Concerning Ito's integral formula,

$$\int_0^t B(s)dB(s) = \frac{1}{2}B^2(t)-\frac{1}{2}t,$$

the MIT lecture notes give a proof that "the standard Brownian motion has a.s. finite quadratic variation which is equal to $t$."

A nice introduction by Karl Sigman notes that "since $$B(t_k)-B(t_{k-1})\sim N(0,t_k-t_{k-1})...E\left[\sum_1^n (B(t_k)-B(t_{k-1})^2 \right] =\sum_1^n (t_k-t_{k-1})=t."$$

My question is, how does the integral formula change if the diffusion constant is no longer 1/2, that is, if the distribution is non-standard normal?

(Can skip). My guess is that the moment generating function would give a different number in the MIT proof above, and the new diffusion constant could be inserted into the proof without affecting the result other than to make it messier and instead of $\sigma^2=2Dt=t$ we would have a coefficient other than 1/2 for $t?$

  • $\begingroup$ Are you asking what would happen if the variance of the white noise was something other than 1? If so, your conclusion is correct, but the level of extra messiness is pretty small, just one parameter-worth. $\endgroup$ – jbowman Oct 5 '17 at 16:46
  • $\begingroup$ @jbowman: I am not sure how the proof (as in the MIT notes) would look. The generating function trick gives a different number... $\endgroup$ – daniel Oct 7 '17 at 13:02

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