I was wondering if someone could recommend a reference that deals with time integrals of diffusion processes.

Suppose $X$ is an Ito diffusion process with dynamics $dX_t = \mu(X_t)dt + \sigma(X_t)dW_t$. The process I'm interested in is $Y_t = \int_0^t X_s ds$. I haven't seen any treatment of the properties of $Y$ in the better-known texts on stochastic analysis - perhaps someone on MO can help.

I'll give a simple example to try to explain part of the reason I'm interested. Suppose $dX^{(1)} = dW_t^{(1)}$ and $dX^{(2)} = \sigma dW_t^{(2)}$, where $W^{(1)}$ and $W^{(2)}$ are independent Brownian motions. $X^{(1)}$ has quadratic variation $t$ almost surely, and $X^{(2)}$ has quadratic variation $\sigma t$. Thus, for $\sigma \neq 1$ the process laws are not equivalent.

I'm wondering what this implies for the laws of $\int^t X^{(1)}_s ds$ and $\int^t X^{(2)}_s ds$. Intuitively, integration should "hide" the small oscillations of the sample paths. Is it possible that the integrated processes have equivalent laws?

  • 1
    $\begingroup$ Does any of the answers below correspond to what you were asking for? $\endgroup$
    – Did
    Commented Apr 11, 2011 at 17:11
  • 1
    $\begingroup$ None of the references were as systematic as I would have liked, but I largely stopped pursuing this line of inquiry when you pointed out that the laws of the integrated processes are still singular. $\endgroup$ Commented Apr 11, 2011 at 23:46
  • $\begingroup$ related: MathSE, math.stackexchange.com/q/4431021/532409 $\endgroup$
    – Quillo
    Commented Apr 19, 2022 at 12:22

4 Answers 4


One can adapt the argument used to show that, for a standard Brownian motion $W$, the laws of $W$ and $\sigma W$ on any interval $[0,t]$ with $t > 0$ and $\sigma^2\ne1$ are singular.

For every positive $v$, let $E_v$ denote the space of $C^1$ real valued functions defined on $[0,t]$ such that the quadratic variation of their first derivative on $[0,t]$ exists and equals $v$. Let $X=(X_s)_{0\le s\le t}$ with $X_s=\displaystyle\int_0^sW_u\mathrm{d}u$. Then $[X\in E_{t}]$ and $[\sigma X\in E_{\sigma^2t}]$ are both almost sure events but $E_t$ and $E_{\sigma^2t}$ are disjoint hence the laws of $X$ and $\sigma X$ are singular.

  • $\begingroup$ Yes, you're absolutely right. Thanks for pointing that out. It's strange that processes of this type do not appear much in the literature though. $\endgroup$ Commented Jan 19, 2011 at 13:13
  • $\begingroup$ Hmmm... A keyword which might help you here is Langevin processes. Bertoin published at least two papers about the influence of reflecting boundaries on these processes. You could also have a look at the recent paper by E. Jacob (Annales IHP Probab. Stat.) dealing with their excursions. And I should have mentioned explicitly in my answer these two related MO questions: mathoverflow.net/questions/51103 and mathoverflow.net/questions/51090. $\endgroup$
    – Did
    Commented Jan 19, 2011 at 13:57

This is not really a full answer, but depending on your needs, can be somewhat helpful. The time integral of a Brownian motion has been studied, for the purposes of a specific problem, in the following paper:

A heuristic, physicist's summary of Sinai's arguments can be found here:

Yet later there was a series of Toufic Suidan's papers on the subject, you can search for this name on the arXiv. Look also for other citations of Sinai's paper.


I believe you can use integration by parts to express \int X_s ds as -\int s dX_s + Boundary terms.

This is then a stochastic integral of the type commonly dealt with.

Edit: I am not 100% sure that what I suggested is correct (though I swear I saw something like this in a class)... but now I would like to make sure I get the correct understanding in my mind.

