"The drift of stock price becomes the risk-free interest rate" under RNP

Assume that the evolution of a stock price is geometric Brownian Motion

$$ dS=\mu Sdt+\sigma SdW(t) $$

where $S$ is the stock price at time $t$ (current time). It says in my book that "under the risk-neutral probability measure, the drift of stock price $\mu$ becomes the risk-free interest rate $r$" and it writes

$$ dS=rSdt+\sigma SdW^{*}(t). $$ where $W^{*}$ is a B.M. under RNP $Q$ (risk neutral probability).

Is the above equation definitely correct? Is there a justification for this?

Answer

Yes, you may as well take this as the definition of the risk-neutral probability $Q$.

I will now try to give you some intuition for that kind of construction.

Assume the risk-free interest rate $r$ is constant and that the world ends at time $T$. Suppose you have a security $B=B_t$ which is riskless, i.e. which follows the dynamics $$ dB/B = r \, dt $$ so that, since $ dB/B = d \ln B $, you can easily see that $B_t = B_0 e^{rt}$. In other words, the process $B_t$ grows at the same speed of the risk-free rate. For this security, the price at time zero is $B_0$, which coincides with the discounted value of its expected payoff: $$ e^{-rT} E[B_T] = e^{-rT} E[B_0 e^{rT}] = e^{-rT} B_0 e^{rT} = B_0 \,.$$

Now consider a stock which is risky, as it follows the dynamics $$ dS/S = \mu \, dt + \sigma\,dW $$ with $\mu > r$ and $\sigma$ constant and $dW$, a standard Brownian Motion, being the source of risk. This time the process $S$ grows in expectation with speed $\mu$, and its discounted expected payoff $$ e^{-rT}E[S_T] = e^{-rT} S_0 e^{\mu T} = S_0 e^{(\mu-r) T} > S_0 $$ is bigger that its current value $S_0$. Why is it so? Well, because there's some risk involved in holding $S$, so that its price should be lower w.r.t. a riskless security! This way the investor who buys the stock at time $0$ will be compensated for bearing this risk, i.e. he will pocket a risk premium. The risk-neutral probability $Q$ is the one which gives the right price when you look at the discounted expected payoff, i.e. $$ S_0 = e^{-rT}E^Q[S_T]\,. $$ If you followed my reasoning so far, it should now be clear that $Q$ is that probability for which $$ dS/S = r\, dt + \sigma\, dW^Q $$ with $dW^Q$ being a Brownian Motion under $Q$.