0

I am a beginner in Ito process, and I encountered a question as follows. Can anyone convince me (may be less rigorous) that $\int_0^t \sigma_t \mathrm{d}W_t$ is a martingale? Here $(W_t)$ is a Wiener process and $\sigma_t$ is a predictable process. (Just consider 1-dimension is Ok)

I have heard that it is just a local martingale unless we assume some properties of $\sigma_t$. So what conditions can we impose? It is enough to assume that $\sigma_t$ is bounded?

Dreamworld2001
  • 73
  • Make your life simple and assume first that $\sigma_t$ is a bounded piecewise continuous function. Then show directly that the finite sum (to which your integral collapses) is a martingale. In a next step you may want to show that the Riemann-Stieltjes-type sums which Ito-san used to define $\int_0^tW_s,dW_s$ are martingales also. For beginners these exercises are much more important than getting headaches when something is a martingale and not just a local martingale. – Kurt G. Oct 16 '22 at 06:37
  • You can have a look here : https://math.stackexchange.com/questions/38908/criteria-for-being-a-true-martingale/38947#38947 – TheBridge Oct 16 '22 at 07:54

0 Answers0