I'm teaching myself Malliavin calculus and Skorohod integrals and with this kind of math I find myself following the logic through but lacking solid intuition about what is going on.

In particular take the Skorohod integral which generalises the Ito integral to possibly anticipating integrands (it reduces to an Ito integral when the integrand is non-anticipating)

So for example take the Skorohod integral, of the Brownian motion $B(t)$, $B(0)=0$ $$ \int_0^T B(T)\delta B(t) = B^2(T)-T$$ (see for example http://www.nhh.no/Files/Filer/institutter/for/dp/1996/wp0396.pdf)

It is emphasised that $$ \int_0^T B(T)\delta B(t) \neq B(T)\int_0^T\delta B(t)=B(T)\int_0^T dB(t)=B^2(T)$$

and I am trying to understand exactly why this is the case. I appreciate that one can't just view it as a Riemann sum which is what gives the intuition that the above is trying to ward off. So the only other kind of intuition I have is that it is because the integrand and integrator are correlated in the same way that the integrand and integrator are correlated in the Stratonovich integral which means it is not a Martingale. I.e. analogously to $$E[f(B)\circ dB]\neq E[f(B)]E[dB]=E[f(B)]\cdot 0=0\quad\left(=E[f(B)dB]\right)$$

However, in which case I don't understand why the difference between the naive and correct interpretations above is $T$ so that the difference grows independently of the correlation.

In other words I would expect that (given $T_2>T_1$) $$\lim_{(T_2-T_1)\to\infty}\int_0^{T_1}B(T_2)\delta B(t)=B(T_2)\int_0^{T_1}\delta B(t)=B(T_2)\int_0^{T_1}dB(t).$$

Or that I would expect the difference $$\int_0^T B(T)\delta B(t) - B(T)\int_0^T\delta B(t)=\int_0^T\xi(t-T)dt$$

where $\xi(t-T)\to 0$ as $T\to\infty$.

But all of this is completely at odds with the definition of the Skorohod integral above. Can someone explain this in a relatively simple with intuitive ideas?

A secondary question it raises is how to numerically simulate such an integral (say given a pre computed Wiener process)


Unfortunately, calling the Skorohod integral an "integral" is a bit of a misnomer, as it doesn't really have many of the properties which you would naturally associate with integrals, except for the fact that it "magically" coincides with the Itô integral when its integrand is adapted. The best way (I think) to get some kind of intuition for the Skorohod integral is to consider the case where the unit interval is partitioned into intervals of length $\delta t_k$, the integrand $F$ is constant and equal to $F_k$ on the $k$th interval, and it depends on the Wiener process $B$ only through its increments $\delta B_k$. In this case, an elementary calculation shows that one has $$ \int_0^1 F(t)\ \delta B(t) = \sum_k \bigl(F_k\ \delta B_k - \partial_k F_k\ \delta t_k\bigr)\;. $$ Here, $\partial_k F_k$ is the derivative of $F_k$ (viewed as a function of the increments $\delta B_j$) with respect to $\delta B_k$. The example you give follows immediately as a special case and many properties of the Skorohod integral can be "guessed" from that formula. (See for example Section 3 in this article.)

You see that the first term looks just like what you would expect from the definition of the Itô integral, but the second term is strange: while it looks again like a Riemann sum, the "integrand" is given by the derivative of the original integrand with respect to the underlying Wiener process. In particular, this shows that there is no way whatsoever to build approximations to the Skorohod integral from a single realisation of $F$ and $B$, since this gives no information at all on the derivative of $F$ with respect to $B$... In the case where $F$ is adapted, this second term vanishes of course (since then $F_k$ is only allowed to depend on the $\delta B_j$ with $j < k$) and you get Itô's integral back.

One natural reaction might be to say "since the second term behaves strangely, let's just throw it away" and to define $\int_0^1 F(t)\ \delta B(t) = \sum_k F_k\ \delta B_k$. The problem with this is that, as an unbounded linear operator on the space of all square integrable processes (not just adapted ones), this is not closable, so there's no "reasonable" way to build a nice space of integrands for which that integral makes sense. The reason why the Skorohod integral makes sense is that the second term introduces stochastic cancellations that cause the sum to converge for far more integrands than what one might naively expect.

There are other ways of introducing such cancellations. For example, in Sections 2-5 of my book with P. Friz available to download from my homepage, we discuss how the theory of controlled rough paths developed by Lyons and Gubinelli provides a notion of stochastic integration based on a different Riemann-type approximation that has the desirable feature that $\int GF_t\ dB_t = G \int F_t\ dB_t$ for any random variable $G$. Better: if the integrand depends on a parameter $a$ and one sets $\hat F(a) = \int_0^1 F_t(a)\ dB_t$, then one has $\hat F(G) = \int_0^1 F_t(G)\ dB_t$ for any random variable $G$, which definitely fails to hold for the Skorohod integral.

This "rough path integral" isn't quite an extension of the Itô integral though since it isn't defined for all square integrable adapted processes, but it coincides with the Itô integral on the intersection of their domains of definition and it admits anticipative integrands.

  • $\begingroup$ A lot to think about there. I was making some process by relating skorohod integrals to wick calculus and thus decomposing them as wick-Riemann sums. As I understand it $A\diamond dx=Adx-E[A]dx-AE[dx]+2E[A]E[dx]-E[Adx]$. Is this what you mean by there being no way to approximate the Skorohod integral with a SINGLE realisation (my emphasis) as you require statistical properties of the integrand/integrator etc? Is what I'm saying remotely correct? $\endgroup$ – user3353819 Nov 25 '15 at 1:54
  • $\begingroup$ See for example books.google.com.au/… $\endgroup$ – user3353819 Nov 25 '15 at 2:00
  • $\begingroup$ and mathoverflow.net/questions/172061/… $\endgroup$ – user3353819 Nov 25 '15 at 6:50
  • $\begingroup$ @ Martin Hairer : The best explanation I have read ever on this subject. Brilliant. Thank's for sharing your thoughts. NB: I know that such type of comments are to be avoided but nonetheless I couldn't help it. Best regards. $\endgroup$ – The Bridge Nov 26 '15 at 10:56

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