finance - 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

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 $P_{T_i}$.