0
$\begingroup$

I was reading George Lowthers notes on Stochastic Calculus and , he says the following but I cannot figure out what it exactly means?

http://almostsure.wordpress.com/2009/11/08/filtrations-and-adapted-processes/#comment-4727

He writes the following after defining a Previsible Process and a Predictable $\sigma$-algebra

Given any set of real-valued functions on a set, which is closed under multiplication, the set of functions measurable with respect to the generated sigma-algebra can be identitified as follows. They form the smallest set of real-valued functions containing the generating set and which is closed under taking linear combinations and increasing limits. So, for example, the predictable processes form the smallest set containing the adapted left-continuous processes which is closed under linear combinations and such that the limit of an increasing sequence of predictable processes is predictable.

$\endgroup$
1
$\begingroup$

It is an identification. Call the generated $\sigma$-algebra $A$, the set of functions being used for the generation $F_0$, and the set of $A$-measurable real-valued functions $F$. Then $A$ is the smallest $\sigma$-algebra such that:

  1. Any function in $F_0$ is $A$-measurable.
  2. Any linear combination of members of $F$ is $A$-measurable.
  3. Any increasing limit of a sequence of members of $F$ is $A$-measurable.

The last sentence is an application of this identification.

$\endgroup$
  • $\begingroup$ Thanks for the explanation. I do not understand why just restrict A-measurability to only Linear combinations of members of F. I mean if $f^-1(B) \in A$, where $is say a borel set$ wouldnt $f^k$ be measurable too. I mean I can easily see the logic behind the definition of a predictable process in a discrete case but the above definition rationale still eludes. I do understand what is being said but I can't make out the big picture. And how is 3 implied from the identification? $\endgroup$ – user3503589 Oct 27 '14 at 0:38
  • 1
    $\begingroup$ @user3503589 I meant "the last sentence of your paragraph is an application of this identification". Part 3 is built into the identification. $\endgroup$ – Ian Oct 27 '14 at 0:51
  • $\begingroup$ @Sorry if this is a stupid question but what's an identification? $\endgroup$ – user3503589 Oct 27 '14 at 1:01
  • 1
    $\begingroup$ You have a mathematical object. It is usually abstractly defined and difficult to understand at first glance. To help understand it, you identify it with a simpler object, meaning that you prove that your given object is actually the same as the simpler object. I would say the Riesz representation theorem is the most famous result of this sort. It takes the dual space of a Hilbert space $H$, which is completely abstractly defined, and identifies it with $H$, modulo a very simple, explicitly defined isomorphism. $\endgroup$ – Ian Oct 27 '14 at 1:06
  • $\begingroup$ @Thank you so much Ian for the nice explanation. $\endgroup$ – user3503589 Oct 27 '14 at 1:14

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.