Interest rate swaps: a comparison of compounded daily versus discrete reference rates

Abstract

This paper studies the hedging effectiveness of interest rate swaps using different reference rates for eliminating interest rate risk from floating rate loans. Two reference rates are studied. The first rate’s maturity, $\Delta$, matches the payment interval of floating rate loans. The second has an incompatible maturity $\Delta /N$. The prime examples are LIBOR and SOFR, respectively. We show that the $\Delta$-based swap provides a good static hedge, but the $\Delta /N$-based swap does not. Although dynamic hedging with the $\Delta /N$-based swap is possible under some conditions, it both introduces model risk and increases transaction costs, making it a less practical alternative.

Appedix: Rrelevant proofs and calculations

1.1 Proof of Theorem 1

Proof

First, we compute the arbitrage-free value of a risky floating rate loan, i.e., claim (2) in theorem 1. For use in the proof, define

$$\begin{aligned} x(t,k\Delta ):=\mathbb {E}_{t}\left[ \frac{\mathbbm {1}_{\tau >k\Delta }}{B_{k\Delta }}\right] B_{t} \end{aligned}$$

to be the present value of a dollar paid at time $k\Delta$, but only if the firm has not defaulted before that date. Here, $\mathbb {E}_{t}[\cdot ]:=\mathbb {E}^{\mathbb {Q}}[\cdot |\mathcal {F}_{t}]$. The calculation consists of three parts.

(i) The payment at time $\Delta$

The payment at time $\Delta$ is $\left( \frac{1}{D(0,\Delta )}-1\right) \mathbbm {1}_{\tau >\Delta }+ \delta _{\tau }\frac{1}{D(0,\Delta )}\mathbbm {1}_{\tau \le \Delta }$.

This is the floating rate payment if no default, and a recovery on the floating rate payment plus notional if default occurs prior to time $\Delta$.

The value at time $t\in (0,\Delta )$ is

$$\begin{aligned} L_{t}^{\Delta }= & {} \mathbb {E}_{t}\left[ \frac{\left( \frac{1}{D(0,\Delta )}-1\right) \mathbbm {1}_{\tau>\Delta }+\delta _{\tau }\left( \frac{1}{D(0,\Delta )}\right) \mathbbm {1}_{\tau \le \Delta }}{B_{\Delta }}\right] B_{t}\\= & {} \mathbb {E}_{t}\left[ \frac{\frac{\mathbbm {1}_{\tau>\Delta }+\delta _{\tau } \mathbbm {1}_{\tau \le \Delta }}{D(0,\Delta )}-\mathbbm {1}_{\tau>\Delta }}{B_{\Delta }}\right] B_{t}\\= & {} \frac{1}{D(0,\Delta )}\mathbb {E}_{t}\left[ \frac{\mathbbm {1}_{\tau>\Delta }+\delta _{\tau } \mathbbm {1}_{\tau \le \Delta }}{B_{\Delta }}\right] B_{t}-\mathbb {E}_{t} \left[ \frac{\mathbbm {1}_{\tau >\Delta }}{B_{\Delta }}\right] B_{t}\\= & {} \frac{D(t,\Delta )}{D(0,\Delta )}-x(t,\Delta ) \end{aligned}$$

(ii) The payment at time $k\Delta$ for $k\in \{2,...,K\}$

The payment at time $k\Delta$ is

$$\begin{aligned} \mathbbm {1}_{\tau>(k-1)\Delta }\left[ \left( \frac{1}{D((k-1)\Delta ,k\Delta )}-1\right) \mathbbm {1}_{\tau >k\Delta }+\delta _{\tau }\frac{1}{D((k-1)\Delta ,k\Delta )}\mathbbm {1}_{\tau \le k\Delta }\right] \end{aligned}$$

