Intuition for consistent Derivative Prices under different Numeraires and Measures

Question

This is essentially the Fundamental Theorem, however I am not asking for a thorough proof, I am more interested in the general intuition.

In words, it makes sense that whatever your unit of account (Numeraire), your derivative price should be the same. However, let's take the Libor Market Model as an example:

Under the $T_i$-forward measure (i.e. using a zero-coupon bond that matures at time $T_i$ as numeraire), the Libor $L(t,T_{i-1},T_i)$ is a martingale and has zero drift. We can write the process as: $$dL(t,T_{i-1},T_i) = \sigma L(t,T_{i−1},T_{i}) dW^{T_i}(t).$$ The solution is the Black 76 formula.

Shifting the measure to $T_{i-1}$-forward measure (i.e. using a zero-coupon bond that matures at time $T_{i-1}$, as numeraire), the Libor acquires a drift term and the process becomes: $$dL(t,T_{i-1},T_i) = rL(t,T_{i−1},T_{i})dt + \sigma L(t,T_{i−1},T_{i}) dW^{T_{i-1}}(t)$$ where $dW^{T_i}(t)$ under $T_{i-1}$ is a different process to the $dW^{T_{i-1}}(t)$ under the $T_i$ measure.

The solution to the second equation above is no longer the Black-76 formula, but a Black-Scholes formula with a drift term (the drift $r$ is a complicated term but we don't need to worry about it for the sake of this example).

Where is the logical mistake here? My understanding is that Caplets and Floorlets on forward Libors are always priced using the Black 76 formula, and so should always have no drift.

Thank you so much, J.

Answer 1

Your dynamics under the $T_{i-1}$-forward measure is wrong.

Specifically, let $P_{i-1}$ and $P_i$ be, respectively, the $T_{i-1}$- and $T_i$-forward probability measures. Moreover, let $\Delta_i = T_i-T_{i-1}$. Then, for $0\le t \le T_{i-1}$, \begin{align*} \eta_t &\equiv \frac{dP_{i-1}}{dP_i}\big|_t \\ &= \frac{P_i(0, T_i)}{P_{i-1}(0, T_{i-1})}\frac{P_{i-1}(t, T_{i-1})}{P_i(t, T_i)}\\ &=\frac{1+\Delta_i L(t, T_{i-1}, T_i)}{1+\Delta_i L(0, T_{i-1}, T_i)}. \end{align*} Furthermore, \begin{align*} d\eta_t &= \frac{ \sigma\Delta_i L(t, T_{i-1}, T_i)}{1+\Delta_i L(0, T_{i-1}, T_i)}dW_t\\ &=\frac{\sigma \Delta_i L(t, T_{i-1}, T_i)}{1+\Delta_i L(t, T_{i-1}, T_i)} \eta_t dW_t. \end{align*} Consequently, under $P_{i-1}$, \begin{align*} dL(t, T_{i-1}, T_i) &= L(t, T_{i-1}, T_i)\left(\frac{\sigma^2 \Delta_i L(t, T_{i-1}, T_i)}{1+\Delta_i L(t, T_{i-1}, T_i)} dt + \sigma d\widehat{W}_t\right), \end{align*} where $\{\widehat{W}_t, 0\le t \le T_{i-1}\}$ is a standard Brownian motion under $P_{i-1}$.