My understanding at the moment is:

To make proper sense of a Reimann-Stieltjes integral of the form \int f dg, you need that one of f and g be continuous and the other be of bounded variation. Which is which doesn't matter because you can define the other from the first via integration by parts.

Since W_t is continuous a.s., you can then define \int h(t) dW_t omega-wise a.s. provided h(t) is of bounded variation. This way of defining "stochastic integration" fails however for \int W_t dW_t since neither of the pieces is of bounded variation... hence the need for more advanced notions of stochastic integration.

However, I believe that the different notions of stochastic integration coincide when the integrand is of bounded variation. And so, provided your integrand was of bounded variation, you could think in terms of RS-integration. And therefore the integration-by-parts I suggested would be legitimate.

Zhoraster's observation does raise some concern (although it could be that the sum of two non-abs continuous functions is abs continuous) so now I am curious if my mental picture is wrong.

  • 1
    $\begingroup$ This is not very helpful. $\int_0^t X_s ds$ is absolutely continuous, while both $t X_t$ and $\int_0^t s dX_s$, which one gets by ibp, are not. So they in fact say nothing about the initial integral. $\endgroup$
    – zhoraster
    Commented Jan 19, 2011 at 7:05

My Stochastic calculus professor always used to say "When in doubt use Ito"

So let $f(t,x) = t x $ and compute $\partial_t f(t,x) = x$, $\partial_x f(t,x) = t $ and $ \partial_{xx} f(t,x) = 0$

Now the Ito lemma says for $f$ twice differentiable with respect to $x$ and once differentiable with respect to $t$ then the following formula holds for any Ito process:

$f(t, X_t) = f(0,X_0) + \int_{0}^{t}\partial_s f(s,X_s) ds + \int_{0}^{t}\partial_x f(s,X_s) dX_s$ $+ \int_{0}^{t}\partial_{xx} f(s,X_s) d \left< X,X \right>_s $

So applying the above fact to the function $f(t,x) = tx$ gives:

$t X_t = 0 + \int_{0}^{t}X_s ds + \int_{0}^{t} s dX_s + 0$ or to give you a starting answer $\int_{0}^{t}X_s ds = t X_t - \int_{0}^{t} s dX_s$

Edit 3: The following only holds now if $X_t$ is a Gaussian process which is not true in general... So in vague words we have that The (Riemann) integral of an ito process is equal to the difference of two Gaussian processes which should again be Gaussian .... (if all this logic is correct it should then suffice to characterize the processes covariance structure in order to have a complete understanding of the law of the process.)

For example one can compute the variance if $X_t$ is standard Brownian motion:

$\mathbb{E}[(\int_{0}^{t}X_s ds)^2] = t^2 \mathbb{E}[(X_t)^2] -2t \mathbb{E}[X_t \int_{0}^{t} s dX_s] + \mathbb{E}[(\int_{0}^{t}sdX_s)^2] $

By the Ito isometry we have $\mathbb{E}[(\int_{0}^{t}sdX_s)^2] = \int_{0}^{t}s^2ds = t^{3}/3$.

To compute $\mathbb{E}[X_t \int_{0}^{t} s dX_s]$ notice first that the Ito integral of a deterministic function is always a Gaussian process. EDIT: Shavi has given that $\mathbb{E}[X_t \int_{0}^{t} s dX_s] = t^2/2$

$\mathbb{E}[(\int_{0}^{t}X_s ds)^2] = t^3 - t^3 + \frac{t^{3}}{3} = \frac{t^3}{3}$

Computing the covariance $\mathbb{E}[\int_{0}^{t}X_s ds \int_{0}^{u}X_s ds]$ involves dealing with terms $\mathbb{E}[\int_{0}^{t}s dX_s \int_{0}^{u}s dX_s]$ and $\mathbb{E}[X_t X_u]$ which are again probably well known in certain cases (the second term is obviously equal to $min(t,u)$ when $X$ is b.m.) but may be difficult to handle in your general case.

