$\newcommand{\dd}{\, \mathrm{d}}$ Consider the exponential martingale, $$ \xi_t^\lambda = \exp \left\{ - \int_0^t \lambda_s \dd z_s - \frac 12 \int_0^T \lambda_s^2 \dd s \right\}, $$ that is used in the statement of Girsanov's theorem (this martingale represents the Radon-Nykodym derivative $\frac{\dd \mathbb Q^\lambda}{\dd \mathbb P}$.).

Exercise 2.4 in Munk's book Financial Asset Pricing Theory deals with applying Ito's lemma to this process.

Suppose $$ X_t = \frac 12 \int_0^t \lambda _s^2 \dd s + \int_0^t \lambda_s \dd z_s. $$ Part (a) asks us to argue that $\dd X_t = \frac 12 \lambda_t^2 \dd t + \lambda_t \dd z_t.$ Part (b) asks, "Suppose that the continuous-time stochastic process $\xi = (\xi_t)$ is defined as $\xi_t = \exp\{-X_t\}$. Show that $\dd \xi_t = -\lambda_t \xi_t \dd z_t$."

Informally, we can argue part (a) by applying Ito's lemma, \begin{align*} \dd X_t &= \left( \frac 12 \lambda_t^2 + \lambda_t \dd z_t\right) \dd t + \lambda_t \dd z_t \\ &= \frac 12 \lambda_t^2 \dd t + \lambda_t \dd z_t, \end{align*} and arguing that $\dd z_t \cdot \dd t = 0$.

How do we solve part (b)?


1 Answer 1


$\newcommand{\dd}{\, \mathrm{d}}$ If we apply Ito's lemma, then \begin{align*} \dd \xi_t &= -\xi_t \dd X_t + \frac 12 \xi_t (\dd X_t)^2\\ &= -\xi_t \left(\frac 12 \lambda_t^2 \dd t + \lambda_t \dd z_t\right) + \frac 12 \xi_t \lambda_t^2 \dd t \\ &= -\xi_t \lambda \dd z_t. \end{align*}


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