$S_t$ is the random variable representing the risky asset price at time $t$.

M_t is the riskless asset. They are governed by the equations

$\frac{dS_t}{dt}=\mu dt + \sigma dZ_t$ and

$dM_t = rM_t dt$

where $Z_t$ is Brownian motion. If we define the discounted risky asset by $S_t^{*}=S_t/M_t$. How does the process $S_t^{*}$ become governed by

$\frac{dS_t^{*}}{dt}=(\mu-r) dt + \sigma dZ_t$ ?

I cannot see why you subtract $rdt$.

  • $\begingroup$ Do you mean S(t)udt in the GBM equation? $\endgroup$
    – SmallChess
    Commented Aug 18, 2014 at 4:30

1 Answer 1


$$ \textbf{Preface} $$ I am assuming log normal asset but this is not clear from the question? Or rather I have misinterpreted the question!

Well as I see it from a a purely mathematical exercise $$ d\left(\dfrac{S_t}{M_t}\right) =\frac{1}{M_t}dS_t - \frac{S_t}{M_t^2}dM_t +O(dt^2) $$

using Ito's lemma.

Then we can sub in the original processes yields

\begin{align} d\left(\dfrac{S_t}{M_t}\right)&=&\frac{1}{M_t}S_t\left(\mu dt + \sigma dZ_t\right) - \frac{S_t}{M_t}\frac{1}{M_t}\left(M_t r dt\right)\\ &=& S^{*}_t\left(\mu dt +\sigma dZ_t\right) - S^{*}_trdt \\ &=& S^*_t\left[(\mu-r)dt+\sigma dZ_t\right] \end{align}

or finally $$ \frac{dS^*_t}{S^*_t} = (\mu-r)dt+\sigma dZ_t $$


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