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Background - Consider the Ito process $X_t$ defined by

$$ dX_t = a(t,X_t) dt + b(t,X_t) dW_t $$

where $W_t$ is the standard continuous-time Wiener process, see this answer for a precise discussion of the notation used. Let's define the process $Y_t$ to be some integral of $X_t$, namely

$$ Y_t =\int_0^t f(t , s , X_s) ds $$

where $f$ is a deterministic function. I am looking for references that deal with time integrals of this kind, or even of the simpler kind:

$$ Y_t =\int_0^t X_s ds \qquad \Rightarrow \qquad dY_t =X_t dt $$

Unfortunately, I haven't seen any treatment of the properties of $Y_t$ in the better-known texts on stochastic analysis. There is also this question on Math Overflow, but the references therein are not very useful. In particular, I am interested in the following questions.

Question - Is $Y_t$ an Ito process? I would say no, but the couple $\mathbf{X}_t=(X_t,Y_t)$ probably yes, because we can consider it as a particular two-dimensional Ito process. However, are there particular cases for which $Y_t$ follows the stochastic dynamics $$ dY_t = A(t,Y_t) dt + B(t,Y_t) dW_t \, \, \, ? $$

Finally, is $Y_t$ even Markovian? Maybe this property can shed light on why we can write down a Fokker-Planck when we consider the couple $\mathbf{X}_t$ but not when we consider $Y_t$ alone? Any reference that is specific to "integrated processes" is appreciated (e.g. how to find its statistical properties like the autocovariance of $Y_t$).

Note: for the special case of the time integral of an Ornstein-Uhlenbeck process, see this MO question and "Time integral of an Ornstein-Uhlenbeck process". Regarding the definition of Ito process, see the Wikipedia link above or this, this and this interesting questions.

Edit (after the useful comments of @KurtG. ): consider $Y_t =\int_0^t f(t , s , X_s) ds$, by applying the Ito's lemma we may find the expression for $dY_t$. At this point, we can start to restrict the generic expression of $f$ in order to try to have something of the form $dY_t=A dt+BdW$ (see e.g. this question for an application of Ito's lemma to a similar case). However, it is not clear to me how to apply Ito's lemma to this kind of "integral function". Do we need some "extension" of Ito's lemma to differentiate $Y_t$?

Quillo
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  • Related: https://quant.stackexchange.com/q/30484 (Differential of integral of a stochastic process), https://math.stackexchange.com/q/1298990/532409 (Stochastic Differential Equation for Time Integral of Stochastic Process), https://mathoverflow.net/q/52448/333546 (Time integrals of diffusion processes) , https://mathoverflow.net/q/126478/333546 (Time integral of a diffusion) – Quillo Apr 19 '22 at 10:17
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    How do you define an Ito process ? If you want it to be of Markovian form, i.e., $a,b$ be deterministic functions of the process itself then the time integral is not an Ito process because the time integration breaks the Markov property. If you allow more general processes for $a,b$ then the time integral is also a more general Ito process. Just use Ito's lemma on $f(s,X_s)$ and "plug" in what this gives you. – Kurt G. Apr 19 '22 at 13:02
  • @KurtG. as I understand, an Ito process is a process such that its differential can be written as a drift-diffusion SDE like $dX = a dt+b dW$, https://en.wikipedia.org/wiki/It%C3%B4_calculus#It%C3%B4_processes – Quillo Apr 19 '22 at 13:44
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    When you say SDE, you mean $a,b$ are both deterministic functions of $t$ and $X_t$ ? It is OK to use that definition but then $Y_t=\int_0^tf(s,X_s),ds$ does not satisfy such a Markovian SDE. – Kurt G. Apr 19 '22 at 13:48
  • @KurtG. yes, $a,b,A,B$ are all deterministic. I have added an edit, I hope that now my doubts are a bit more clear. I guess that such a time-integral of an Ito process is not an Ito process anymore, right? Maybe it's possible to see it directly by calculating the differential $dY$. – Quillo Apr 19 '22 at 14:01

2 Answers2

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This is handwaivy, but might be better to read in an answer than in the comments:

Assume $$Y_t = \int_0^t f(s,X_s) ds$$ for some process $X_t$. Your first question now is whether we can find functions $a,b$ such that

$$Y_t = \int_0^t a(s,Y_s) ds + \int_0^t b(s,Y_s) dW_s$$ for Brownian motion $W_t$. Subtracting both equations yields

$$ 0 = \int_0^t f(s,X_s)-a(s,Y_s) ds - \int_0^t b(s,Y_s) dW_s$$ $$\implies \int_0^t b(s,Y_s) dW_s= \int_0^t f(s,X_s)-a(s,Y_s) ds$$ $$\implies \left[\int_0^t b(s,Y_s) dW_s\right]= \left[\int_0^t f(s,X_s)-a(s,Y_s) ds\right]$$ $$ \implies \int_0^t b(s,Y_s)^2 ds = 0$$

where $\left[\cdot,\cdot\right]$ is quadratic variation. So, $b$ would have to be $0$ almost surely. But then we would have

$$Y_t = \int_0^t a(s,Y_s) ds,$$

but this is an ODE. So I would conclude that the only way this could work is if $X_t$ actually only fulfills an ODE, rather than an SDE.

a_student
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Applying Ito's formular(lemma) leads to 2nd order PDE, but solving it isn't easy.

So I think this keyword will be helpful:

Klein-Kramer's equation

and see 3rd citation of the page; In the Risken's book, you may find how to solve the equation.

Patche
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  • It would help this answer if some parts of the paper being mentioned are displayed on screen. Otherwise it a link only answer and may end up being removed. – Leucippus Jan 22 '25 at 06:07
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    @Leucippus I know this answer is too short; but I don't have enough reputation for comment and sharing this :D – Patche Jan 22 '25 at 07:11
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    Also, see https://math.stackexchange.com/questions/74882/joint-distribution-of-the-stochastic-process – Patche Jan 22 '25 at 12:24
  • The Klein-Kramer's equation is just a special case of the Fokker–Planck equation, but it's for Hamiltonian systems. Now, it is true that my $Y_t$ may resemble "position" if $X_t$ is formally interpreted as "velocity", but here I do not necessarily have a potential that depends on the "position" $Y_t$. However, this is an interesting suggestion, thank you. – Quillo Jan 25 '25 at 15:56