1
$\begingroup$

In stochastic calculus, a rule of thumb for computations is $(dW_t)^2 = dt$ for a Wiener process $W_t$.

Say we have a diffusion process $dX_t = dW^1_t + X_t dW^2_t$, with $W_t^1, W_t^2$ independent (of each other) 1-dimensional Wiener processes. Under what conditions, using the above rule of thumb, can this computation be made rigorous:

$(dX_t)^2 = (dW^1_t + X_t dW^2_t)^2 = (1+X_t^2)dt \implies dX_t = \sqrt{1+X_t^2}dW_t^3$ for some Wiener process $W_t^3$.

$\endgroup$
1
$\begingroup$

Hint: Show that

$$W_t^3 := \int_0^t \frac{1}{\sqrt{1+X_s^2}} \, dW_s^1 + \int_0^t \frac{X_s}{\sqrt{1+X_s^2}} \, dW_s^2$$

defines a Brownian motion. To this end, use e.g. Lévy's characterization of Brownian motion.

$\endgroup$
  • $\begingroup$ @Did Thanks for your upvote + correcting the typo. :) $\endgroup$ – saz Sep 30 '14 at 14:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.