The value at time $t\in (0,\Delta )$ is

$$\begin{aligned} L_{t}^{k\Delta }= & {} \mathbb {E}_{t}\left[ \frac{\mathbbm {1}_{\tau>k\Delta }+\delta _{\tau }\mathbbm {1}_{\tau \le k\Delta }}{D((k-1)\Delta ,k\Delta )}\frac{\mathbbm {1}_{\tau>(k-1)\Delta }}{B_{k\Delta }}-\frac{\mathbbm {1}_{\tau>k\Delta }}{B_{k\Delta }}\right] B_{t}\\= & {} \mathbb {E}_{t}\left[ \frac{\mathbbm {1}_{\tau>k\Delta }+\delta _{\tau }\mathbbm {1}_{\tau \le k\Delta }}{D((k-1)\Delta ,k\Delta )}\frac{\mathbbm {1}_{\tau>(k-1)\Delta }}{B_{k\Delta }}\right] B_{t}-\mathbb {E}_{t}\left[ \frac{\mathbbm {1}_{\tau>k\Delta }}{B_{k\Delta }}\right] B_{t}\\= & {} \mathbb {E}_{t}\left[ \frac{\mathbbm {1}_{\tau>(k-1)\Delta }}{D((k-1)\Delta ,k\Delta )}\mathbb {E}_{(k-1)\Delta }\left[ \frac{\mathbbm {1}_{\tau>k\Delta }+\delta _{\tau }\mathbbm {1}_{\tau \le k\Delta }}{B_{k\Delta }}\right] \right] B_{t}-x(t,k\Delta )\\= & {} \mathbb {E}_{t}\left[ \frac{\mathbbm {1}_{\tau>(k-1)\Delta }}{D((k-1)\Delta ,k\Delta )}\frac{D((k-1)\Delta ,k\Delta )}{B_{(k-1)\Delta }}\right] B_{t}-x(t,k\Delta )\\= & {} \mathbb {E}_{t}\left[ \frac{\mathbbm {1}_{\tau >(k-1)\Delta }}{B_{(k-1)\Delta }}\right] B_{t}-x(t,k\Delta )\\= & {} x(t,(k-1)\Delta )-x(t,k\Delta ). \end{aligned}$$

In the third equality, we used the law of iterated expectations and the fact that $\mathcal {F}_{(k-1)\Delta }\supseteq \mathcal {F}_{t}$ since $t<\Delta <(k-1)\Delta$.

(iii) The notional payment at time T

The notional of 1 dollar at time T pays $1_{\tau >K\Delta }$ dollars, whose value at time $t\in (0,\Delta )$ is

$$\begin{aligned} N_{t}=E_{t}\left[ \frac{1_{\tau >K\Delta }}{B_{K\Delta }}\right] B_{t}=x(t,T) \end{aligned}$$

Thus, the value of a risking floating rate loan at time $t\in (0,\Delta )$ is

$$\begin{aligned} L_{t}&=L_{t}^{\Delta }+\sum _{k=2}^{K}L_{t}^{\Delta k}+N_{t}\\&=\left[ \frac{D(t,\Delta )}{D(0,\Delta )}-x(t,\Delta )\right] +\sum _{k=2}^{K}\left[ x(t,(k-1)\Delta )-x(t,k\Delta )\right] +x(t,T)\\&=\frac{D(t,\Delta )}{D(0,\Delta )}. \end{aligned}$$

Now, the case for the default-free floating rate loan just follows, i.e., claim (1) in theorem 1. If there is no default risk, then $p(t,k\Delta )=D(t,k\Delta )$ and thus

$$\begin{aligned} L_{t}=\frac{p(t,\Delta )}{p(0,\Delta )}. \end{aligned}$$

The proof concludes. $\square$

1.2 Proof of Theorem 3

Proof

First, we compute the value of a floating leg of $\Delta$-swap on a credit risky reference rate at time $t\in (0,\Delta )$. It consists of two parts.

(i) The payment at time $\Delta$

The payment at time $\Delta$ is $S_{\Delta }=\frac{1}{D(0,\Delta )}-1$, whose value at $t\in (0,\Delta )$ is

$$\begin{aligned} S_{t}^{\Delta }&=\mathbb {E}_{t}\left[ \frac{\frac{1}{D(0,\Delta )}-1}{B_{\Delta }}\right] B_{t}=\frac{1}{D(0,\Delta )}\mathbb {E}_{t}\left[ \frac{1}{B_{\Delta }}\right] B_{t}-\mathbb {E}_{t}\left[ \frac{1}{B_{\Delta }}\right] B_{t}\\&=\frac{p(t,\Delta )}{D(0,\Delta )}-p(t,\Delta ) \end{aligned}$$

