Why do we have zero drift when switching to a martingale measure?

Question

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?

Answer 1

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.