Using basic techniques from Malliavin calculus it can be shown that $$ \int_0^T W_T dW_t = W_T^2 - T $$ As can be seen the above integral is a non-adapted stochastic integral.

We also know using Ito that $$ 2 \int_0^T W_t dW_t = W_T^2 - T $$ since $$ dW_t^2 = 2W_t dW_t + (dW_t)^2 $$

Question 1:

Is there a direct way to show, by which I mean without using Malliavin calculus, i.e. only using more classical techniques, that $$ \int_0^T W_T dW_t = 2 \int_0^T W_t dW_t $$ ?

Question 2: Why $$ \int_0^T W_T dW_t \neq W_T \int_0^T dW_t $$ ? I am having trouble understanding intuitively why you cannot just take $W_T$ out of the integral.

In the above, $W_t$ denotes standard Brownian motion.


Please see Montero & Kohatsu-Higa, An application of Malliavin calculus to finance for more details on Malliavin calculus. In particular, I have used formula (1) from their paper to derive my first expression above, where to follow their notation I have set $F = W_T$ and $u_t = 1$.

  • $\begingroup$ @UBM please see my edit with a link to a paper and the formula I used to derive my first expression. $\endgroup$
    – user34971
    Commented Oct 26, 2020 at 5:37
  • $\begingroup$ see related post quant.stackexchange.com/questions/58927/… $\endgroup$
    – develarist
    Commented Oct 26, 2020 at 6:20
  • $\begingroup$ @develarist Yes I saw. I believe these type of integrals are called Skorokhod integrals. $\endgroup$
    – user34971
    Commented Oct 26, 2020 at 6:48
  • $\begingroup$ Are you sure you mean $∫W_T dW_t$ and not $∫W_t dW_t$? $\endgroup$
    – develarist
    Commented Oct 26, 2020 at 8:53
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    $\begingroup$ Re question 2, I think @UBM's answer is all there is to it. Your stochastic integral is not defined in the Itô sense, which by definition integrates over processes adapted to the filtration generated by $W_t$. $W_T$ is not adapted, and it is not independent from the increment $\text{d}W_t$: Itô integrals are defined over the left-point of the integrand. $\endgroup$ Commented Oct 26, 2020 at 16:45

2 Answers 2


So we are seeking interpretation in terms of the Ito's integral, whose definition, as we know from the comments below, is in the sense of adapted process. This is not the end though, one can extend the Ito for non-adapted processes- e.g., Skorokhod which replaces the adaptability by regularity condition, and one can understand this integral intuitively in terms of Riemann sum and step processes. In essence one can extend Ito's integral to non-adaptive processes, the processes have to satisfy some conditions, but not going to go there!

The answer to one could vary depending on the interpretation one uses. Here is one way to go about it:




I think it should equal $2\int_0^TW_t\,dW_t+T$ in the Ito's sense. On the other hand, if one tries a slightly different interpretation when approximating the integral via finite sum (think $n \to \infty$ in the partition sense etc.)

$\int_0^TW_TdW_t=\int_0^T\left(W_T-W_t\right)dW_t+\int_0^T W_tdW_t$

$={ \sum_{k=1}^{n}{\left( W_{t_{n}} - W_{t_{k}} \right) \Delta W_{t_{k}} }}+\int_0^T W_tdW_t$

$={ \sum_{k=1}^{n}{\left( W_{t_{n}} -W_{t_{k}}+W_{t_{k-1}}-W_{t_{k-1}} \right) \Delta W_{t_{k}} }}+\int_0^T W_tdW_t$

$={ \sum_{k=1}^{n}{\left( W_{t_{n}} -\Delta W_{t_{k}}-W_{t_{k-1}} \right) \Delta W_{t_{k}} }}+\int_0^T W_tdW_t$

$= W_{t_{n}}\sum_{k=1}^n{\Delta W_{t_{k}} }-\sum_{k=1}^n{\Delta W_{t_{k}}^2} -\sum_{k=1}^n W_{t_{k-1}}\Delta W_{t_{k}}+\int_0^T W_tdW_t$

$= W_{t_{n}}^2-\sum_{k=1}^n{\Delta W_{t_{k}}^2} $

$= W_{T}^2-T=2\int_0^TW_t\,dW_t$

For intuitive understanding of the non-adapted (and adapted!)integral, it helps to think of approximating the integrand by a sequence of step functions, and then multiplying the process values in each interval by the Brownian increment, and summing across the intervals.

Q2 can be rephrased as follows, and answer should follow from the above:

$$\int_0^T\int_0^TdW_s\,dW_t \neq \int_0^TdW_s \int_0^TdW_t?$$

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    $\begingroup$ Re second equality, it is just rectangle equals two triangles! $\endgroup$ Commented Oct 27, 2020 at 8:17
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    $\begingroup$ Ah, thanks for the answer. $\endgroup$ Commented Oct 27, 2020 at 8:58
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    $\begingroup$ I like your expanded answer of approximating the integral by finite sums. Makes sense, thank you. $\endgroup$
    – user34971
    Commented Oct 27, 2020 at 9:06
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    $\begingroup$ @UBM, I don’t think there is any disagreement that Ito will need some modification to work for anticipative integral. $\endgroup$ Commented Oct 27, 2020 at 12:26
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    $\begingroup$ Surprisingly this is a very well researched topic. Like you said few times, Ito works for adapted processes, so not going to repeat what I said before, but the question then is: if one were to hypothetically apply the interpretation to non adapted processes, with a view to extension (outside its remit of course in the original Ito sense) but the interpretation were to give the same result as the traditional Ito when applied to adapted processes, would that make it less ‘wrong’? $\endgroup$ Commented Oct 28, 2020 at 12:31

Ok, based on Magic is in the chains answer, this is how I interpret it intuitively. We have the expression $\int_0^TW_TdW_t$ which is not defined as a ordinary Ito integral since the integrand $W_T$ is not adapted. Therefore we split the integrand as the sum of two parts, one which is based the past and the present, $W_t$, and one which is based on future events, $W_T - W_t$.
The integral $\int_0^TW_t \,dW_t$ gives us no trouble since the integrand is adapted. The other integral $\int_0^TW_T - W_t \, dW_t$ still don't make sense as an Ito integral since it is not adapted.

However, we know that Brownian motion is a predictable process. So it makes sense to use that fact to split the difference $W_T - W_t$ into a telescope sum where each term makes sense in the limit (Just as Magic in the chain is doing with the step functions). The expression we get is something that is the proper approximation of an Ito integral and converge in the limit.

Yes, I know what I just wrote is a bit vague, but this is how I interpret the situation intuitively.

  • $\begingroup$ I believe that Brownian motion is NOT predictable? I am going by the answer here: "The process $(A_t)_{t\geqslant1}$ is $(\mathcal F_t)_{t\geqslant0}$-predictable if and only if, for every $t\geqslant0$, $A_{t+1}$ is $\mathcal F_{t}$-measurable. In other words, at each time t, one can predict the next value $A_{t+1}$ of the process using only the information available at time t, that is, the sigma-algebra $\mathcal F_t$." I don't think we can predict the value of $W_{t^+}$ at time $t$? $\endgroup$ Commented Oct 27, 2020 at 8:47
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    $\begingroup$ This should follow from the continuity of the Brownian motion, if I am not mistaken. We have that $ \mathcal{F}_{t^-} = \cup_{s<t} \mathcal{F}_s = \mathcal{F}_{t}$. Since there is no jump at $t$, the information up to, but not including, time t is enough to predict the value at time t. I agree the terminology can be a bit confusing though as it sounds like we can predict future values of the Brownian motion even though it is only in a infinitesimal sense. $\endgroup$ Commented Oct 27, 2020 at 9:21
  • $\begingroup$ yeh, ok, I recall that working in continuous time gets a bit tricky in terms of predictability definitions. I will have a read through some old lecture notes and then try to link it in my mind to your answer above. I personally find it important to understand things intuitively: otherwise I feel I just end up working with symbols that don't translate into a real-world application. $\endgroup$ Commented Oct 27, 2020 at 9:53
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    $\begingroup$ This is very close to an alternative interpretation of the anticipated integral. You are right $B_t$ is adapted so no problem there. The second component- the one containing $B_T-B_t$- is what is known as ‘instantly independent process’. And it does not take much to apply Ito’s logic to such processes. $\endgroup$ Commented Oct 28, 2020 at 17:35
  • $\begingroup$ Of course just like adapted, instant independence is wrt the filtration. $\endgroup$ Commented Oct 28, 2020 at 17:41

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