# 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.

## 1 Answer

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 $$t\rightarrow \infty$$.