Why is it possible to substitute in differential form?

Question

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!

Answer 1

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.

Answer 2

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.

Answer 3

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.