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The expectation of an Itô stochastic integral is zero

$$ E[\int_0^t X(s)dB(s)\,]=0 $$

if

$$ \int_0^t E[X^2(s)]ds\,<\infty $$

It is sometimes possible to check this condition directly if the Itô integrand is simple enough but how would you do it if the integrand is the process itself? For example consider the linear SDE

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

where W(t) is the standard Brownian motion and a, b are constants. How to show that this condition is satisfied for the Itô integral in this process?

Thanks!

1 Answers1

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Before you can even ask this question, you need to know that there exists a solution to the linear SDE you have written. Without the existence of a solution, the question does not make sense. The standard theorem about the existence of (strong) solutions to SDEs is proved using Picard iteration. As a byproduct of the proof, one obtains an estimate on $E[X^2(s)]$ that can be used to answer your question.

A good reference is Theorem 5.2.9 in Karatzas & Shreve. In the example you gave, the theorem states that a strong solution exists, and for each fixed $t>0$, there exists a constant $C$ that depends only on $t$, $a$, and $b$, such that \[ E[X^2(s)] \le C(1 + E[X^2(0)])e^{Cs}, \quad 0\le s\le t. \]

Jason Swanson
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  • @Kirillo You're welcome. And welcome to the Mathematics Stack Exchange. In case you were unaware, I thought I would mention that if you are satisfied with an answer to a question you post here, then you should "accept" it by clicking on the check mark next to the answer. There is no rush, you can continue to wait for better answers as long as you want. And if none of the answers satisfy you, then you do not have to accept any of them. For more info, see here: http://math.stackexchange.com/help/someone-answers – Jason Swanson Jun 04 '14 at 21:46