I'm working my way through an elementary stochastic calculus textbook. I'm having trouble with one of the questions:

Bachelier type stock price dynamics. Let the SDE for stock price $S$ be given by $dS(t) = \mu dt + \sigma dB(t)$, where $\mu$ and $\sigma$ are constant. Derive the SDE for $S$ under the money market account as the numeraire.

I interpret as follows: they want $dS^*$, where $S^* = S e^{-rt} $, where $r$ is the market rate. Applying the usual rules for stochastic differentials I arrive at:

$dS^* = (\mu e^{-rt} - r S^*) \ dt + \sigma dB $


$dS^* = (\mu - r S) e^{-rt} dt + \sigma dB$

Then turn this into a martingale by a suitable Girsanov transformation, and specify the expression for the Radon–Nikodym derivative.

Here I'm unsure how to proceed. Help much appreciated. What I've tried so far was defining a new Brownian that depended on the stock price but that was inconsistent / did not make sense in the end.

What I've worked with previously either had the stock price on every term (so that it could be taken out) or not at all. Rest of the question, if it helps below:

Thereafter derive the SDE for the undiscounted stock price and solve that SDE. Finally, using this latest stock price, derive the expected value of max[S(T ) − K, 0] where K is a constant.


1 Answer 1


You forget the $e^{-rt}$ term in the diffusion: $$dS^*=-rS^*dt+e^{-rt}dS=(\mu e^{-rt}-rS^*)dt+e^{-rt}\sigma B^\mathbb{P}(t)$$ Using Girsanov we can write $B^\mathbb{P}(t)=B^\mathbb{Q}(t)+\phi(t)dt$, where $\phi(t)$ is the Girsanov kernel and the superscript denotes the measure for the Brownian Motion. Under $\mathbb{Q}$ the dynamics are $$dS^*=(\mu e^{-rt}-rS^*)dt+e^{-rt}\sigma (B^\mathbb{Q}(t)+\phi(t)dt)=e^{-rt}(\mu-rS+\sigma\phi(t))dt+e^{-rt}\sigma B^\mathbb{Q}(t)$$ We know that the drift must be zero for the discounted price process to be a martingale under $\mathbb{Q}$, so as $e^{-rt}>0$ $$\mu-rS+\sigma\phi(t)=0\quad\iff\quad \phi(t)=\frac{rS-\mu}{\sigma}$$ This depends on the stock price itself, so it is not straight forward to find the Radon-Nikodym derivative (as in Black-Scholes where the Girsanov kernel is constant). $$S(t)=S(0)+\int_0^t dS(u)=S(0)+\mu t+\sigma W(t)$$ It seems like $S$ starts in 0 as this is not part of the original SDE, so $S(0)=0$. The Radon-Nikodym derivative is given by $$L(t)=\exp\left(\int_0^t \phi(u)dW(u)-\frac{1}{2}\int_0^t(\phi(u))^2du\right)$$ with $\phi(u)=\frac{r\mu u+r\sigma W(u)-\mu}{\sigma}$. It is definitely possible to calculate the above integrals, but it will be tedious. There might be an easy way to calculate the two integrals using some trick, but the complexity of the expression could also be due to an error on my side :)

The last question can be answered by using the distribution of $S$ under $\mathbb{Q}$ which depend on the Girsanov kernel.

  • $\begingroup$ I see. So you are allowed to have S itself in the transform. A couple of questions though: 1) The answer I reach is $\frac{d \phi}{d t} = \frac{\mu -r S}{ \sigma} $. Is this what you meant? Thanks for catching the missing exponent. $\endgroup$
    – nonseqseq
    Commented Apr 28, 2021 at 20:10
  • $\begingroup$ Actually disregard the last comment -- just comes down to a slight notation difference between you and I. I get the following for the Radon-Nikodym derivative: $Z = \exp [-\frac{B}{\sigma} ( \mu - \frac{r}{t} \int_0^t S dt) -\frac{1}{2 \sigma^2}( \mu - \frac{r}{t} \int_0^t S dt)^2 ]$. Does this look correct? $\endgroup$
    – nonseqseq
    Commented Apr 29, 2021 at 2:09
  • $\begingroup$ Looking at it again, it is not trivial to find the Radon-Nikodym derivative. I have edited my answer to reflect this. $\endgroup$
    – mmencke
    Commented Apr 29, 2021 at 21:45

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