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EDIT: corrected the problem thesis.

Suppose $(W_s)_{s \geq 0}$ is a Wiener process. Define $$ V_t := \frac{1}{\sqrt{t}}\log \left[\int_{0}^{t}\exp(W_s)ds\right] $$ Show that $$ V_t \xrightarrow{t \rightarrow \infty} \sup_{s \in[0, 1]}W_s $$ in distribution.

tosi3k
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1 Answers1

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$$ V_t := \frac{1}{\sqrt{t}}\int_{0}^{t}\exp(W_s)ds = \sqrt{t} \int_{0}^{1} \exp(W_{tu}) du $$

The process $(X^{[t]}_s)_{s\ge 0}$ defined by $$ X^{[t]}_s = \frac 1{\sqrt t} W_{ts} $$is a Brownian motion for each $t$, and

\begin{align} V_t &= \sqrt{t} \int_{0}^{1} \exp(\sqrt t X^{[t]}_u) du\\ &\overset{(dist.)}= \sqrt{t} \int_{0}^{1} \exp(\sqrt t W_u) du\\ \end{align}

Now I am not convinced that the result holds, but you have: $$ \left(\frac{V_t}{\sqrt t}\right)^{\frac 1{\sqrt t}} \overset{(dist.)}= \left(\int_{0}^{1} \exp(\sqrt t W_u) du\right)^{\frac 1{\sqrt t}} \to \sup_{0\le u \le 1} \exp W_u $$

With the second version: $$V_t=\frac 1 {\sqrt t} \log ∫_0^t \exp W_s ds =\log \left[ ∫_0^t \exp W_s ds\right] ^{ \frac 1 {\sqrt t} } \\\to \log \sup_{0\le u\le 1} \exp \exp W_u = \sup_{0\le u\le 1}W_u $$

mookid
  • 28,236
  • Hmm, I think you're right that the problem's thesis is not true. Moreover, from your reasoning we can imply that $V_t$ does not converge weakly to any probability distribution. – tosi3k Apr 12 '15 at 12:20
  • OK, I received a corrected version of the problem - $V_t = \frac{1}{\sqrt{t}} \log \left[ \int_{0}^{t} \exp(W_s)ds \right]$ and the thesis stays as it was. – tosi3k Apr 13 '15 at 10:26
  • @tosi3k can you take it from here? – mookid Apr 13 '15 at 16:55