Brownian Scaling and the Logarithm of an Exponential Integral

Summary

The document proves a distributional limit for the logarithm of the time integral of exponentiated standard Brownian motion, scaled by the square root of time. Brownian scaling rewrites the integral over a fixed unit interval using a standard Brownian path. The scaled logarithm then becomes the logarithm of an Lp norm of the path’s exponential, with the norm order growing as time increases. The vanishing contribution from the time factor does not affect the limit.

As the norm order tends to infinity, the Lp norm converges to the supremum norm for a continuous path on the compact unit interval. Taking the logarithm therefore yields the maximum of that Brownian path. Because the rescaled process has the same law as standard Brownian motion, this establishes convergence in distribution to the Brownian maximum on the unit interval. The argument relies on Brownian scaling and the norm limit; it is a probability result rather than a trading strategy or empirical market finding.

Key ideas

• Brownian scaling converts the time integral into an integral over a fixed unit interval.
• The scaled logarithm of the integral can be expressed through an Lp norm of exponentiated Brownian motion.
• As the norm order grows, the Lp norm converges to the path’s supremum norm.
• The resulting limit has the distribution of the maximum of standard Brownian motion on the unit interval.

Tags

# Invariance Scaling of Brownian Motion


# Invariance Scaling of Brownian Motion












Prove $\frac{1}{\sqrt{t}}\log\left(\int_0^t \exp(B_s)\mathrm{d}s\right)$ converges to $\sup\limits_{t\in [0,1]}B_t$ in distribution as $t\to\infty$. I have a sense to use scaling invariance, but no idea how to derive this whole thing.

## Answer by Gordon (score 5)

https://quant.stackexchange.com/a/49056

Note that \begin{align*} \int_0^t e^{B_s}ds &= t\int_0^1 e^{B_{tu}}du\\ &=t\int_0^1 e^{\sqrt{t}\frac{1}{\sqrt{t}}B_{tu}}du\\ &=t\int_0^1 e^{\sqrt{t}W_u}du, \end{align*} where $\{W_u=\frac{1}{\sqrt{t}}B_{tu}, \, u\ge 0\}$ is a standard Brownian motion. Then \begin{align*} \frac{1}{\sqrt{t}} \ln \int_0^t e^{B_s}ds &= \frac{\ln t}{\sqrt{t}} + \frac{1}{\sqrt{t}}\ln \int_0^1 e^{\sqrt{t}W_u}du\\ &= \frac{\ln t}{\sqrt{t}} + \ln \bigg(\int_0^1 \left(e^{W_u}\right)^{\sqrt{t}}du \bigg)^{\frac{1}{\sqrt{t}}}\\ &= \frac{\ln t}{\sqrt{t}} + \ln\, \big\lVert{e^{W_u}}\big\rVert_{\sqrt{t}}\\ &\rightarrow\ln\, \big\lVert{e^{W_u}}\big\rVert_{\infty}\\ &= \ln\Big( \max_{0\le u \le 1} e^{W_u}\Big)\\ &=\max_{0\le u \le 1} W_u. \end{align*} That is, $\frac{1}{\sqrt{t}} \ln \int_0^t e^{B_s}ds$ converges to $\max_{0\le t \le 1} B_t$ in distribution, as $t\rightarrow \infty$.

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.