(ii) The payment at time $k\Delta$ for $k\in \{2,...,K\}$

The payment at time $k\Delta$ is $S_{k\Delta }=\frac{1}{D\left( (k-1)\Delta ,k\Delta \right) }-1$, whose value at time $t\in (0,\Delta )$ is

$$\begin{aligned} S_{t}^{k\Delta }= & {} \mathbb {E}_{t}\left[ \frac{\frac{1}{D((k-1)\Delta ,k\Delta )}-1}{B_{k\Delta }}\right] B_{t}\\= & {} \mathbb {E}_{t}\left[ \frac{1}{D((k-1)\Delta ,k\Delta )}\mathbb {E}_{(k-1)\Delta }\left[ \frac{1}{B_{k\Delta }}\right] \right] B_{t}-\mathbb {E}_{t}\left[ \frac{1}{B_{k\Delta }}\right] B_{t}\\= & {} \mathbb {E}_{t}\left[ \frac{1}{D((k-1)\Delta ,k\Delta )}\frac{p((k-1)\Delta ,k\Delta )}{B_{(k-1)\Delta }}\right] B_{t}-p(t,k\Delta ) \end{aligned}$$

Again, in the second inequality, we used the law of iterated expectation and the fact that $\mathcal {F}_{(k-1)\Delta }\supseteq \mathcal {F}_{t}$, and that $\frac{1}{D((k-1)\Delta ,k\Delta )}$ is $\mathcal {F}_{(k-1)\Delta }$-adapted.

Now, we are ready to compute the value of a risky floating leg at time $t\in (0,\Delta )$ as given by

$$\begin{aligned} S_{t}&=S_{t}^{\Delta }+\sum _{k=2}^{K}S_{t}^{k\Delta }\\&=\frac{p(t,\Delta )}{D(0,\Delta )}-p(t,\Delta )+\sum _{k=2}^{K}\left( \mathbb {E}_{t}\left[ \frac{p((k-1)\Delta ,k\Delta )}{D((k-1)\Delta ,k\Delta )}\frac{1}{B_{(k-1)\Delta }}\right] B_{t}-p(t,k\Delta )\right) \\&=\frac{p(t,\Delta )}{D(0,\Delta )}+\sum _{k=2}^{K}\mathbb {E}_{t}\left[ \frac{p((k-1)\Delta ,k\Delta )}{D((k-1)\Delta ,k\Delta )}\frac{1}{B_{(k-1)\Delta }}\right] B_{t}-\sum _{k=1}^{K}p(t,k\Delta ) \end{aligned}$$

This proves claim (2) in theorem 3.

Furthermore, note that claim (1) directly follows from claim (2). Specifically, when there is no default risk, $p(t,k\Delta )=D(t,k\Delta )$, and thus

$$\begin{aligned} S_{t}= & {} \frac{p(t,\Delta )}{p(0,\Delta )}+\sum _{k=2}^{K}\mathbb {E}_{t}\left[ \frac{p((k-1)\Delta ,k\Delta )}{p((k-1)\Delta ,k\Delta )}\frac{1}{B_{(k-1)\Delta }}\right] B_{t}-\sum _{k=1}^{K}p(t,k\Delta )\\= & {} \frac{p(t,\Delta )}{p(0,\Delta )}+\sum _{k=2}^{K}\mathbb {E}_{t}\left[ \frac{1}{B_{(k-1)\Delta }}\right] B_{t}-\sum _{k=1}^{K}p(t,k\Delta )\\= & {} \frac{p(t,\Delta )}{p(0,\Delta )}+\sum _{k=2}^{K}p(t,(k-1)\Delta )-\sum _{k=1}^{K}p(t,k\Delta )\\= & {} \frac{p(t,\Delta )}{p(0,\Delta )}-p(t,T) \end{aligned}$$

The proof concludes. $\square$

1.2.1 A lower bound on $S_{t}$

Lemma 1

$S_{t}\ge \frac{p(t,\Delta )}{D(0,\Delta )}-p(t,T)$.

