Let
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space
- $\mathbb F=(\mathcal F_t)_{t\ge 0}$ be a filtration on $(\Omega,\mathcal A)$
- $B=(B_t)_{t\ge 0}$ be a Brownian motion on $(\Omega,\mathcal A,\operatorname P)$ with respect to $\mathbb F$
- $H=(H_t)_{t\ge 0}$ be $\mathbb F$-adapted and $\mathbb F$-progressively measurable with $$\operatorname E\left[\int_0^\infty H_t^2\right]<\infty$$
Under this conditions, the Itô integral of $H$ with respect to $B$ $$\int_0^\infty H_s\;dB_s$$ is well-defined. Now, let
- $b,\sigma:[0,\infty)\times\mathbb R\to\mathbb R$ be Borel measurable
and suppose we are considering a "process" $X=(X_t)_{t\ge 0}$ whose local behavior in time can be described by a differential equation $$dX_t=b\left(t,X_t\right)dt\;.\tag{1}$$ However, maybe $(1)$ is not accurate cause the process is disturbed by a random influence.
The Brownian motion at time $t$ is $\mathcal N_{0,\;t}$-distributed and the increments $B_t-B_s$ are $\mathcal N_{0,\;t-s}$-distributed. So, since the normal distribution occurs almost naturally in many practical problems, it makes sense to me, that we somehow want to integrate $X$ by $B$ and add this term to $(1)$.
Lebesgue–Stieltjes integration would make perfectly sense to me. The integration interval would be weighted by a normally distributed factor (the increments of the Brownian motion).
However, we all know that Lebesgue–Stieltjes integration with the Brownian motion as the integrator is impossible.
So, we use the Itô integral and model our problem by $$dX_t=b(t,X_t)dt+\sigma(t,X_t)dB_t\tag{2}\;.$$ While I know, that the Itô integral has beautiful properties like being a continuous $\mathbb F$-martingale, I ask myself in which terms $(2)$ is still appropriate for our problem.
Let $X$ be the solution of $(2)$. How can we motivate, that $X$ is really a solution for our problem? $X$ should be the solution of $(1)$ (which is the perfect model function without any disturbance) plus some normally distributed distortion whose intensity depends on $\sigma$.
Honestly, I don't see that $X$ has this property.