# Is an SDE really equal to an integral equation, or is it rather “its integral” that is?

Ive been told and been reading in some textbooks on SDE's that an SDE really is an integral equation. In other words, that

$$dX= \beta dt + \sigma dW$$ $$\,$$ "really means" $$\,$$ $$X_{t}= X_{0} +\int_{0}^{t} \beta dt +\int_{0}^{t} \sigma dW$$

However I am getting the impression that this is not the case and that $$dX$$ really is a measure type object which we in turn can "take the integral off", giving us the latter above. This in turn implies that the SDE really has a meaning maybe not as a differental equation in the usual sense but as something else. This impression was obtained for instance from comments in the following post,

If $dX_{t} = X_{t}\,dt + \,dB_{t}$, why does $e^{- t}dX_{t} = e^{-t} X_{t} \,dt + e^{-t} \,dB_{t}$?

In particular the discussion involving $$dX=dY$$, $$AdX=AdY$$ and then taking the integral of that.

With that background I ask the following;

Is an SDE then really equal to an integral equation? This would be the same as saying or declaring rather that $$\mu= d\mu$$, which is fine if one sticks to that convention, but that dosnt seam to be the case in some calculations with differentials.

I also wonder if it is possible to make sense of the $$dX$$ if one uses more advanced tools? Similar to the case of $$dx$$ in a ODE, that indeed justify some "informal" operations. My impression is that in general the $$dx$$ can be tought of as a measure, a differential form etc..

Found these;

Are SDE's really "differential"?

that looks like similar questions, but which are not answering my question.

• what kind of answer are you looking for? Would a definition of the first as the second from a text book satisfy you? – Calvin Khor Oct 11 '18 at 8:19
• @CalvinKhor that "Ive been told" means that I have been reading it in a text, in particular Oksendals intro text on SDE's. However when I start follow calulations etc it donst seem that ppl treat the objects as the same. If they are equal it means that we do not have to integrate the former to get the latter. To clarify; a measure is not equal to an integral. We say $\mu(A)=\int_{A} d\mu$ – Vlad Oct 11 '18 at 8:22
• for instance read the commets in the following post, math.stackexchange.com/questions/1647732/… – Vlad Oct 11 '18 at 8:23
• You may want to clarify in the question that you don't mean you overheard the definition from across the canteen, say. My feeling is that your question is something like 'does that post generalise, and how much?' – Calvin Khor Oct 11 '18 at 8:36
• @CalvinKhor thanks for the input, I fixed it now. I think. – Vlad Oct 11 '18 at 8:38

Continuing the comments, more precisely, $$dB_t$$ is an $$L^2$$-valued measure: To every Borel set $$A\subseteq \mathbb R$$ with finite Lebesgue measure, it assigns an element of $$L^2$$. If $$A$$ is the interval $$[a,b]$$, the corresponding element of $$L^2$$ is the random variable $$B_b-B_a$$. See page 8 of Nualart's book.
It seems true that similarly, for any Ito process $$X_t$$, one can regard $$dX_t$$ as a $$L^2$$-valued measure. So, an SDE is an equality between three $$L^2$$-valued measures.
Update: For a deterministic measurable function $$f(t)$$, the Ito integral $$\int_a^b f(s)dB_s$$ is just the integration of $$f$$ against the $$L^2$$-valued measure $$dB_t$$. However, for general $$L^2$$-valued functions of $$t$$ like $$f(t,\omega)$$, the integral $$\int_a^b f(s)dB_s$$ is not well defined. So the integral of $$f$$ as an $$L^2$$-valued function against the $$L^2$$-valued measure $$dB_t$$ might also be undefined. One at least requires that $$f$$ is adapted to the filtration.