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I'm trying to understand the example problem on the Wikipedia page for Ito's Lemma and need it dumbed down a little bit.

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

$$ f(S) = log(S) $$

Given Ito's lemma, below:

$$ df(t, X_t) = (\frac{\partial f}{\partial t} + \mu_t \frac{\partial f}{\partial x} + \frac{1}{2} \frac{\partial ^2 f}{\partial x^2}) dt + \sigma_t \frac{\partial f}{\partial x} dB_t $$

I don't understand where the first line of the result comes from:

$$d\log(S) = f'(S)dS + \frac{1}{2}f''(S)S^2 \sigma^2 dt$$

I can see that $$...\frac{1}{2} f''(S)S^2 \sigma ^2$$

Comes from

$$...\frac{1}{2} \frac{\partial ^2 f}{\partial x^2}) dt$$

But everything else I can't make sense of. I imagine there are some simple intermediate steps that I am not quite getting. Any help with this would be greatly appreciated, I feel like I'm close to having Ito's lemma "click" or me.

Soo
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1 Answers1

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Firsly we cannot apply Ito's lemma since $f(s):= \log s$ is not $C^2(\mathbb{R})$ (Wikipedia is unfortunately wrong on a lot of pages in stochastic analysis). It does turn out to give the correct answer anyway though.

What Wikipidea wants you to do is to apply it to $f(t,x) = \log x$. In this case $\partial_t f = 0$.

You should note that for a continuous local semimartingale $X_t= X_0 + M + B$ where $M$ is a martingale and $B$ is bounded variation, the notation $dX$ is shorthand for $dM + dB$ where the first is an Ito integral and the second is a Lebesgue-Stieljes integral. Using this shorthand, $\langle X\rangle_t = \langle M \rangle_t$ and so Ito's formula can be phrased as

$$ d f(t,X_t) = \partial_t f(t,X_t)dt + \partial_x f(t,X_t)dX_t + \frac{1}{2}\partial_{xx}f(t,X_t)d\langle X \rangle_t. $$ This form is preferable because of its similarity to Taylor's formula. In the case where $dM=R_tdW_t$ is is an integral with respect to brownian motion, so $d\langle M\rangle_t = R_t^2dt$, is is convenient to simplify further $$ = \partial_t f(t,X_t)dt + \partial_x f(t,X_t)(R_tdW_t+dB_t) + \frac{1}{2}\partial_{xx}f(t,X_t)R_t^2dt $$ $$ = (\partial_t f(t,X_t)+\frac12\partial_{xx}f(t,X_t)R_t^2)dt + \partial_x f(t,X_t)R_tdW_t + \partial_x f(t,X_t)dB_t. $$

When $f$ does not depend on $t$, and $B$ is absolutely continuous w.r.t. the Lebesgue measure, i.e. $dB_t = b_tdt$ this simplifies even further to $$ =(b_tf'(X_t)+\frac12f''(X_t)R_t^2)dt+ f'(X_t)R_tdW_t $$ which is the form you were quoting.

Applying Ito's lemma blindly anyway we have $$ d(\log S_t) = \frac{1}{S_t}dS_t + \frac12(\frac{-1}{S_t^2})d\langle S \rangle_t = \frac{1}{S_t}dS_t - \frac12 \sigma^2dt $$ $$ \sigma dB_t + \mu dt - \frac12 \sigma^2dt. $$

Thus assuming $S_0=1$, $\log(S_t)-\log(S_0) = \log(S_t) = \sigma B_t + (\mu-\sigma^2/2)t$ and hence $$ S_t = \exp( \sigma B_t + (\mu-\sigma^2/2)t). $$ To make this rigorous. We can start with that guess of $S_t$, and use $g(t,x)=e^{\sigma x+(\mu-\sigma^2/2)t}$ to apply Ito's formula and verify that $g(t,B_t)$ satisfies the original SDE.

jeffery_the_wind
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nullUser
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  • there is no point to restrict to $C^2(\Bbb R)$, the proof also applies to $C^2(U),U \subset \Bbb R$ , U open. – quallenjäger Mar 19 '19 at 12:34
  • @quallenjäger I believe you can only apply Ito on $C^2(U)$ if the process in known to a.s. live inside the open set $U$, correct? Since in this case we are dealing with an unknown solution $S$, it is not known whether $S$ lives inside $(0,\infty)$, although a fortiori this is the case. I suppose you could just assume this is the case and then note at the end that indeed $S$ lives on the set we assumed it lives on. Although I also don't remember whether the existence and uniqueness theorems apply to $U$ or only to $\mathbb{R}$. – nullUser Mar 24 '19 at 00:19
  • Yes you are right on this point. Usually we use a stopping time to keep the process in $U$ and then decide if the local martingale is a true martingale. The same concept is used for the inverse Bessel Process. And existence and uniqueness holds indeed for $U$, as the proof is first shown locally and extended to globally.(The extension is guaranteed by linear growth condition). – quallenjäger Mar 24 '19 at 00:29