# Properties of the time integral of Ito process

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. Let's define the process $$Y_t$$ to be some integral of $$X_t$$, namely

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

I was wondering if someone could recommend a reference that deals 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 references there are not useful. In particular:

Is $$Y_t$$ an Ito process? I would say no, but the couple $$(X_t,Y_t)$$ probably yes, because we can consider it as a particular two-dimensional Ito process $$(dX_t,dY_t) = a(t,X_t,Y_t) dt + b(t,X_t,Y_t) dW_t$$ where now $$a$$ and $$W_t$$ are 2-dimensional vectors and $$b$$ is 2 by 2 matrix. Are there particular cases (i.e. choices of $$a$$ and $$b$$) for which $$dY_t$$ alone is still an Ito process? I mean, given $$a$$ and $$b$$, can we find $$A$$ and $$B$$ such that $$dY_t = A(t,Y_t) dt + B(t,Y_t) dW_t \, \, \, ?$$

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 $$(X_t,Y_t)$$ but not when we consider $$Y_t$$ alone? Any reference that is specific on "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 the Ito's lemma to a similar case). However, it is not clear to me how to apply the Ito's lemma to this kind of "integral function". Do we need some "extension" of Ito's lemma to differentiate $$Y_t$$?

• Related: quant.stackexchange.com/q/30484 (Differential of integral of a stochastic process), math.stackexchange.com/q/1298990/532409 (Stochastic Differential Equation for Time Integral of Stochastic Process), mathoverflow.net/q/52448/333546 (Time integrals of diffusion processes) , mathoverflow.net/q/126478/333546 (Time integral of a diffusion) Apr 19, 2022 at 10:17
• 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. Apr 19, 2022 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 $dX = a dt+b dW$, en.wikipedia.org/wiki/It%C3%B4_calculus#It%C3%B4_processes Apr 19, 2022 at 13:44
• 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. Apr 19, 2022 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 $dY$. Apr 19, 2022 at 14:01

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.

• Thank you, your answer contains some interesting ideas. Apr 20, 2022 at 0:35