $t_n=T$ the swap has reached expiry and needs to be settled regardless. Let us go back one (termination) date to $t_{n-1}$ then $A$ can choose between (1) receiving right now the amount: \begin{align} v_A(t_{n-1}) &=(S_{n-1}-S_0)-rS_0t_{n-1}\\ &=S_{n-1}-S_0(1+rt_{n-1}), \end{align} (this is the definition of the breakable clause) or (2) waiting until $T$ and cashing in $(S_{T}-S_0)-rS_0T$. The value at $t_{n-1}$ of this second payoff is: \begin{align} v_A(t_{n-1},T) &=E^\mathbb{Q}_{n-1}\left(\frac{B_{n-1}}{B_T}\left((S_{T}-S_0)-rS_0T\right)\right)\\ &=(S_{n-1}-P_{n-1}(T)S_0)-P_{n-1}(T)rS_0T\\[4pt] &=S_{n-1}-P_{n-1}(T)S_0(1+rT) \end{align} Party $A$ wants to maximize value hence the value $\widetilde{V}_A$ of the trade to him (excl. the clause from $B$) is: \begin{align} \widetilde{V}_A(t_{n-1}) &=\max\left\{S_{n-1}-S_0(1+rt_{n-1}), S_{n-1}-P_{n-1}(T)S_0(1+rT)\right\}\\ &=\max\left\{v_A(t_{n-1}),v_A(t_{n-1},T)\right\} \end{align}

If $v_A(t_{n-1})\geq v_A(t_{n-1},T)$ then party $A$ will terminate now. Suppose the contrary is true then $A$ would rather terminate at expiry $T$ because the continuation value is higher than breaking now. Take the point of view from $B$ then we obviously have (again ignoring the other party's clause): \begin{align} \widetilde{V}_B(t_{n-1}) &=\max\left\{S_0(1+rt_{n-1})-S_{n-1},P_{n-1}(T)S_0(1+rT)-S_{n-1}\right\}\\ &=\max\left\{v_B(t_{n-1}),v_B(t_{n-1},T)\right\}\\ &=\max\left\{-v_A(t_{n-1}),-v_A(t_{n-1},T)\right\} \end{align} But by assumption $v_A(t_{n-1})<v_A(t_{n-1},T)$ therefore $v_B(t_{n-1})>v_B(t_{n-1},T)$ hence the optimal decision for $B$ is to break now.

Hence in any case there is one counterparty for which the optimal decision is to terminate at the current time $t_{n-1}$. Knowing this, the actual value $V_A$ to $A$ accounting for both termination clauses is: $$V_A(t_{n-1})=S_{n-1}-S_0(1+rt_{n-1})=v_A(t_{n-1})$$ The opposite is true for $B$.

The argument is then applied backwards. By recursion we reach the same conclusions for every $t_k\geq t$, where each time the counterparty $X$ needs to choose between (1) the current payout $v_X(t_{k})$ and (2) the previous iterated value for time $t_{k+1}$, hence by making the partition $\{t_1,\dots,t_n\}$ infinitesimally fine, the value today of the breakable swap must be: $$V_A(t)=S_t-S_0(1+rt)=v_A(t)$$

The financial intuition

Another way to look at the previous formula is by assuming a diffusion framework for prices and applying Itô's Lemma: $$\text{d}V_A(t)=\text{d}S_t-rS_0\text{d}t$$ You clearly see you are getting the "pure" performance of the stock against a financing cost, that is you have no dependency on the stock forward price, which can be impacted by changes in rate or dividend expectations (e.g. the dividend derivative market might move due to supply-demand dynamics, which affects the stock forward and therefore the value of a non-breakable swap).

Hence the breakable swap is basically equivalent to holding the stock itself except that there are capital advantages i.e. the swap is an unfunded trade and therefore allows for leverage. In particular you can terminate the trade at any point - just as you would do with an outright stock holding. Hence it makes sense that the trade is priced by simply taking today's stock value.

Note that breakable swaps would also allow banks to manage counterparty risk more effectively by being able to terminate the transaction if they become wary of the creditworthiness of the other party (in particularly considering the most common counterparty in these deals are hedge funds).


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