Proof

To get a lower bound on $S_{t}$, we use the following inequality

$$\begin{aligned} \frac{\mathbbm {1}_{\tau >k\Delta }+\delta _{\tau }\mathbbm {1}_{\tau \le k\Delta }}{D((k-1)\Delta ,k\Delta )}-1\le \frac{1}{D((k-1)\Delta ,k\Delta )}-1 \end{aligned}$$

for $k\in \{2,...,K\}$. This further implies that

$$\begin{aligned} S_{t}^{k\Delta }&\ge \mathbb {E}_{t}\left[ \left( \frac{\mathbbm {1}_{\tau>k\Delta }+\delta _{\tau }\mathbbm {1}_{\tau \le k\Delta }}{D((k-1)\Delta ,k\Delta )}-1\right) \frac{1}{B_{k\Delta }}\right] B_{t}\\&=\mathbb {E}_{t}\left[ \frac{\mathbbm {1}_{\tau>k\Delta }+\delta _{\tau }\mathbbm {1}_{\tau \le k\Delta }}{D((k-1)\Delta ,k\Delta )}\frac{1}{B_{k\Delta }}\right] B_{t}-\mathbb {E}_{t}\left[ \frac{1}{B_{k\Delta }}\right] B_{t}\\&=\mathbb {E}_{t}\left[ \frac{1}{D((k-1)\Delta ,k\Delta )}\mathbb {E}_{(k-1)\Delta }\left[ \frac{\mathbbm {1}_{\tau >k\Delta }+\delta _{\tau }\mathbbm {1}_{\tau \le k\Delta }}{B_{k\Delta }}\right] \right] B_{t}-p(t,k\Delta )\\&=\mathbb {E}_{t}\left[ \frac{1}{D((k-1)\Delta ,k\Delta )}\frac{D((k-1)\Delta ,k\Delta )}{B_{(k-1)\Delta }}\right] B_{t}-p(t,k\Delta )\\&=\mathbb {E}_{t}\left[ \frac{1}{B_{(k-1)\Delta }}\right] B_{t}-p(t,k\Delta )\\&=p\big (t,(k-1)\Delta \big )-p(t,k\Delta ). \end{aligned}$$

Consequently,

$$\begin{aligned} S_{t}&=S_{t}^{\Delta }+\sum _{k=2}^{K}S_{t}^{k\Delta }\\&\ge \frac{p(t,\Delta )}{D(0,\Delta )}-p(t,\Delta )+\sum _{k=2}^{K}\left[ p(t,(k-1)\Delta )-p(t,k\Delta )\right] \\&=\frac{p(t,\Delta )}{D(0,\Delta )}-p(t,T) \end{aligned}$$

The proof concludes. $\square$

1.3 Proof of Theorem 4

Proof

The computation consists of two parts.

(i) The payment at time $\Delta$. The payment at time $\Delta$ is $e^{\int _{0}^{\Delta }r_{s}ds}-1=B_{\Delta }-1$, and its value at time $t\in (0,\Delta )$ is

$$\begin{aligned} S_{t}^{\Delta }=\mathbb {E}_{t}\left[ \frac{B_{\Delta }-1}{B_{\Delta }} \right] B_{t}=B_{t}-\mathbb {E}_{t}\left[ \frac{1}{B_{\Delta }}\right] B_{t}=B_{t}-p(t,\Delta ). \end{aligned}$$

(ii) The payment at time $k\Delta$ for $k>1$. The payment at time $k\Delta$ is

$$\begin{aligned} e^{\int _{(k-1)\Delta }^{k\Delta }r_{s}ds}-1=\frac{B_{k\Delta }}{B_{(k-1)\Delta }}-1, \end{aligned}$$

and its value at time $t\in (0,\Delta )$ is

$$\begin{aligned} S_{t}^{k\Delta }&=\mathbb {E}_{t}\left[ \frac{\frac{B_{k\Delta }}{B_{(k-1)\Delta }}-1}{B_{_{k\Delta }}}\right] B_{t}=\mathbb {E}_{t}\left[ \frac{1}{B_{(k-1)\Delta }}\right] B_{t} -\mathbb {E}_{t}\left[ \frac{1}{B_{_{k\Delta }}}\right] B_{t}\\&=p(t,(k-1)\Delta )-p(t,k\Delta ). \end{aligned}$$

