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A similar question have been posted earlier but one part has remained unanswered. Let us define: $$X_t = \int_0^t W_s ds,$$

where $W_t$ is a standard Brownian Motion. Is $X_t$ an Itô process or a Riemann integral? How to write the Itô form of: $$\int_0^tW_sds\text{ ?}$$ Is the following formula correct? Why? $$d\biggl(\int_0^tW_sds\biggl) = W_tdt $$

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As usual with those kind of integrals, another way to reach the result is to:

  • Express $W_s$ in integral form as $\int_0^s dW_u$
  • Use Fubini theorem to change the integration bounds of the resulting double integral

More specifically, \begin{align} \int_0^t W_s ds &= \int_0^t \int_0^s dW_u ds \\ &= \int_0^t \int_u^t ds dW_u \\ &= \int_0^t (t-u) dW_u \end{align} which is indeed an Ito integral and in this case a Gaussian r.v. with mean zero and variance given by Ito isometry.

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  • $\begingroup$ Hello, thank you so much for the explanation, I agree that it cannot be a Reimann integral, and also the previous answer in the end finds the $\int_0^t(t-s)dW_s$ which shows it is an Ito integral by definition. could you please verify the following formula and also could you please give an explanation if it is correct? $$d\biggl(\int_0^tW_sds\biggl)=W_tdt$$ $\endgroup$ – agassi Oct 24 '19 at 8:17
  • $\begingroup$ This formula is correct. It's as if you were to apply Itô to $I_t = f(t) = \int_0^t W_s ds$. Since $$\partial f(t)/\partial t = W_t$$ from Leibniz integral rule (cf. en.wikipedia.org/wiki/Leibniz_integral_rule), we do have that $$dI_t = \partial f(t)/\partial t dt = W_t dt$$. $\endgroup$ – Quantuple Oct 24 '19 at 8:27
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    $\begingroup$ Thank you very much for your response, indeed very informative and helpful. $\endgroup$ – agassi Oct 24 '19 at 9:07
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It is indeed Riemann integrable, so you don't need stochastic integration. For a given path, you can interpret the integral in the Riemann sense. For a given t, the paths are random, so it is a random variable.

You can also express it as an Ito’s process. To see the connection, just apply ito's lemma to $tW_t$:

$d \left(tW_t\right)=tdW_t+W_tdt$

$W_tdt=d \left(tW_t\right)-tdW_t$

Then integrate:

$X_t=\int_0^t{W_sds}=tW_t-\int_0^t{sdW_s}$

$\quad =t\int_0^t{dW_s}-\int_0^t{sdW_s}$

$\quad =\int_0^t{\left(t-s\right)dW_s}$

So it is normally distributed. Easy to check mean is zero, and variance is:

$V\left[X_t\right]=\int_0^t{\left(t-s\right)^2ds}=\frac{1}{3}t^3$

Please see more detailed discussion here: Integral of Brownian motion w.r.t. time

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  • $\begingroup$ I can follow until ..."so it is normally distributed". How do you know this? $\endgroup$ – Permian Oct 24 '19 at 19:23
  • $\begingroup$ How is it easy to check that the mean is 0 and variance is $\frac{1}{3}t^3$? $\endgroup$ – Permian Oct 24 '19 at 19:24
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    $\begingroup$ This is because the integrand is non random and the integrator is brownian. Think of the discrete version: integral will be sum, so it will be a linear combination of changes in brownian, where the change in brownian has a normal distribution with mean zero, and hence the linear combination will also be normal with mean zero. Variance is just the Ito’s isometry, essentially you square the integrand and change integrator to time. $\endgroup$ – Magic is in the chain Oct 24 '19 at 19:36
  • $\begingroup$ $$Var[X_t] = E\big[X_t^2\big]-\big(E[X_t]\big)^2$$second term in the right is zero because expectation is zero. and by Itô's isometry we can write, $$E\big[\big(\int_0^t(t-s)dW_s\big)^2\big] = E\big[\int_0^t(t-s)^2ds\big]$$ $\endgroup$ – agassi Oct 24 '19 at 21:27

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