Edit 2: To give an approach to answer the question "Is it possible that the integrated processes have equivalent laws?"

Since $\int_{0}^{t}X_{s}^{(1)}ds$ and $\int_{0}^{t}X_{s}^{(2)}ds$ are Gaussian processes (we proved this using ito) it suffices to check if there covariance functions $g_{1}(t,u)=\mathbb{E}[\int_{0}^{t}X_{s}^{(1)}ds \int_{0}^{u}X_{s}^{(1)}ds]$ and $g_{2}(t,u) = \mathbb{E}[\int_{0}^{t}X_{s}^{(2)}ds \int_{0}^{u}X_{s}^{(2)}ds]$ are equal for all $t,u >0$ to show that the two processes have equivalent laws.

Now applying the result we got above from the ito calculation lets us start computing the covariance:

$\mathbb{E}[\int_{0}^{t}X_s ds \int_{0}^{u}X_s ds]$ = $\mathbb{E}[( t X_t - \int_{0}^{t} s dX_s)( u X_u - \int_{0}^{u} s dX_s)]$ $ = t u \mathbb{E}[ X_t X_u ] - t \mathbb{E}[X_t \int_{0}^{u} s dX_s ] -u \mathbb{E}[X_u \int_{0}^{t} s dX_s ]$ $+\mathbb{E}[\int_{0}^{t}s dX_s \int_{0}^{u}s dX_s]$

I refer to my above example on ways to deal with the terms in this expression given certain assumptions on $\mu$ and $\sigma$.Edit 3: Again this is just a way to start and obviously the calculations involving standard Brownian motion are trivial but the point is that the laws of $Y^{(1)}$ and $Y^{(2)}$ are equivalent (as opposed to equal) as soon as you show $g_1(t,u) = g_2(t,u)$ for all $t,u>0$.

  • $\begingroup$ I have partially went over your answer. It should be noted that ${\rm E}[X_t \int_0^t {s\,{\rm d}X_s } ] = t^2 /2$, and ${\rm E}[(\int_0^t {X_s \,{\rm d}s} )^2 ] = t^3 /3$ (as is well known, $\int_0^t {X_s \,{\rm d}s} \sim {\rm N}(0,t^3/3)$). $\endgroup$
    – Shai Covo
    Commented Jan 19, 2011 at 21:31
  • $\begingroup$ Thank you for the help I will make the appropriate edits. $\endgroup$
    – jzadeh
    Commented Jan 19, 2011 at 22:48
  • $\begingroup$ I don't see the connection with the initial question. What is your point exactly ? $\endgroup$
    – The Bridge
    Commented Jan 20, 2011 at 7:16
  • $\begingroup$ Thanks for the downvote The Bridge.... I refer you to my above passage "it should then suffice to characterize the processes covariance structure in order to have a complete understanding of the law of the processes". Since that is obviously to vague I have elaborated a little more in edit 2 and I refer you to one of the excellent texts by Robert Adler for the theorems I am citing on Gaussian processes. # R.J. Adler, (1990), , An Introduction to Continuity, Extrema, and Related Topics for General Gaussian Processes, IMS Lecture Notes-Monograph Series, Vol 12, vii + 160 – jzadeh 0 secs ago $\endgroup$
    – jzadeh
    Commented Jan 20, 2011 at 10:16
  • $\begingroup$ +1 for The Bridge's question. Re jzadeh's second edit: indeed, $E(Y^{(1)}_tY^{(1)}_s)=E(Y^{(2)}_tY^{(2)}_s)$ for every $(t,s)$ iff $E(X^{(1)}_tX^{(1)}_s)=E(X^{(2)}_tX^{(2)}_s)$ for every $(t,s)$. This is obvious and general--but not the point. The OP asks for cases when the laws of $Y^{(1)}$ and $Y^{(2)}$ are equivalent (as opposed to equal). $\endgroup$
    – Did
    Commented Jan 20, 2011 at 10:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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