Suppose that $S$ follows a geometric brownian motion

$$dS=S(\mu dt+\sigma dB).$$

It is well understood that

$$S_{T}=S_{0}exp((\mu-\dfrac{\sigma^{2}}{2})T+\sigma B_{T}).$$

Method 1 (I have no problem with this)

Letting $f(S)=log(S)$ and doing a 2nd order Taylor expansion and noting that $(dB)^{2}=dt$. For example:

$$d(log(S)) = f'(S)dS+\dfrac{1}{2}f''(S)S^{2}\sigma^{2}dt=\dfrac{1}{S}(\mu dt+\sigma SdB)-\dfrac{1}{2}\sigma^{2}dt=(\mu-\dfrac{\sigma^{2}}{2})dt+\sigma dB.\quad (*)$$

It follows that

$$log(S_{T})=log(S_{0})+(\mu-\dfrac{\sigma^{2}}{2})T+\sigma B_{T}$$ and hence

$$S_{T}=S_{0}exp((\mu-\dfrac{\sigma^{2}}{2})T+\sigma B_{T}).$$

The above derivation is perfectly fine and can be found on Wikipedia for example.

Method 2 (is this allowed?)

Let's use Ito's lemma

$$dF(t,X(t))=(F_{t}'+a(t)F_{x}'+\dfrac{1}{2}b(t)^{2}F''_{xx})dt+(b(t)F'_{x})dB $$ for a process $dX(t)=a(t)dt+b(t)dB(t)$ and for a function $F(t,X(t))$. Let $F(t,X(t))=log(X(t)).$ Let $dS=S\mu dt+S\sigma dB$, and let $a(t)=S\mu$ and $b(t)=S\sigma$. This is my question: $a(\cdot)$ is a function of $t$ and not $S$, so is this still OK to do?. Then

$$dF=(0+\mu+\dfrac{1}{2}\sigma^{2}S^{2}\times(\dfrac{-1}{S^{2}}))dt+\sigma\dfrac{S}{S}dB=(\mu-\dfrac{1}{2}\sigma^{2})dt+\sigma dB.$$ Here we arrive at (*) from Method 1 and hence the result follows.

Is this derivation technically correct? Just because it gives the right answer, it does not imply the method is sound.


1 Answer 1


The second method you use is correct and, actually, is completely equivalent to the first one. The reason is that the proof of Ito's lemma relies on a Taylor expansion of the second order.

Notice that Wikipedia's formulation of Ito's lemma is a bit misleading, as they write $$ dX_t = \mu_t dt + \sigma_t dB_t $$ but, actually, the functions $\mu$ and $\sigma$ are allowed to depend on $X$, that is $$ \mu_t = \mu(t,X_t) \qquad \text{and}\qquad \sigma_t = \sigma(t,X_t). $$ Hence your application of Ito's lemma is formally correct.

PS: See for instance here for a derivation of the lemma, where the fact that $\mu$ and $\sigma$ can depend on $X_t$ is made explicit after equation (1).

  • $\begingroup$ Yes. But my question is that $a(\cdot)$ and $b(\cdot)$ are not functions of $S$ -- and whether this substitution should be allowed in the form of Ito's lemma that was given. (Indeed, Ito's lemma in general is nothing but Taylor expansion to 2nd order of $dt$ anyway. I get that... and that's fine. But using the lemma given in the original question, surely that's not fine?) $\endgroup$ Commented Mar 25, 2015 at 15:30
  • $\begingroup$ Sorry, I didn't notice this was the point: I updated my answer accordingly. Btw +1, this is the right spirit! $\endgroup$
    – Abramo
    Commented Mar 25, 2015 at 19:19

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