LIBOR Forward Rate Dynamics under the T-Forward Measure
Summary
The document derives the dynamics of a LIBOR forward rate under the measure associated with a zero-coupon bond maturing at the forward rate’s end date. Starting from bond price dynamics under the risk-neutral measure, it applies Itô’s lemma to the ratio of bond prices defining the forward rate. It then uses the change of measure to replace the original Brownian motion with a Brownian motion under the T-forward measure, canceling the drift term.
The resulting volatility depends on the volatilities of both bond prices, rather than simply carrying over the rate’s stated volatility under the physical measure. The derivation illustrates how the chosen numeraire affects drift and measure-specific dynamics. It assumes the stated bond price diffusion model and does not establish that the question’s proposed volatility specification is generally valid; the answer instead derives a rate volatility from the underlying bond volatilities.
Key ideas
- A forward rate can be represented as a ratio of zero-coupon bond prices.
- Applying Itô’s lemma to that ratio gives its dynamics under the risk-neutral measure.
- Changing to the T-forward measure shifts the Brownian motion according to the volatility of the T-maturity bond.
- Under the T-forward measure, the derived rate has zero drift and volatility determined by the two bond volatilities.
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Full text
# Dynamics of LIBOR foward rate under T-forward measure
# Dynamics of LIBOR foward rate under T-forward measure
Assume that under the physical measure $\mathbb{P}$ we have for the LIBOR forward rate $L(t):=L(t;S,T) = \frac{1}{T-S}\left(\frac{P(t,S)}{P(t,T)}-1\right)$ that $$ \mathrm{d}L(t) = L(t)\left(\mu(t)\mathrm{d}t +\sigma(t,T)\mathrm{d}W^{\mathbb{P}}(t)\right) $$ I want to show that under the $T$-forward measure, the dynamics are
$$ \mathrm{d}L(t) = \sigma(t,T)L(t)\mathrm{d}W^{\mathbb{Q^{(T)}}}(t) $$ Unfortunately, I'm unable to show this.
## Answer by Gordon (score 3, accepted)
https://quant.stackexchange.com/a/39180
We assume that, under the risk-neutral measure $Q$, \begin{align*} dP(t, T) = P(t, T)(r_t + \sigma(t, T)dW_t), \end{align*} where $\{W_t, \, t \ge 0\}$ is a standard Brownian motion. Then \begin{align*} dL(t) &= \frac{1}{T-S}\bigg(\frac{dP(t, S)}{P(t, T)} -\frac{dP(t, S)}{P^2(t, T)}dP(t, T) \\ &\qquad + \frac{dP(t, S)}{P^3(t, T)} \langle dP(t, T), \, dP(t, T)\rangle -\frac{1}{P^2(t, T)} \langle dP(t, S), \, dP(t, T)\rangle\bigg)\\ &=\frac{1}{T-S}\frac{P(t, S)}{P(t, T)}\bigg(\big(\sigma^2(t, T) -\sigma(t, S)\sigma(t, T) \big)dt + \big(\sigma(t, S)- \sigma(t, T)\big) dW_t \bigg).\tag{1} \end{align*} Let $Q^T$ be the $T$-forward measure. Then \begin{align*} \frac{dQ^T}{dQ}\big|_t &= \frac{P(t, T)}{P(0, T) e^{\int_0^t r_s ds}}\\ &=e^{-\frac{1}{2}\int_0^t \sigma^2(s, T)ds + \int_0^t \sigma(s, T) dW_s}. \end{align*} Moreover, $W^T=\{W_t^T, \, t \ge 0\}$, where \begin{align*} W_t^T = W_t - \int_0^t \sigma(s, T) ds, \end{align*} is a standard Brownian motion under $Q^T$. Furthermore, from (1), \begin{align*} dL(t) &=\frac{1}{T-S}\frac{P(t, S)}{P(t, T)}\bigg(\big(\sigma^2(t, T) -\sigma(t, S)\sigma(t, T) \big)dt + \big(\sigma(t, S)- \sigma(t, T)\big) dW_t \bigg)\\ &=\frac{1}{T-S}\frac{P(t, S)}{P(t, T)}\big(\sigma(t, S)- \sigma(t, T)\big) dW_t^T\\ &=L(t)\frac{1+(T-S)L(t)}{(T-S)L(t)}\big(\sigma(t, S)- \sigma(t, T)\big) dW_t^T\\ &\equiv L(t)\sigma_L(t, T) dW_t^T, \end{align*} where \begin{align*} \sigma_L(t, T) = L(t)\frac{1+(T-S)L(t)}{(T-S)L(t)}\big(\sigma(t, S)- \sigma(t, T)\big). \end{align*}Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
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