2
$\begingroup$

I am told that this is a consequence of the Girsanov theorem, yet I do not see how it it is.

Consider some standard model with $dS_i = \mu S_i dt + \sigma S_i dW^P$. Let $Q$ be an equivalent martingale measure. Then, it is claimed that due to the Girsanov theorem, $dS_i = \sigma S_i dW^Q$.

However, the Girsanov theorem only proves this for a particular measure $Q$ which it defines by first introducing a particular variable $L$ defined in terms of another process called the kernel. The $Q$ which is defined through this process may be very different than the $Q$ we have given above, so I don't understand how the Girsanov theorem can be used?

Should we not instead prove that given any martingale measure $Q$, then we can always determine a kernel which can be used to define this measure $Q$ using the recipe in the Girsanov theorem, and THEN we can use Girsanov?

$\endgroup$
2
$\begingroup$

As you didn't explain your notation: First note that if $S$ denotes the price process of a spot asset (such as a stock), then it is not a martingale under the risk-neutral probability measure $\mathbb{Q}$. Instead, the discounted price process is a martingale under $\mathbb{Q}$.

Girsanov's theorem is more general than just for finding the risk-neutral probability measure. In case of a Brownian motion, it defines how the drift changes under an equivalent change of measure. It states that when $W^{\mathbb{P}}$ is a $\mathbb{P}$ Brownian motion, then the process

\begin{equation} W^{\mathbb{Q}}_t = W_t^{\mathbb{P}} - \int_0^t \lambda_u \mathrm{d}u \end{equation}

is a Brownian motion under $\mathbb{Q}$ defined through the Radon-Nikodym derivative process

\begin{equation} \left. \frac{\mathrm{d}\mathbb{Q}}{\mathrm{d}\mathbb{P}} \right| \mathfrak{F}_t = \mathcal{E}_t \left( \int_0^\cdot \lambda_u \mathrm{d}W_u^{\mathbb{P}} \right). \end{equation}

If I understand your question correctly then you refer to the process $\lambda$ as the "kernel". I agree with you that Girsanov's theorem does not by itself yield a risk-neutral probability measure but only under an appropriate choice for the process $\lambda$. And yes - when you want to construct a martingale, you often first search for the process $\lambda$ and then invoke Girsanov's theorem to define the corresponding measure change.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.