What is the formula for a likelihood ratio $L$ that transforms martingale Geometric BM given by $dS = \sigma S \,dW_t$ to Geometric BM with positive growth $dS = \mu S \, dt + \sigma S \, dW_t$?
My attempt is to use the fact that regular brownian motion can be changed to brownian motion with drift $a$ using the likelihood ratio $$ L = \exp\left( \int_0^T a \, dW_t - \frac{1}{2} \int_0^T a^2 \,dt\right) $$ so $dW_t \to d\hat W_t = a \, dt + dW_t$. Setting $a = \mu / \sigma$ should provide the proper likelihood ratio. However, in my numerical tests, this doesn't seem to work. Is it correct?
Attempt 2
If we approximate the geometric brownian motion using discrete time intervals, we get conditional probabilities between intervals that are Gaussian. Evaluating the exponent in the limit as $\Delta t \to 0$, we find $$ L = \exp \left( \int_0^T \frac{\mu}{S_t \sigma^2} dS_t - \frac{1}{2} \int_0^T \frac{\mu^2}{\sigma^2} dt \right). $$ I think this likelihood ratio should work for our purposes.