I would like to establish the equations of forward libors under risk neutral measure. Here is how I do it, and what I get :

Under the $P_{T_j} $ measure, forward Libor $L_j$ is martingale. Thus:

$$ dL_j = L_j \times \sigma_j(t) \times dW^{T_j} $$

The change of numéraire implies that:

$$ \frac{dQ^{T_j}}{dQ^*} = \frac{P(t,T)\times\exp{\left(-\int_0^T{r(s)ds}\right)}}{P(0,T)} $$

Under $Q^*$, $P(t,T)$ discounted is martingale, which means that:

$$ \frac{dP(t,T)}{P(t,T)} = r(t) dt + \eta(t) dW^* $$

Solving this gives:

$$ P(t,T) = P(0,T) \times \exp\left(\int_0^T\left(r(s)-\frac{1}{2} \eta(s)^2\right)ds + \int_0^T{\eta(s)}dW^{T_j}\right) $$


$$ \frac{dQ^{T_j}}{dQ^*} = \exp\left(-\int_0^T{\frac{1}{2}} \eta(s)^2ds + \int_0^T{\eta(s)}dW^{T_j}\right) $$

Girsanov shows that:

$$ dW^{T_j} = dW^* - \eta(t) dt $$

Under $Q^*$:

$$ \frac{dL_j}{L_j} = \alpha(t) dt + \sigma_j(t) dW^{*} $$

Writing $P(t,T) $ as $\exp(-\int_t^T{f(t,s)ds}) = \exp(Y_t)$:

$$ \frac{dP(t,T)}{P(t,T)} = dY_t +\frac{1}{2}<Y_t>dt $$

with $ dY_t = f(t,t) dt -\int_t^T{\alpha(t) dt ds } - \int_t^T{\sigma(t)dt dW_s}$. Identifying, I conclude that:

$$ \eta(t) = \sigma(t) $$


$$ dL_j(t) = -L_j(t) \sigma_j(t) \int_t^{T_j}{\sigma_j(s)ds} dt + L_j\sigma_j(t) dW^*.$$

Is it correct ?

  • $\begingroup$ It is not clear why $dY_t = f(t,t) dt -\int_t^T{\alpha(t) dt ds } - \int_t^T{\sigma(t)dt dW_s}$. Note that $f(t, T)$ is the instantaneous forward rate, while $L_j$ is a forward rate over a given time period, for example, $[T_{j-1}, \, T_j]$. $\endgroup$
    – Gordon
    Sep 7, 2018 at 18:02

1 Answer 1


We assume that, under the $T_j$-forward probability measure $P_{T_j}$, \begin{align*} \frac{dP(t, T_j)}{P(t, T_j)} = \mu_P(t, T_j) dt + \sigma_P(t, T_j) dW_t^{T_j}, \end{align*} where $\mu_P(t, T_j)$ and $\sigma_P(t, T_j)$ are the respective drift and volatility functions. Let $Q$ be the risk-neutral probability measure. Then \begin{align*} \frac{dQ}{dP_{T_j}}\big|_t &= \frac{e^{\int_0^t r_s ds}P(0, T_j)}{P(t, T_j)}\\ &=e^{\int_0^t \big(r_s -\mu_P(s, T_j)+\frac{1}{2} \sigma_P(s, T_j)^2 \big) ds - \int_0^t \sigma_P(s, T_j) dW_s^{T_j}}. \end{align*} Since $\frac{dQ}{dP_{T_j}}\big|_t$ is a martingale under $P_{T_j}$, \begin{align*} \int_0^t \Big(r_s -\mu_P(s, T_j)+\frac{1}{2} \sigma_P(s, T_j)^2 \Big) ds = -\frac{1}{2}\int_0^t \sigma_P(s, T_j)^2 ds. \end{align*} That is, \begin{align*} \mu_P(t, T_j) = r_t + \sigma_P(t, T_j)^2, \end{align*} and \begin{align*} \frac{dQ}{dP_{T_j}}\big|_t &= e^{-\frac{1}{2} \int_0^t\sigma_P(s, T_j)^2 ds - \int_0^t \sigma_P(s, T_j) dW_s^{T_j}}. \end{align*} Then, under the risk-neutral probability measure $Q$, $\{W_t, \, t \ge 0\}$, where, for $t \ge 0$, \begin{align*} W_t = W_t^{T_j} + \int_0^t \sigma_P(s, T_j) ds, \end{align*} is a standard Brownian motion. Moreover, \begin{align*} \frac{dL_j}{L_j} = -\sigma_j(t) \sigma_P(t, T_j) dt + \sigma_j(t) d W_t. \end{align*}


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