I'm repeatedly having problems with the differential notion of SDEs. For example I don't get why it is possible to kind of "substitute" in the short form. I will try to give an easy example of that.

Let $(X_t)_{t \in \mathbb{R}_+}$ be a geometric brownian motion, that is a stochastic process that follows the dynamic

$dX_t = X_t b dt + X_t \sigma dB_t$.

If I'm correct that is short notion for

$X_t = X_0 + \int_0^t X_s b ds + \int_0^t X_s \sigma dB_s$.

One example for a substitution that confuses me is

$\frac{1}{X_t} \color{red}d\color{red}X_\color{red}t = \frac{1}{X_t} (\color{red}X_\color{red}t \color{red}b \color{red}d\color{red}t \color{red}+ \color{red}X_\color{red}t \color{red}\sigma \color{red}d\color{red}B_\color{red}t) = b dt + \sigma B_t$.

I understand this as character by character substitution. But why is this allowed?

Is it correct that the following is the same statement in integral notion?

$\int_0^t \frac{1}{X_s}dX_s = \frac{1}{X_s} (\int_0^t X_s b ds + \int_0^t X_s \sigma dB_s) = \int_0^t b ds + \int_0^t \sigma dB_s$.

If it is correct, why is it allowed to take $\frac{1}{X_s}$ out of the integral like that (and put it back in)? I think this should not be possible, because the term depends on s.

I'm sorry if my problem is not very clear. I'm struggling to formate it precisely, as english is not my first language.

Thank you for your help in advance!

  • $\begingroup$ Taking in and out $1/X_{s}$ is not allowed, but apart from that - why would you want that in the first place? $\endgroup$
    – Tobsn
    Sep 28, 2022 at 16:50

3 Answers 3


Take the integral form more generally as $$ X_c-X_a=\int_a^cX_tb\,dt+\int_a^cX_tσ\,dB_t. $$ Then you can combine several instances of this formula to get the "substitution" formula for piecewise constant factors. In the limit you get all factors that are integrable in $t$ and adapted to the filtration, even those that are functions of $X_t$ itself.

Or you could try to make the infinitesimal version more meaningful, writing $$ X_{t+dt}=X_t+X_tb\,dt+X_tσ\,dB_t=X_t(1+σ\,dB_t+b\,dt) =X_t\exp(\ln(1+σ\,dB_t+X_tb\,dt))\\ =X_t\exp(σ\,dB_t+b\,dt-\tfrac12σ^2\,dB_t^2+o(dt)) $$ and with $dB_t^2=dt$ a.s. in the sum of many infinitesimal segments the usual solution formula for the geometric Brownian motion follows.

  • $\begingroup$ If I understand you correctly does that mean, that I cannot show $\frac{1}{X_t} dX_t = b dt + \sigma d W_t$ formally by using the integral notion, and that I have to use one of your two statements to show the "usual solution formula for geometric Brownian motion"? $\endgroup$ Sep 28, 2022 at 9:09
  • 1
    $\begingroup$ The correct way is to use the Ito transformation theorem using $Y=\ln(X)$ inspired by $dX/X$. In general it does not work that way, the transformation, if it exists, is not that obvious. $\endgroup$ Sep 28, 2022 at 9:21
  • $\begingroup$ And you get full well $\int_a^c\frac{dX_t}{X_t}=σ(B_c-B_a)+b(c-a)$, only that the left side is not a "complete" integral. While the intuitive view is that $σdB$ is a perturbation, some small add-on, locally the $dB$ term dominates the $dt$, as in the small numbers $\sqrt{dt}$ is huge against $dt$. Thus the need for corrections like in the Ito formula. $\endgroup$ Sep 28, 2022 at 10:13

The differential form $$ dX_t = f(t, X(t)) dt + \sigma(t, X(t)) d W(t), $$ is a shorthand notation for the rigorous integral form $$ X(t) = X(0) + \int_0^t f(s, X(s)) ds + \int_0^t \sigma(s, X(s)) d W(s). $$ Manipulations such as $$ \frac{dX(t)}{X(t)} = bdt + \sigma dW(t), $$ should be always considered at a formal level and are not rigorous. In particular, the form above is employed to "make the guess" that $$ d\log(X(t))= bdt+\sigma dW(t), $$ and hence derive applying Itô formula to $f(x)=\log(x)$ the solution of the SDE - the geometric BM.

  • $\begingroup$ I'm afraid I still don't quite understand. Could you maybe elaborate, what you mean with "Manipulations such as [...] should always be considered at a formal level and are not rigorous"? Does that mean that there is no integral form such that $\frac{dX(t)}{X(t)} = bdt + \sigma dW(t)$ is a mathematically correct statement? $\endgroup$ Sep 28, 2022 at 9:03
  • $\begingroup$ Well dividing by $X(t)$ is a formal operation. Who told you $X \neq 0$? $\endgroup$
    – G. Gare
    Sep 30, 2022 at 7:05

The validity of this kind of differential substitution must be proven. For example, see Corollary 3.2.20 in Karatzas and Shreve, which says (informally) that if $$ N(t) = \int_0^t X(s)\,dM(s), $$ then $$ \int_0^t Y(s)\,dN(s) = \int_0^t X(s)Y(s)\,dM(s). $$ Not every integral satisfies this property. There are in fact ordinary integrals having nothing to do with stochastic processes for which this property fails. For instance, one can define a Stratonovich-type integral for cadlag, bounded variation functions. Such an integral does not satisfy the above property. For such an integral, the act of making the differential substitution introduces a correction term analogous to that in the Ito formula. For details, see Theorem 3.36 in these notes.


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