Reordering a Nested Itô Integral with a Deterministic Kernel

Summary

The document poses a stochastic-calculus problem: evaluate a time integral whose integrand is itself an Itô integral of Brownian motion with kernel equal to the difference between the outer and inner times. The questioner splits the kernel into two terms and applies integration by parts to simplify one of them, asking whether the resulting expression is correct and how to handle the other term.

This sets up the use of stochastic integration by parts and interchange of integration order over a triangular time region, which can reduce the nested expression to a single Itô integral. However, the document contains only the question and the asker’s partial derivation; it provides no answer, final identity, or verification. It is a mathematical prompt rather than a trading application, and readers must independently check the stochastic integration conditions and complete the calculation.

Key ideas

• The problem asks for a time integral containing a nested Itô integral driven by Brownian motion.
• The inner integrand is a deterministic time kernel formed from the difference of its integration bounds.
• The questioner rewrites part of the expression using stochastic integration by parts.
• Interchanging the order of integration over the triangular domain is a natural way to simplify the remaining term.
• No answer or completed derivation is included, so the proposed manipulation is not verified in the document.

Tags

# How to integrate Itô integral w.r.t time?


# How to integrate Itô integral w.r.t time?












Let $W_t$ be a Brownian motion. How to calculate the following integral $$ I:=\int_0^t\left( \int_0^u(u-s)dW_s\right) du? $$

My attempt so far is: First note that $$ \int_0 ^u (u-s)dW_s = \int_0^u udW_s - \int_0^u sdW_s =uW_u - \int_0^u sdW_s, $$ so the integral $I$ becomes \begin{equation} I= \int_0^t uW_u du - \int_0^t\left( \int_0^u sdW_s\right) du. \tag{1} \end{equation} Remembering that $$ d(u^2W_u) = u^2dW_u + 2u W_u du, \quad \text{i.e.}\quad u W_u du = \frac{1}{2} d(u^2W_u)-\frac{1}{2}u^2dW_u, $$ we can write the first term in equation (1) as \begin{align*} \int_0^t uW_u du &= \int_0^t \frac{1}{2}d(u^2W_u) - \int_0^t \frac{1}{2}u^2 dW_u\\ &=\frac{1}{2}t^2W_t - \int_0^t \frac{1}{2}u^2dW_u\\ &= \int_0^t \frac{1}{2}(t^2-u^2)dW_u. \end{align*} Is this correct? And how should I calculate the latter term in equation (1), i.e. the integral $$ \int_0^t\left( \int_0^u sdW_s\right) du? $$

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