# Pricing of compounded swaps

As far as I understand, a compounded swap rolls up individual payments into one final payment which becomes: $$V(t_n) = N \prod_{i = 0}^{n-1}(1 + d_i L_i)-N$$

where $$d_i$$ is the day fraction for period $$t_i$$ to $$t_{i+1}$$ and $$L_i$$ is the index for the same period and where $$N$$ is deducted at the end because we assume no exchange of notional.

Now, to value this we need to calculate the expectation of $$V(T)$$ under some appropriate numéraire and measure, but we are dealing with products of various $$L_i$$'s which are, in general, not mutually independent, so it's not a simple matter of replacing with them forwards.

How is this then done? An internet search only revealed simple formulas using forwards. A good reference text would be welcome.

Following suggestions in the comments, if I use the adjusted forward numéraire with maturity equal to the payment date $$t_n$$ and using $$P(t_i, t_{i+1}) = \frac{1}{1 + d(t_i,t_{i+1}) L(t, t_{i+1})}$$, then I get: $$V(t) = P(t, t_n) \Bbb{E}^{Q^{t_n}} [V(t_n)|F_t] = N P(t, t_n) \left(\Bbb{E}^{Q^{t_n}} \left[\prod_{i=0}^{n-1} \frac{1}{P(t_i, t_{i+1})} | F_t \right]-1\right)$$

but I'm not sure that this gets me anywhere.

• Check out "A quant perspective on IBOR fallback proposals" by M. Henrard. In Section 5.2 he refers to Th. 2.4 of Interest Rate Modelling in the Multi-curve Framework by the same author, where he proves a formula for the case of a daily rate $r_i$. See also Section 5.4, option 4, of the same paper where he derives a formula very close to your case using nested conditional expectations, and coming up with a multiplicative convexity adjustment based on a 1F HJM Gaussian model. – Daneel Olivaw Nov 19 '19 at 16:48
• Write the Libors in terms of bonds, then the product can be simplified. – Gordon Nov 19 '19 at 16:49
• [Note: my comment refers to the case of overnight rates and is related to incoming SOFR-related changes, where this kind of compounded rates are common] – Daneel Olivaw Nov 19 '19 at 16:54
• Hint: 1. Write the price as zc bond (for discount) multiplied by an expectation under the forward measure (of the payment date) 2. Write the compounding factor in terms inverse of ZC bonds: $1 + d L(T, T + d) = \frac{1}{P(T, T + d)}$ and 3. Use conditional exportations inside the expectation. If you are stuck after this, we might be able to help here. – byouness Nov 19 '19 at 19:14
• @byouness Thank you for your hints but, alas, I didn't get far. – Confounded Nov 19 '19 at 23:21

Let us start from your last equation, and focus specifically on the expectation. Assuming that the end date of each period is the start period of the next, the idea is to simplify it using conditional expectations.

Since $$t < t_{n-2}$$, we can write using the tower property of conditional expectations: \begin{aligned} \Bbb{E}_{t}^{Q^{t_n}} \left[\prod_{i=0}^{n-1} \frac{1}{P(t_i, t_{i+1})} \right] &= \Bbb{E}_{t}^{Q^{t_n}} \left[\prod_{i=0}^{n-2} \frac{1}{P(t_i, t_{i+1})} \times \frac{1}{P(t_{n-1}, t_{n})} \right]\\ &= \Bbb{E}_{t}^{Q^{t_n}} \left[ \Bbb{E}_{t_{n-2}}^{Q^{t_n}} \left[ \underbrace{\prod_{i=0}^{n-2} \frac{1}{P(t_i, t_{i+1})}}_{\mathcal{F}_{t_{n-2}}-\text{measurable}} \times \frac{1}{P(t_{n-1}, t_{n})} \right] \right]\\ &= \Bbb{E}_{t}^{Q^{t_n}} \left[\prod_{i=0}^{n-2} \frac{1}{P(t_i, t_{i+1})}\times \Bbb{E}_{t_{n-2}}^{Q^{t_n}} \left[ \underbrace{\frac{P(t_{n-1}, t_{n-1})}{P(t_{n-1}, t_{n})}}_{\mathbb{Q}^{t_n}\text{martingale}} \right]\right]\\ &= \Bbb{E}_{t}^{Q^{t_n}} \left[\prod_{i=0}^{n-2} \frac{1}{P(t_i, t_{i+1})}\times \frac{P(t_{n-2}, t_{n-1})}{P(t_{n-2}, t_{n})} \right]\\ \end{aligned}

