3
$\begingroup$

Let $X_t,t\in T $ continuous time stochastic process. What is the meaning of $dX_t$ which is differentiation of $X_t$? Does that mean $X_{t+dt}$ and $X_t$ are random variables so $dX_t \approx X_{t+dt}-X_t$ is also random variable?

$\endgroup$

1 Answer 1

0
$\begingroup$

In the context of stochastic differential equations, the meaning of $dX_t$ is "derived" from the definition of corresponding integrals $\mathcal I[f](\omega)=\int_S^Tf(t,\omega)dX_t(\omega)$. The integral is defined by suitable approximation of $f$ by elementary functions $\phi$ of the form $\phi(t,\omega)=\sum_j e_j(\omega)\cdot\chi_{[t_j,t_{j+1})}(t)$.

However, the problem is that the actually used suitable approximation strongly influences the definition of the resulting integral and its properties. The Ito integral corresponds more or less to using $e_j(\omega)=f(t_j,\omega)$, while using $e_j(\omega)=f(\frac{t_j+t_{j+1}}{2},\omega)$ leads to the Stratonovich integral.

If I try to translate these definitions back into your language/notation, the Ito integral would more or less correspond to your suggestion $dX_t \approx X_{t+dt}-X_t$, while the Stratonovich integral would correspond more or less to $dX_t \approx X_{t+dt/2}-X_{t-dt/2}$.

$\endgroup$
2
  • $\begingroup$ thank you for you answer, in classic calculus $dX_t=X'dt$ represents the amount of the tangent line rises or falls when $t$ changes by amount $dt$. in stochastic calculus how can we interperet $dX_t$? $\endgroup$
    – cabri61
    Dec 19, 2013 at 13:04
  • $\begingroup$ @cabri61 Because Brownian motion is (in a suitable sense) nowhere differentiable, thinking about tangent lines won't help. However, if you understand the Stieltjes integral and the corresponding Stieltjes measure, then you might be able to appreciate an interpretation of $dX_t$ as a (stochastic) measure. (I say stochastic measure, because it's $dX_t(\omega)$, if we are careful about notation. I don't know whether $dX_t(\omega)$ is almost surely a Stieltjes measure for fixed $\omega$, but I guess not.) $\endgroup$ Dec 19, 2013 at 13:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .