Interpreting an Anticipating Brownian Stochastic Integral

# Anticipating stochastic integral $\int_0^T W_T dW_t$


# Anticipating stochastic integral $\int_0^T W_T dW_t$












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.

EDIT:

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 $u_t = 1$.

## Answer by Magic is in the chain (score 4, accepted)

https://quant.stackexchange.com/a/58931

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:

$\int_0^TW_TdW_t=\int_0^T\int_0^TdW_s\,dW_t$

$=2\int_0^T\int_0^tdW_s\,dW_t-\int_0^T{dW_s^2}$

$={2\int_0^T\int_0^t{dW_s\,dW_t}}-T$

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:

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

## Answer by Jesper Tidblom (score 0)

https://quant.stackexchange.com/a/58946

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.