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I have a queston about the convergence in law of the following stochastic processe:

$\left\{I_t=\left(\int_0^te^{B_s}ds\right)^{1/\sqrt{t}}\right\}_{t\geq 0}$

with $\{B_t\}_{t\geq 0}$ is a standrad brownian motion.

Prove that $I_t\rightarrow e^{|N|}$ in law, where $N$ has the gaussian distribution $N(0,1)$.

I have tried by scaling property of brownian motion, but it does not work. I try also with the Laplace tranformation, but it is really difficult to discrible $I_t$'s transformation. Does someone have an idea? Thanks a lot!

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    @ Higgs88 : You should think about Laplace Method (the one that talk about $L^P$ convergence when $p\to\infty$ of a suitable random variable (or measurable function), by the way scaling property is clearly a step in the process. Best regards2012-10-13

1 Answers 1

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This is an answer based on the comment given by TheBridge above.

Basically we note by the scaling property that $(B_s)_{s\in[0,t]} \stackrel{d}{=} t^{1/2} (B_{s/t})_{s \in [0,t]}$. Therefore $\int_0^t e^{B_s}ds \stackrel{d}{=} \int_0^t e^{t^{1/2} B_{s/t}}ds = t\int_0^1 e^{t^{1/2}B_u}du,$ where we make a change of variable $s=tu$ in the final equality. Now using the fact that $L^p$ norms converge to the $L^{\infty}$ norm as $p \to \infty$, and setting $p = \sqrt t$ we see that $\lim_{t \to \infty} \bigg(t\int_0^1 e^{t^{1/2}B_u}du \bigg)^{1/\sqrt{t}} = 1 \cdot e^{\max_{t \in [0,1]} B_t}.$ We used the fact that $t^{1/\sqrt{t}} \to 1$.

But (see Revuz & Yor Chapter 3) it is well known that $\max_{[0,1]} B_t \stackrel{d}{=} |N|$ where $N \sim N(0,1)$. This completes the proof.