Apparent inconsistency in Skorohod integration

Question

Let $W$ be a standard $1$-dimensional Wiener process. Fix time points $t$ and $\tau$, with $t<\tau$, and consider the Itô integral $$\int_{t}^{\tau}W(t)~\mathrm{d}W(s).$$ Of course, this is a standard Itô integral of a simple process (as $W(t)$ is free of $s$), and equals $$W(t)\int_t^\tau 1~\mathrm{d}W(s) = W(t)\big(W(\tau) - W(t)\big) = W(t)W(\tau) - W(t)^2.$$ Now write the integral on top as $$\int_{0}^{\tau}W(t)~\mathrm{d}W(s) - \int_{0}^{t}W(t)~\mathrm{d}W(s).$$ This can be problematic since the second term isn't even defined in the Itô sense. Thankfully, it is defined in the Skorohod sense, which generalizes the Itô integral to certain classes of non-adapted integrands. In fact, it can be shown that (see, e.g., di Nunno, Øksendal, Proske: Malliavin Calculus for Lévy Processes with Applications to Finance) $$\int_{0}^{t}W(t)~\mathrm{d}W(s) = W(t)^2 - t.$$ The first term is a regular Itô integral, and equals $W(t)W(\tau)$. But, this would mean $$\int_{0}^{\tau}W(t)~\mathrm{d}W(s) - \int_{0}^{t}W(t)~\mathrm{d}W(s) = W(t)W(\tau) - W(t)^2 + t$$ $$\not= W(t)W(\tau) - W(t)^2 = \int_{t}^{\tau}W(t)~\mathrm{d}W(s),$$ which is strange!!

So, my question would be, where did I go wrong? All of the individual steps make sense in the Skorohod sense (and the Itô sense, when applicable). So, my guess is that the decomposition $$\int_{t}^{\tau}W(t)~\mathrm{d}W(s) = \int_{0}^{\tau}W(t)~\mathrm{d}W(s) - \int_{0}^{t}W(t)~\mathrm{d}W(s)$$ is not valid for the Skorohod integral, but I am not exactly sure.

Answer 1

The first term is a regular Itô integral, [...]

That's not true, the term

$$\int_0^{\tau} W(t) \, dW(s)$$

is not a regular Itô integral since the integrand $W(t)$ is clearly not $\mathcal{F}_s$-adapted for $s \leq t$. Since

$$\int_0^{\tau} X(t) \, dW(s) = X(t) \int_0^{\tau} dW(s)$$

fails, in general, to hold for Skorohod integrals (see the Remark at the end of Example 2.4 in the book by di Nunno et al.), your computation of this integral is not correct.

Remark: Note that both the Skorohod integral and the Itô integral have zero expectation. If you are unsure about calculations of stochastic integrals (...no matter whether Itô or Skorohod...), then it is a good idea to take the expectation and to see whether it equals zero. For instance,

$$\int_0^{\tau} W(t) \, dW(s) - \int_0^t W(t) \, dW(s) = W(t) W(\tau)-W(t)^2 +t$$

cannot hold true since the left-hand side has zero expectation, but the right-hand side does not.

Answer 2

I think you should use distinct notations for the Itô integral and the Skorohod Integral.

By Proposition 2.6 of Malliavin Calculus for Lévy Processes with Applications to Finance (the book you refered), the equality $$ \int_t^{\tau}W(t)\delta W(s) = \int_0^{\tau}W(t)\delta W(s) - \int_0^t W(t)\delta W(s) $$ is correct. As you write, $$ \int_0^t W(t)\delta W(s) = W(t)^2 - t $$ is also correct. However, $$ \int_0^{\tau} W(t) \delta W(s) \neq W(t)W(\tau). $$

Let us calculate this Skorohod integral by following the definition. Write $u(s) := W(t)$ for the considering (constant) stochastic process (note that $t$ is constant, and the time variable is $s$). Since $$ W(t) = u(s) = \int_0^t 1 dW(u_1) = \int_0^{\tau}\chi_{[0,t]}(u_1)dW(u_1) = I_1(\chi_{[0,t]}(u_1)), $$ the Wiener-Itô chaos expansion $$ W(t) = u(s) = \sum_{n\geq 0}I_n(f_n(u_1,\cdots,u_n; s)) $$ of $u(s)$ is obtained by $$f_0 = f_2 = f_3 = \cdots = 0, \ \ f_1(u_1;s) = \chi_{[0,t]}(u_1).$$ The symmetrization of $f_1(u;s)$ is $\frac{1}{2}(\chi_{[0,t]}(u) + \chi_{[0,t]}(s))$. Hence the Skorohod integral is calculated as the following multiple Itô integral: $$ \begin{align*} \int_0^{\tau}W(t)\delta W(s) &= \frac{1}{2}\int_0^{\tau}\int_0^{\tau}(\chi_{[0,t]}(u) + \chi_{[0,t]}(s))dW(u)dW(s) \\ &= \int_0^{\tau}\int_0^s(\chi_{[0,t]}(u) + \chi_{[0,t]}(s))dW(u)dW(s) \\ &= \int_0^{\tau}\left(\int_0^s\chi_{[0,t]}(u)dW(u) + \chi_{[0,t]}(s)W(s)\right) dW(s) \\ &= \int_0^{\tau} \left( W(\min(t,s)) + \chi_{[0,t]}(s)W(s)\right) dW(s) \\ &= \int_0^{\tau} W(\min(t,s)) dW(s) + \int_0^{\tau}\chi_{[0,t]}(s)W(s)dW(s) \\ &= \left( \int_0^t W(s) dW(s) + \int_t^{\tau}W(t)dW(s)\right) + \int_0^t W(s)dW(s) \\ &= \int_0^t 2W(s) dW(s) + W(t)(W(\tau)-W(t)). \end{align*} $$ By Itô's formula, $d(W(s)^2) = 2W(s)dW(s) + dt$. Hence we can continue the above calculation: $$ \begin{align*} &= W(t)^2 - t + W(t)(W(\tau) - W(t)) \\ &= W(t)W(\tau) - t \\ &\neq W(t)W(\tau). \end{align*} $$

On the other hand, by using Example 2.4 and Proposition 2.6 of Malliavin Calculus for Lévy Processes with Applications to Finance, we can calculate the above integral as follows: $$ \begin{align*} \int_0^{\tau}W(t) \delta W(s) &= \int_0^t W(t)\delta W(s) + \int_t^{\tau}W(t) dW(s) \\ &= W(t)^2 -t + W(t)(W(\tau) - W(t)) \\ &= W(t)W(\tau) - t. \end{align*} $$ This is expected conclusion.