We can see that the product is getting smaller, since the term $$P(t_{n-2}, t_{n-1})$$ that appeared in the numerator will simplify with the the last term of the product. $$\Bbb{E}_{t}^{Q^{t_n}} \left[\prod_{i=0}^{n-1} \frac{1}{P(t_i, t_{i+1})} \right] = \Bbb{E}_{t}^{Q^{t_n}} \left[\prod_{i=0}^{n-3} \frac{1}{P(t_i, t_{i+1})}\times \frac{1}{P(t_{n-2}, t_{n})} \right]$$ By repeating this operation, the product disappears (assuming that at pricing date $$t$$, the swap didn't start yet, i.e.: $$t < t_0$$), and you get: \begin{aligned} \Bbb{E}_{t}^{Q^{t_n}} \left[\prod_{i=0}^{n-1} \frac{1}{P(t_i, t_{i+1})} \right] &= \Bbb{E}_{t}^{Q^{t_n}} \left[\frac{P(t_0, t_0)}{P(t_{0}, t_{n})} \right]\\ &= \frac{P(t, t_0)}{P(t, t_n)}\\ \end{aligned}

This ratio can also be written as a product of capitalization factors using Libor forwards as follows: $$\frac{P(t, t_0)}{P(t, t_n)} = \prod_{i=0}^{n-1} \frac{P(t, t_i)}{P(t, t_{i+1})} = \prod_{i=0}^{n-1} 1 + d(t_i, t_{i+1}) L(t, t_i, t_{i+1})$$

• Thank you for posting the answer. I am not sure, however, I understand the steps in the first equation. It seems that they assume that $P(t_{n-2},t_{n-1}) P(t_{n-1},t_{n}) = P(t_{n-2},t_{n})$, but I don't see why this has to be the case. Could you elaborate? – Confounded Nov 20 '19 at 11:36
• I simply used the fact that the expectation of $\frac{P(t, t_{n-1})}{P(t, t_n)}$ under $\mathbb{Q}^{t_n}$, conditional on $\mathcal{F}_{t_{n-2}}$ is $\frac{P(t_{n-2}, t_{n-1})}{P(t_{n-2}, t_n)}$. In other words, $\left(\frac{P(t, t_{n-1})}{P(t, t_n)}\right)_t$ is a $\mathbb{Q}^{t_n}$-martingale. I hope this clarifies things. – byouness Nov 20 '19 at 13:05
• Ah okay, its simply the tower property of conditional expectations, since $t < t_{n-2}$, you can write: $\mathbb{E}_t [ AB ] = \mathbb{E}_t \left[ \mathbb{E}_{t_{n-2}} [ AB] \right]$. Next, if $A$ is $\mathcal{F}_{t_{n-2}}$-measurable then : $\mathbb{E}_t [ AB ] = \mathbb{E}_t \left[ A \mathbb{E}_{t_{n-2}} [ B] \right]$. I will clarify in my answer. – byouness Nov 20 '19 at 14:46
• $(P(s, t_n))_s>0$ is the numéraire, so the expectation of $\frac{X(T)}{P(T, t_n)}$ (with $T > t$) under the associated measure (conditional to $\mathcal{F}_t$) is equal to $\frac{X(t)}{P(t, t_n)}$. This is what is written with $T = t_{n-1}$ and $X(s) = P(s, t_{n-1})$. – byouness Nov 20 '19 at 16:12
• You can't just put anything in $X$. it must be a self financed asset, so $X(T) = 1$ doesn't work. 1 (or any constant for that matter) is a martingale under the T-forward measure, but $(\frac{1}{P(s, T)})_s$ is not. – byouness Nov 20 '19 at 17:58