Pricing Floating-Rate Bond Coupons with Zero-Coupon Bond Replication

# How do I price $P(t)=P(t,T_{n})+\sum_{i=1}^{n}[P(t,T_{i-1})-P(t,T_{i})]$?


# How do I price $P(t)=P(t,T_{n})+\sum_{i=1}^{n}[P(t,T_{i-1})-P(t,T_{i})]$?












Derive the pricing formula $$P(t)=P(t,T_{n})+\sum_{i=1}^{n}[P(t,T_{i-1})-P(t,T_{i})]$$directly, by constructing a self-financing portfolio which replicates the cash flow of the floating rate bond.

$P(t,T_{i-1}) $ means Buy, at time $t$, one $T_{i−1}$-bond. This will cost $P(t, T_{i−1})$ $P(t)$ means price at $t$ time

This question is related to the Arbitrage Theory in Continuous Time book by Tomas Bjork.

## Answer by Gordon (score 3, accepted)

https://quant.stackexchange.com/a/18886

In another solution, the answer is based on replication approach. Here, we provide some other approaches for the valuation of the LIBOR rate, \begin{align} L(T_{i-1}; T_{i-1}, T_i) = \frac{1}{\Delta T_i}\left(\frac{1}{P(T_{i-1}, T_i)}-1\right), \end{align} set a $T_{i-1}$ and paid at $T_i$, where $\Delta T_i =T_i-T_{i-1}$.

Let $E$ be the expectation operator under the risk-neutral measure $P$, which has the money market account value process $B_t$ as the numeraire. Then the value at time $t$ of the float payment $L(T_{i-1}; T_{i-1}, T_i)\Delta T_i$ made at $T_i$ is given by \begin{align*} B_t E\left(\frac{L(T_{i-1}; T_{i-1}, T_i)\Delta T_i}{B_{T_i}}\mid\mathcal{F}_t \right) &= B_t E\left(\frac{L(T_{i-1}; T_{i-1}, T_i)\Delta T_i}{B_{T_{i-1}}} E\left(\frac{B_{T_{i-1}}}{B_{T_i}} \mid \mathcal{F}_{T_{i-1}}\right)\mid\mathcal{F}_t \right)\\ &=B_t E\left(\frac{L(T_{i-1}; T_{i-1}, T_i)\Delta T_i}{B_{T_{i-1}}} P(T_{i-1}, T_i)\mid\mathcal{F}_t \right)\\ &=B_t E\left(\frac{1}{B_{T_{i-1}}} \Big[1 - P(T_{i-1}, T_i)\Big]\mid\mathcal{F}_t \right)\\ &= B_t E\left(\frac{1}{B_{T_{i-1}}}\mid\mathcal{F}_t \right) - B_t E\left(\frac{P(T_{i-1}, T_i)}{B_{T_{i-1}}}\mid\mathcal{F}_t \right)\\ &=P(t, T_{i-1}) - B_t \times \frac{P(t, T_i)}{B_t}\\ &= P(t, T_{i-1}) - P(t, T_i). \end{align*}

$$ $$

Alternatively, let $E_{T_i}$ be the expectation operator under the $T_i$-forward measure $P_{T_i}$, which has the bond price process $\{P(t, T_i)\mid t \geq 0\}$ as the numeraire. Then the LIBOR rate process $\{L(t; T_{i-1}, T_i) \mid 0\leq t \leq T_{i-1} \}$ is a martingale under $P_{T_i}$. Moreover, for $0 \leq t \leq T_{i-1}$, let \begin{align} \eta_t &\triangleq \frac{dP}{dP_{T_{i}}}\big|_t\\ &=\frac{B_t P(0, T_{i})}{P(t, T_i)}. \end{align} By Bayes formula, for $0 \leq t \leq T_{i-1}$, the value at time $t$ of the float payment $L(T_{i-1}; T_{i-1}, T_i)\Delta T_i$ made at $T_i$ is given by \begin{align*} B_t E\left(\frac{L(T_{i-1}; T_{i-1}, T_i)\Delta T_i}{B_{T_i}}\mid\mathcal{F}_t \right) &= B_t E_{T_i}\left(\frac{\eta_{T_i}}{\eta_t}\frac{L(T_{i-1}; T_{i-1}, T_i)\Delta T_i}{B_{T_i}}\mid\mathcal{F}_t \right)\\ &=P(t, T_i)E_{T_i}\left(L(T_{i-1}; T_{i-1}, T_i)\Delta T_i\mid\mathcal{F}_t \right)\\ &=P(t, T_i)L(t; T_{i-1}, T_i) \Delta T_i\\ &=P(t, T_{i-1}) - P(t, T_i), \end{align*} from the martingale property of $L$ under the $T_i$-forward measure $P_{T_i}$.

## Answer by user16891 (score 1)

https://quant.stackexchange.com/a/18872

Edit for Gordon. First, fix point in time $T_0,...,T_n$ whereas $T_1,...,T_n$ are the coupon dates and $T_0$ is interpreted as the emission date of the bond. At time $T_i$, $i = 1,...,n$ the owner of the bond receives $c_i$.At time $T_n$ the owner receives the face value K.We now go on to compute the price of this bond, and it is obvious that the coupon bond can be replicated by holding a portfolio of zero coupon bonds with maturities $T_i$, $i = 1,...,n$.So the price,$P(t)$, at a time $t < T_1$, of the coupon bond is given by $$P(t)=KP(t,T_{n})+\sum_{i=1}^{n}c_i P(t,T_{i})$$

If the coupon rate $r_i$ is set to the spot LIBOR rate $L(T_{i−1}, T_i)$ ,then \begin{align} c_i=(T_i-T_{i−1})L(T_{i−1}, T_i)K \end{align} We now go on to compute the value of this bond at some time $t < T_0$, in the case when the coupon dates are equally spaced, with $T_i−T_{i−1}=\delta$, and to this end we study the individual coupon $c_i$. Without loss of generality we may assume that $K = 1$, and inserting the definition of the LIBOR rate \begin{align} c_i=\frac{1}{P(T_{i−1}, T_i)}-1 \end{align} The value at $t$, of the term $−1$ , is of course equal to $-P(t, T_i)$ and it remains to compute the value of the term$\frac{1}{P(T_{i−1}, T_i)}$ which is paid out at $T_i$.This is, however, easily done through the following argument.

- Buy, at time $t$, one $T_{i−1}$-bond. This will cost $P(t, T_{i−1})$.

- At time $T_{i−1}$ you will receive the amount $1$.

- Invest this unit amount in $T_{i−1}$-bond. This will give you exactly $\frac{1}{P(T_{i−1}, T_i)}$ bonds.

- At $\,T_i$ the bonds will mature, each at the face value $1$. Thus, at time $T_i$, you will obtain the amount $\frac{1}{P(T_{i−1}, T_i)}$

Thus the value at $t$, of obtaining $\frac{1}{P(T_{i−1}, T_i)}$ at $T_i$, is given by $P(t, T_{i-1})$, and the value at t of the coupon $c_i$ is $P(t, T_{i−1}) − P(t, T_i)$. then, by summing up all values of $c_i$ and the value of the notional amount (i.e. 1) at $T_n$, we have $$P(t)=P(t,T_{n})+\sum_{i=1}^{n}P(t,T_{i-1})-P(t,T_{i})=p(t,T_0)$$