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I have limited knowledge of the theory of stochastic processes. While working on a problem I've stumbled upon some expected values of time integrals of Gaussian stochastic processes. Before starting to delve into the literature on stochastic processes, I'd like to know if one can say anything (exact or approximate) about quantities like

$$ \left\langle \int_0^Tdt_1\int_0^{t_1}dt_2 B(t_1)B(t_2) \right\rangle $$

or

$$ \left\langle \left(\int_0^Tdt_1\int_0^{t_1}dt_2 B(t_1)B(t_2)\right)\left(\int_0^Tdt_3\int_0^{t_3}dt_4 B(t_3)B(t_4)\right) \right\rangle $$

where $B(t)$ is a realization of a stationary Gaussian process with zero mean and a known autocorrelation function $K(t_1-t_2) = \langle B(t_1)B(t_2) \rangle $.

Did
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Pincopallino
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1 Answers1

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The question is asking about the first and second moments of the random variable $$X=\int_0^Tdt_1\int_0^{t_1}dt_2 B(t_1)B(t_2).$$ By symmetry, $$2X=\int_0^Tdt_1\int_0^Tdt_2 B(t_1)B(t_2)=Y^2,$$ where $$Y=\int_0^Tdt_1B(t_1).$$ Since the family $(B(t))$ is centered gaussian, $Y$ is centered gaussian hence $$X=\langle X\rangle Z^2,$$ where $Z$ is standard gaussian. Note that $$\langle X\rangle=\int_0^Tdt_1\int_0^{t_1}dt_2 \langle B(t_1)B(t_2)\rangle=\int_0^Tdt_1\int_0^{t_1}dt_2 K(t_1-t_2),$$ hence the first expectation you are asking about is $$\langle X\rangle=m(T),\qquad m(T)=\int_0^T(T-t)K(t)dt.$$ Finally, $\langle Z^2\rangle=3$, hence the second expectation you are asking about is $$\langle X^2\rangle=\langle X\rangle^2\langle Z^2\rangle=3m(T)^2.$$

Did
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  • Thank you for your answer. However, I don't understand why you say that $\langle B(t_1)B(t_2)\rangle=t_2$. For me, $\langle B(t_1)B(t_2)\rangle$ the quantity is the autocorrelation function of the stochastic process $B(t)$ and is the defining property of the stochastic process considered. E.g., for an Ornstein-Uhlenbeck process, $\langle B(t_1)B(t_2)\rangle = \exp(-\gamma |t_1-t_2|)/\gamma$. – Pincopallino Dec 20 '14 at 16:56
  • Suggestion: check a definition of Brownian motion. – Did Dec 20 '14 at 20:15
  • Ok, but I'm not dealing with Brownian motion. I'm dealing with a stationary Gaussian process with a given autocorrelation function $K(t_1-t_2) = \langle B(t_1)B(t_2) \rangle. – Pincopallino Dec 21 '14 at 09:29
  • Oops, somehow I convinced myself that your B stood for Brownian motion. Sorry about that. The edited version deals with the setting you describe. – Did Dec 21 '14 at 18:17
  • OK, thank you very much for your help! – Pincopallino Dec 22 '14 at 08:10
  • I just edited the question with another integral that is related to the other ones. Could you please have a look at it? – Pincopallino Dec 22 '14 at 15:40
  • Not good. New problem? Then, new question. – Did Dec 23 '14 at 00:10
  • Sorry, I thought that another question would be closed being very similar to this one. Here it is http://math.stackexchange.com/questions/1078510/expected-value-of-time-integral-of-a-gaussian-process – Pincopallino Dec 23 '14 at 07:28