# convergence in distribution of exponential of a brownian motion

If $(B_t)_{t≥0}$ is a standard Brownian motion, show that, as $t \to \infty$, $$\left(\int_0^t e^{B_s} \, ds\right)^{1/\sqrt{t}} \text{ converges in distribution to} \ e^{M_1},$$ where $M_1 = \sup_{0 \leq s \leq1} B_s$.

I know that $M_1$ has the same distribution as $|N|$ where $N$ $\sim \mathcal{N}(0,1)$. Any hints on how to get started.

1. Change variables $s=tu$:

$$\int_0^t e^{B_s} \, ds = \int_0^1 e^{B_{t u} }t \, du=(*)$$

1. Use scaling: $(B_{tu}:u \ge 0) \overset{\mbox{dist}}{=}\sqrt{t}( B_{u}:i \ge 0)$. Therefore

$$(*) \overset{\text{dist}}{=} \sqrt{t} \int_0^t e^{\sqrt{t} B_u} \, du$$

1. Take $\sqrt{t}$-th root:

$$t^{1/\sqrt{t}} \left(\int_0^1 e^{\sqrt{t} B_u} \, du\right)^{1/{\sqrt{t}}}.$$

2. Take limits.

a. $t^{1/\sqrt{t}}= \left(\sqrt{t}^{1/\sqrt{t}}\right)^2 \to 1$ as $t\to\infty$.

b. For any continuous nonnegative function $f$ on $[0,1]$ we have $\left(\int f^r(u) \, du \right)^{1/r} \to \max_{u \in [0,1]} f(u)$ as $r\to\infty$. This is basic calculus, or also something you know from measure and integration ($L^p$ norm on a probability space increase to $L^\infty$ norm). Here $f$ is the random function $e^{B_u}$ and $r=\sqrt{t}$.

Result follows.