8
$\begingroup$

What's the difference between optional quadratic variation (which sometimes is denoted by $ [M]$) and predictable quadratic variation (i.e $\ < M > $) of a stochastic process?

$\endgroup$
1

1 Answer 1

13
$\begingroup$

Hi here are my two cents :

$<>$ is the predictable compensator of $[]$ process ( so the difference of both is a local martingale).

The difference is best illustrated by examining a Poisson process of intensity $\lambda$.

Then the $[N]_t=\sum_{s\le t}(\Delta N_s)^2=\sum_{s\le t}(\Delta N_s)=N_t$ this process is not predictable as its jumps are inaccessible.

But $<N>_t=\lambda.t$ (comes from standard calculations) this process is predictable as it is deterministic.

And finally $[N]_t-<N>_t=N_t -\lambda.t$ is a (local) martingale.

You should take a look at George Lowther fantastic Blog "almost sure"

Best regards

$\endgroup$
2
  • $\begingroup$ Why is $\sum_{s\le t}(\Delta N_s)^2=\sum_{s\le t}(\Delta N_s)$? $\endgroup$
    – abc
    Commented Jan 10, 2023 at 23:40
  • $\begingroup$ Because Poisson process jump sizes are equal to 1. $\endgroup$
    – TheBridge
    Commented Jan 11, 2023 at 13:19

You must log in to answer this question.

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