Pricing of compounded swaps

Question

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.

Add 1

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.

Answer 1

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: $$ \frac{P(t, t_0)}{P(t, t_n)} = \prod_{i=0}^{n-1} \frac{P(t, t_i)}{P(t, t_{i+1})} = \prod_{i=0}^{n-1} 1 + d(t_i, t_{i+1}) L(t, t_i, t_{i+1}) $$