Ito, Stochastic Exponential and Girsanov

Question

This is a two-part question relating to the change of measure density used in Girsanov and secondly to the Stochastic Exponential.

Whilst reading notes relating to Girsanov it is stated that the change of measure density martingale may be written: \begin{align} \rho_t = \exp \left[- \int_{0}^{t} \lambda_s \, dW_s - \tfrac{1}{2}\int_{0}^{t} \lambda_{s}^{2} \, ds \right] \end{align}

It is stated that using Ito's Lemma it is straightforward to verify that the stochastic differential of $\rho_t$ is given by \begin{align} d\rho_t = -\rho_t \, \lambda_t \, dW_t \end{align}

I think I've solved this and applied ito as follows: \begin{align} \rho_t &= exp\left[-\lambda_t W_t - \tfrac{1}{2} \, \lambda_{t}^{2} \, t \right]\\ d\rho_t &= \frac{\partial\rho_t}{\partial t} dt + \frac{\partial \rho_t}{\partial W} dW_t + \tfrac{1}{2} \frac{\partial^2 \rho_t}{\partial W^{2}} (dW_{t})^2\\ &= -\tfrac{1}{2} \lambda_{t}^{2}\exp\left[\dots\right] - \lambda_t\exp\left[\dots\right] dW_{t} + \tfrac{1}{2} \lambda_{t}^2 \exp\left[\dots\right] (dW_{t})^{2}\\ &= -\tfrac{1}{2}\,\lambda_{t}^{2}\,\rho_{t}\,dt - \lambda_{t}\,\rho_{t}dW_{t} + \tfrac{1}{2}\,\lambda_{t}^{2}\,\rho_{t} dt\\ &= -\lambda_{t}\rho_{t}\,dW_{t} \end{align}

$\textbf{Question 1}$ - Is this correct?

The Stochastic Exponential is stated as: \begin{align} \mathcal{E}_t(X) = \exp\left[ X_t - \tfrac{1}{2} \langle X,X \rangle_{t} \right] \end{align}

$\textbf{Question 2}$ - I believe $\langle X,X \rangle_t$ is the quadratic variation, should I interpret this the same way I do the $\int_{0}^{t} \lambda_{t}^{2}\,ds$ term in the change of measure density above?

It is stated (Filipovic - Term-Structure Models) that if $X_t$ is a continuous local martingale with $X_0 = 0$, then using Ito one can see $d\mathcal{E}_t(X) = \mathcal{E}_t(X)dX_t$.

The solution I have is as follows: \begin{align} d\mathcal{E}_t(X) &= \exp\left[X_t - \tfrac{1}{2}\langle X \rangle_t \right]\left(dX_t - \tfrac{1}{2}d\langle X \rangle_t \right) + \tfrac{1}{2}\exp\left[X_t - \tfrac{1}{2}\langle X \rangle_t \right]d\langle X \rangle_t\\ &= \exp\left[X_t - \tfrac{1}{2}\langle X \rangle_t \right] dX_t\\ &= \mathcal{E}_{t}dM_t\\ \text{obviously $\mathcal{E}_0(X) = 1.$} \end{align}

$\textbf{Question 3}$ - I have difficulty understanding the notation here and cannot see how Ito has been applied in this case (as I cannot see the $dt$ and $dW$ terms). I'd appreciate any help showing me how ito has been applied in this case (and why it is obvious that $\mathcal{E}_{0}(X) = 1$).

Many thanks,

John

Answer 1

For question I, the identity \begin{align*} \rho_t = \exp\big(-\lambda_t W_t - \frac{1}{2} \lambda_t^2t\big) \end{align*} does not appear correct, unless $\lambda_t$ is a constant.

For question II, yes. If $X_t = -\int_0^t \lambda_s dW_s$, then $\langle X \rangle_t = \int_0^t \lambda_s^2 ds$.

For question III, you need to note that \begin{align*} \langle X \rangle_t = \int_0^t \frac{\partial\langle X \rangle_s}{\partial s}ds. \end{align*} Then $d\langle X, \langle X \rangle \rangle_t = 0$ and $d\langle \langle X \rangle, \langle X \rangle \rangle_t = 0$.

Answer 2

It might be easier to go the other way: start with $$ d\mathcal{E}_t = \mathcal{E}_t dX_t $$ apply Ito to the $\log$ function $$ d\log(\mathcal{E})_t = \frac{1}{\mathcal{E}_t}d\mathcal{E}_t - \frac{1}{2} \frac{1}{\mathcal{E}_t^2}d\langle\mathcal{E},\mathcal{E}\rangle_t = \frac{1}{\mathcal{E}_t}\mathcal{E}_tdX_t - \frac{1}{2} \frac{1}{\mathcal{E}_t^2}\mathcal{E}_t^2d\langle X,X\rangle_t $$ in other words $$ d\log(\mathcal{E})_t = dX_t - \frac{1}{2} d\langle X,X\rangle_t $$ $$ \log(\mathcal{E})_T = \log(\mathcal{E}_0) + X_T - \frac{1}{2} \langle X,X\rangle_T $$ $$ \mathcal{E}_T = \mathcal{E}_0\exp(X_T - \frac{1}{2} \langle X,X\rangle_T) $$ In case $X_t = \int_0^t \lambda_s dW_s$, you get $dX_t = \lambda_t dW_t$ and $\langle X,X\rangle_T =\int_0^T d\langle X,X\rangle_t =\int_0^T \lambda_t^2 d\langle W,W\rangle_t = \int_0^T \lambda_t^2 dt$.