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I previously asked a question (Expectation of Ito integral). I have additional questions on the same subject.

Let's say that we have an Ito process such as

$$ X(t)=X(0) + \int_0^t a ds + \int_0^t b X(s) dW(s) $$

where a and b are constants and W(t) is the standard Brownian motion.

Using Itô's formula for $X^2$, we have

$$ X(t)^2=X(0)^2 + \int_0^t (2 a X(s) + b^2 X(s)^2 )ds + 2ab\int_0^t X(s)^2 dW(s) $$

In my calculations I need the fact that $E[\int_0^t X(s)^2 dW(s)]=0$, which requires that $E[\int_0^t |X(s)|^4 s]<\infty$. We know that $E[\int_0^t |X(s)|^2 s]<\infty$ from the existence of the strong solution for the first SDE. Is there a link between them?

The second question is the justification of the application of Fubini as follows

$$ E[\int_0^t (2 a X(s) + b^2 X(s)^2 )ds] = \int_0^t (2 a E[X(s)] + b^2 E[X(s)^2] )ds $$

Fubini can be applied if $E[\int_0^t |X(s)|ds]<\infty$ and $E[\int_0^t |X(s)^2|ds]<\infty$. This very much resembles the first condition, however it is not entirely clear how I can tie them up (for both $X$ and $X^2$ processes). Perhaps Jensen's inequality can be used $E(X)^2\leq E(X^2)$ (but I can only use it for a partition within the integral)?

Thanks and kind regards!

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In my answer to your previous question, I mentioned Theorem 5.2.9 in Karatzas & Shreve. The proof of that theorem, part of which is relegated to Problem 5.2.10 (which has a worked solution), can be modified to yield an improved result on moment estimates. It is mostly just an exercise in using the Burkholder-Davis-Gundy inequalities, but I have worked out the details and posted them to the "Lecture notes" section of my website. Here is a direct link to the short note. In the note, you will want to use (4) with $m=2$ and $m=1$, and Corollary 3 with $r=1$. The end result is that in your example, everything will work out as long as $E|X(0)|^4<\infty$.

  • Is the post-scriptum in good taste, or even necessary? – Did Jun 04 '14 at 19:49
  • @Did I thought so. The user is obviously new. From his comment in response to my earlier answer and his post here, he seems to consider his earlier question answered to his satisfaction. I therefore considered it likely that he did not know about clicking the check mark. But thanks for mentioning it. I moved the remark to a comment in his other question. – Jason Swanson Jun 04 '14 at 21:49
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    In case it's helpful: It seems that Springer make the entire book of Karatzas & Shreve available for free online download. http://link.springer.com/book/10.1007%2F978-1-4612-0949-2 – Julian Newman Dec 28 '15 at 01:59
  • @JulianNewman: Are you sure? It looks like only the table of content and index are free to download. – Hans Jun 16 '16 at 16:20
  • @Hans: Yes, it looks like you're right. I guess that either I made a mistake, or the whole book was available for free at the time I wrote the comment. – Julian Newman Jun 17 '16 at 17:26