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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.

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  • $\begingroup$ Although this is not a full proof, you have shown that L(t) is the ratio of two assets, with the denominator equal to the value of a T-ZCB. Therefore L(t) is a martingale in the forward measure, removing the drift term. $\endgroup$ – dm63 Apr 8 '18 at 14:31
  • $\begingroup$ Yes, I had that in mind, too. But I was unable to prove it rigorously. $\endgroup$ – lbf_1994 Apr 8 '18 at 14:50
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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*}

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