Now, we can compute the floating leg of a dt-swap, and the value at time $t\in (0,\Delta )$ is

$$\begin{aligned} S_{t}&=S_{t}^{\Delta }+\sum _{k=2}^{K}S_{t}^{k\Delta }\\&=B_{t}-p(t,\Delta )+\sum _{k=2}^{K}\left[ p(t,(k-1)\Delta )-p(t,k\Delta )\right] \\&=B_{t}-p(t,T). \end{aligned}$$

The proof concludes. $\square$

1.4 Proof of Theorem 5

Proof

The calculation consists of two parts.

(i) The payment at time $\Delta$. Note that the payment at time $\Delta$ is

$$\begin{aligned} S_{\Delta }=\prod _{i=1}^{N}\frac{1}{p\left( \frac{(i-1)\Delta }{N},\frac{i\Delta }{N}\right) }-1 \end{aligned}$$

For $t\in (0,\frac{\Delta }{N})$, the time t value of $S_{\Delta }$ is

$$\begin{aligned} S_{t}^{\Delta }={ \frac{p\big (t,\Delta /N\big )}{p\Big (0,\Delta /N\Big )}-p(t,\Delta )} \end{aligned}$$

(12)

More generally, for $t\in \left[ \frac{(i-1)\Delta }{N},\frac{i\Delta }{N}\right)$ with $i\in \{1,...,N\}$,

$$\begin{aligned} S_{t}^{\Delta }=\frac{p(t,i\Delta /N)}{\prod _{n=1}^{i}p\Big (\frac{(n-1) \Delta }{N},\frac{n\Delta }{N}\Big )}p(t,\Delta ). \end{aligned}$$

(13)

(ii) The payment at time $k\Delta$ for $k>1$. The payment at time $k\Delta$ is

$$\begin{aligned} S_{k\Delta }=\prod _{i=1}^{N}\frac{1}{p\left( (k-1) \Delta +\frac{(i-1)\Delta }{N},(k-1)\Delta +\frac{i\Delta }{N}\right) }-1 \end{aligned}$$

For $t<(k-1)\Delta$, the time t value of $S_{k\Delta }$ is

$$\begin{aligned} S_{t}^{k\Delta }=p(t,(k-1)\Delta )-p(t,k\Delta ) \end{aligned}$$

(14)

With the aid of eqs. (12) and (14), we can compute the floating leg of a $\Delta /N$-based swap as given by

$$\begin{aligned} S_{t}&=S_{t}^{\Delta }+\sum _{k=2}^{K}S_{t}^{k\Delta }\\&={ \frac{p\big (t,\Delta /N\big )}{p\Big (0,\Delta /N\Big )} -p(t,\Delta )}+\sum _{k=2}^{K}\left[ p(t,(k-1)\Delta )-p(t,k\Delta )\right] \\&={ \frac{p\big (t,\Delta /N\big )}{p\Big (0,\Delta /N\Big )}-p(t,T)} \end{aligned}$$

Now, it suffices to show equations eqs.(12) and (14) to complete the proof.

$\square$

The proof for equation (12).

The proof for equation (14)

