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I am trying to simplify the following expression:

$X(t)-Y(t)=\int\mu(X\{s\}-Y\{s\})ds+\int\sigma(X\{s\}-Y\{s\})dB\{s\}$

$|X\{t\}-Y\{t\}|\le\int|\mu(X\{s\}-Y\{s\})|ds+\int|\sigma(X\{s\}-Y\{s\})|dB\{s\}$

looking at the second moment gives me:

$E|X\{t\}-Y\{t\}|^2\le\int\mu^2 E|X\{s\}-Y\{s\}|^2ds^2+\int\sigma^2 E|X\{s\}-Y\{s\}|^2ds$

I want to get the expression into the following form:

$E|X\{t\}-Y\{t\}|^2\le A+B\int E|X\{s\}-Y\{s\}|^2ds$

I figured it out, but i cannot figure out how to show that the right hand side is finite

E|X{s}-Y{s}|^2

is finite

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Can that second equation ever be true ? The second term on rhs is not positive. –  mike Aug 22 '12 at 20:47
    
Peter: Is there any chance you might answer @mike's comment? –  Did Aug 25 '12 at 21:32

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