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?


2 Answers 2


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 $Q$.

  • 2
    $\begingroup$ Any other proper way to justify way is by Girsanov Theorem. $\endgroup$
    – SmallChess
    Mar 19, 2015 at 23:27
  • $\begingroup$ @StudentT I learnt girsanov theorem at university but I didn't really understand it intuitively. Thank you. $\endgroup$
    – Kian
    Mar 21, 2015 at 10:44
  • $\begingroup$ What if the stock pays dividend? I think in this case, you need to subtract the dividend from the drift term? $\endgroup$ May 22, 2019 at 21:35

I've written a blog precisely on answering this question if anyone is interested. The basic idea is that you start in the real world probability measure with a risky asset and a risk-free asset modelled by geometric Brownian motion. You then compose the two together to form another function. Ito's lemma then tells you the dynamics of this new composed function. Girsanov's theorem then comes along and takes this stochastic differential equation (SDE), as per Ito in the previous step, and the original real-world probability measure, and outputs a new SDE and a new probability measure. Then, some algebra and substitution back in to the original SDE will give you the new SDE.


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