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Why Brownian Time Integrals Cannot Be Replaced by a Scaled Brownian Motion

Article Quant Q&A · Author: Leonardo S. Vieira
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Summary

The document examines an attempted calculation of the time integral of Brownian motion. The question uses Itô's product rule to rewrite the integral in terms of a Brownian stochastic integral, then incorrectly treats that stochastic integral as a fixed multiple of the terminal Brownian value because their distributions have related variances.

The key issue is that matching marginal distributions does not make two random variables equal or interchangeable in an expression. The rewritten terms are dependent, and their covariance matters when calculating the variance of their difference. The document states the correct variance of the time integral and points to an answer discussing the covariance error, but does not reproduce the derivation or supporting calculations. Its lesson is about stochastic calculus and distributional reasoning rather than a trading strategy.

Key ideas

  • Itô's product rule can express the time integral of Brownian motion using a stochastic integral.
  • A variance match does not justify replacing a stochastic integral with a scaled terminal Brownian value.
  • The covariance between the terms in a stochastic decomposition affects the variance of their sum or difference.
  • The document gives the correct variance but omits the detailed derivation.

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# Integral of Brownian Motion w.r.t Time: what is wrong with this solution?


# Integral of Brownian Motion w.r.t Time: what is wrong with this solution?












My question is about a stochastic integral of brownian motion w.r.t time.

Let $W(t)$ the Wiener process (or brownian motion). I want to calculate this: \begin{eqnarray} X(t)=\int_{0}^t dt' W(t'). \end{eqnarray} My strategy:

1) Itô's Fórmula: \begin{eqnarray} d(tW(t))=tdW(t)+W(t)dt \implies W(t)dt=d(tW(t))-tdW(t). \end{eqnarray} 2) Integrate: \begin{eqnarray} X(t)=\int_{0}^t dt'W(t')=\int_0^t d(t'W(t'))+\int_{0}^tdW(t')t'=tW(t)-\frac{t}{\sqrt{3}}W(t)=\left(1-\frac{1}{\sqrt{3}}\right)tW(t). \end{eqnarray} I used: \begin{eqnarray} \int_{0}^t dW(t')f(t')=\left(\frac{1}{t}\int_{0}^t dt'|f(t')|^2\right)^{1/2}W(t)\implies \int_0^t dW(t')t'=\frac{t}{\sqrt{3}}W(t), \end{eqnarray} because... \begin{eqnarray} \int_{0}^t dW(t')f(t')\sim \mathcal{N}\left(0,\int_0^t dt'|f(t')|^2\right), \hspace{0.5cm} W(t)\sim \mathcal{N}(0,t). \end{eqnarray}

The "problem" is: \begin{eqnarray} \sigma^2_X=\left(1-\frac{1}{\sqrt{3}}\right)^2 t^3 \end{eqnarray} But the correct is: \begin{eqnarray} \sigma^2_X=\frac{t^3}{3} \end{eqnarray} Can some illuminated mind tell me where the error is?

## Answer by Magic is in the chain (score 1, accepted)

https://quant.stackexchange.com/a/41861

Let me know if any of the steps is not clear.

## Answer by Magic is in the chain (score 0)

https://quant.stackexchange.com/a/41908

Re-comment and your second question as to why your transformation leads to a different result, it is because the transformation does not take into account the covariance between the two terms. As you can see below, the covaraince is distorted;

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.