Why a Constant-Volatility Brownian Integral Is Brownian in Law

Summary

The note explains why a stochastic integral of a predictable process with constant absolute magnitude has the same distribution as a scaled Brownian motion. The argument first identifies the integral as a continuous martingale, then computes its quadratic variation: the squared volatility integrates to a constant times elapsed time. After dividing by that constant’s square root, the process has quadratic variation equal to time, so Lévy’s characterization identifies it as Brownian motion in law.

The conclusion concerns distribution, not pathwise equality with a particular Brownian motion. It relies on the stated constant-magnitude condition and the usual conditions needed for the stochastic integral and characterization to apply. If the constant is zero, the integral is identically zero; the normalization step must be handled separately. The note gives a concise proof rather than discussing broader applications or extensions.

Key ideas

• A stochastic integral with predictable integrand is a continuous martingale under the usual integrability assumptions.
• Its quadratic variation is the time integral of the squared integrand.
• Constant absolute volatility makes quadratic variation grow linearly with time.
• Lévy’s characterization then gives Brownian motion in law after normalization.
• Equality in distribution does not imply pathwise equality with a preselected Brownian motion.

Tags

# stochastic calculus - brownian motion


# stochastic calculus - brownian motion












I don't know how to prove this :

let be $X_t = \int_{0}^{t}\sigma_{u}dW_{u}$ where $\sigma_{t}$ is a predictable process.

If $|\sigma_{t}| = c$ a.s. how can I prove that $X_{t}=c*\beta_{t}$ (equality in distribution) ? (obvious if there wouldn't be absolute value..)

Thank you

## Answer by Gordon (score 2, accepted)

https://quant.stackexchange.com/a/17744

Note that $X$ is a continuous martingale. Moreover, the quadratic variation is given by \begin{align*} \langle X_t, \, X_t\rangle = \int_0^t |\sigma_u|^2 du = c^2 t. \end{align*} That is, \begin{align*} \langle X_t/c, \, X_t/c\rangle = t. \end{align*} From Levy's characterization, $X/c$ is by law a Brownian motion, which we denote by $\beta$. Then, by law, \begin{align*} X_t = c\, \beta_t. \end{align*}

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