Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a transition function for a Markov process $X_t$. I want to find a density function for the stochastic process $Y_t := \int_0^t X_s \,ds$. Some questions about this:

  1. Is this the same as the transition function for the stochastic process $Y'_t := \int_0^t 1 \,dX_s$? If not, what is the difference in real-world meaning between $Y_t$ and $Y'_t$?

  2. In normal calculus, one typically integrates by invoking the Fundamental Theorem and taking antiderivatives. Is there a Fundamental Theorem analog in stochastic calculus that might be useful here?

  3. My best lead so far is to use the Feynman-Kac formula on $X_t$ to find the characteristic function of the density function of $Y_t$, then invert it. The transition function for $X_t$ is very long and ugly, so this would be hugely complicated to execute. I've got Mathematica as my integration calculator, so I can handle some amount of ugliness, but this strategy was too complicated for Mathematica to stomach, and it froze up. Is there a known simpler method?

Thanks - advice on any of these three questions is appreciated.

share|improve this question
    
"Transition function" is usually used for a Markov process. Even if $X_t$ is a Markov process, in general $Y_t$ is not. –  Nate Eldredge Jan 2 '13 at 21:04
1  
To answer your first question, $Y_t^{'}$ is just $X_t - X_0$, while $Y_t$ is the (ordinary) integral of $X_t$. The real world meaning is whatever the difference between $X_t$ and its integral is in your application. There is no actual stochastic calculus here, since you are not integrating any nonconstant functions against a stochastic integrator. The ordinary fundamental theorem of calculus would be the correct one to apply here, since all of your integrals are classical (but random). The issue of course is that your functions are random so you don't have a known anti-derivative. –  Chris Janjigian Jan 2 '13 at 21:06
    
Good point, thank you. $X_t$ is indeed a Markov process, and I am only interested in a density function for $Y_t$. I've edited this in. –  GMB Jan 2 '13 at 21:06
    
Note that $\int\limits_0^tu(t)\mathrm dt$ and $\int\limits_0^tu(t)\mathrm dX_t$, used in the post for various functions $u$, are very strange objects. Perhaps one means $\int\limits_0^tu(s)\mathrm ds$ and $\int\limits_0^tu(s)\mathrm dX_s$ instead. –  Did Jan 2 '13 at 22:19

1 Answer 1

For (1), since $dX_s \equiv X_{s+ds}-X_s$, $Y'_t = \int_0^t dX_s = X_t - X_0$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.