Why the Integral of Squared Brownian Motion Has Zero Quadratic Variation

Summary

The document asks for the quadratic variation of the process formed by integrating the square of Brownian motion over time. The stated result is zero. The key reasoning is that this integral has finite variation on bounded time intervals: its increments arise from integrating the continuous process W squared against ordinary time, rather than from a direct Brownian increment.

For an Itô process, quadratic variation is driven by its stochastic differential term, the part proportional to Brownian motion. Since the process here has no such term, its quadratic variation vanishes. This gives a compact example of how to distinguish accumulated drift from stochastic noise when analyzing a process. The response does not show a formal derivation or discuss broader conditions, so readers should treat it as a concise application of the standard quadratic-variation rule for finite-variation processes.

Key ideas

• The time integral of squared Brownian motion has zero quadratic variation.
• Quadratic variation comes from the process’s stochastic increment, not its accumulated time integral.
• An Itô process without a Brownian differential term has zero quadratic variation.

Tags

# Quadratic variation of an integral of a function of a Brownian motion


# Quadratic variation of an integral of a function of a Brownian motion












I'm asked to find the quadratic variation of the integral $\int_{0}^{t} W_s^2 ds$.

## Answer by Bjørn Kjos-Hanssen (score 4)

https://quant.stackexchange.com/a/42097

The quadratic variation of $$X_t=\int_0^t W_s^2\,ds$$ is 0. This is because it's an Ito process with no $dB_s$ term.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.