Girsanov Theorem application to Geometric Brownian Motion

Question

I recently read this from a book on mathematical finance

The important example for finance the (unique) EMM for the geometric Brownian. Let $S_{t}$ be the price of an asset, $${{d{S_t}} \over {{S_t}}} = \mu dt + \sigma d{W_t}$$ and let $r \ge 0$ be the risk-free rate of interest. For the exponential martingale $${Z_t} = \exp \left( { - {t \over 2}{{\left( {{{r - \mu } \over \sigma }} \right)}^2} + {{r - \mu } \over \sigma }{W_t}} \right)$$ the process $W_t^Q \buildrel\textstyle.\over= {{\mu - r} \over \sigma }t + {W_t}$ is Q-brownian motion, and the price process satisfies, $${{d{S_t}} \over {{S_t}}} = rdt + \sigma dW_t^Q$$ Hence, $${S_t} = {S_0}\exp \left( {\left( {r - {1 \over 2}{\sigma ^2}} \right)t + \sigma W_t^Q} \right)$$ and ${S_t}{e^{ - rt}}$ is a Q-martingale

but why is this so important , the result is just the solution to the ${{d{S_t}} \over {{S_t}}} = \mu dt + \sigma d{W_t}$?

Answer 1

Denote $B_t=e^{rt}$ the discount factor. Requiring $S_t/B_t$ to be a martingale it would mean the equation $S_0/B_0=E[S_t/B_t]$ hold. Therefore we can calculate the price of an option by discounting the expectation value at the maturity. If $S_t/B_t$ is not a Q-martingale, then we cannot discount the expectation value, which make calculation of $S_0$ extremely difficult.

Another thing to mention: it is not a solution to $dS_t/S_t=\mu*dt+\sigma*dW$. It is a solution to $dS_t/S_t=r*dt+\sigma*dW^Q$. The first dynamic doesn't satisfy risk-neutral-pricing assumption, namely no-arbitrage assumption. The second dynamic is the right dynamic for risk-neutral-pricing. That's why we need girsanov theorem to transform the dynamic.