Proof

At time $t\in (0,\frac{\Delta }{N})$,

$$\begin{aligned} S_{t}^{\Delta }&=\mathbb {E}_{t}\left[ \frac{S_{\Delta }}{B_{\Delta }}\right] B_{t}\\&=\frac{1}{{{ p(0,\frac{\Delta }{N}}{ )}}}\mathbb {E}_{t}\left[ \prod _{i=2}^{N}\frac{1}{p\left( \frac{(i-1)\Delta }{N},\frac{i\Delta }{N}\right) }\frac{1}{B_{\Delta }}\right] B_{t}-\mathbb {E}_{t}\left[ \frac{1}{B_{\Delta }}\right] B_{t}\\&=\frac{1}{p(0,\frac{\Delta }{N})}\mathbb {E}_{t}\left[ \prod _{i=2}^{N}\frac{1}{p\left( \frac{(i-1)\Delta }{N},\frac{i\Delta }{N}\right) }\frac{1}{B_{\Delta }}\right] B_{t}-p(t,\Delta )\\&=\frac{1}{p(0,\frac{\Delta }{N})}\mathbb {E}_{t}\left[ \prod _{i=2}^{N}\frac{1}{p\left( \frac{(i-1)\Delta }{N},\frac{i\Delta }{N}\right) }\mathbb {E}_{\frac{(N-1)\Delta }{N}}\left[ \frac{1}{B_{\Delta }}\right] \right] B_{t}-p(t,\Delta )\\&=\frac{1}{p(0,\frac{\Delta }{N})}\mathbb {E}_{t}\left[ \prod _{i=2}^{N}\frac{1}{p\left( \frac{(i-1)\Delta }{N},\frac{i\Delta }{N}\right) }\frac{p\left( \frac{(N-1)\Delta }{N},\Delta \right) }{B_{\frac{(N-1)\Delta }{N}}}\right] B_{t}-p(t,\Delta )\\&=\frac{1}{p(0,\frac{\Delta }{N})}\mathbb {E}_{t}\left[ \prod _{i=2}^{N-1}\frac{1}{p\left( \frac{(i-1)\Delta }{N},\frac{i\Delta }{N}\right) }\frac{1}{B_{\frac{(N-1)\Delta }{N}}}\right] B_{t}-p(t,\Delta ) \end{aligned}$$

Continuing as above by applying the law of iterated expectations repeatedly,

$$\begin{aligned} S_{t}^{\Delta }={\frac{p\big (t,\Delta /N\big )}{p\Big (0,\Delta /N\Big )}-p(t,\Delta )} \end{aligned}$$

This completes the proof for eq. (12). $\square$

Proof

To simplify notation, rewrite $S_{k\Delta }=\prod _{i=1}^{N}W(i)$, where

$$\begin{aligned} W(i):=\frac{1}{p\left( (k-1)\Delta +\frac{(i-1)\Delta }{N},(k-1)\Delta +\frac{i\Delta }{N}\right) }. \end{aligned}$$

For $t<(k-1)\Delta$,

$$\begin{aligned} S_{t}^{k\Delta }&=\mathbb {E}_{t}\left[ \frac{S_{k\Delta }}{B_{k\Delta }}\right] B_{t}\\&=\mathbb {E}_{t}\left[ \left( \prod _{i=1}^{N}W(i)\right) \frac{1}{B_{k\Delta }}\right] B_{t}-\mathbb {E}_{t}\left[ \frac{1}{B_{k\Delta }}\right] B_{t}\\&=\mathbb {E}_{t}\left[ \left( \prod _{i=1}^{N}W(i)\right) \frac{1}{B_{k\Delta }}\right] B_{t}-p(t,k\Delta )\\&=\mathbb {E}_{t}\left[ \left( \prod _{i=1}^{N}W(i)\right) \mathbb {E}_{(k-1)\Delta +\frac{(N-1)\Delta }{N}}\left[ \frac{1}{B_{k\Delta }}\right] \right] B_{t}-p(t,k\Delta )\\&=\mathbb {E}_{t}\left[ \left( \prod _{i=1}^{N}W(i)\right) \times \frac{p\left( (k-1)\Delta +\frac{(N-1)\Delta }{N},k\Delta \right) }{B_{(k-1)\Delta +\frac{(N-1)\Delta }{N}}}\right] B_{t}-p(t,k\Delta )\\&=\mathbb {E}_{t}\left[ \left( \prod _{i=1}^{N-1}W(i)\right) \frac{1}{B_{(k-1)\Delta +\frac{(N-1)\Delta }{N}}}\right] B_{t}-p(t,k\Delta ) \end{aligned}$$

Continuing as above by applying the law of iterated expectations repeatedly,

$$\begin{aligned} S_{t}^{k\Delta }=\mathbb {E}_{t}\left[ \frac{1}{B\left( (k-1)\Delta \right) }\right] B_{t}-p(t,k\Delta )=p(t,(k-1)\Delta )-p(t,k\Delta ) \end{aligned}$$

This completes the proof for eq. (14). $\square$

All the proofs conclude for theorem 5. $\square$