Limit of a Wiener integral

Question

How to show that $$ \lim _{\alpha \rightarrow \infty } \sup_{t \in \left [0,T \right]} \left | e^{-\alpha t} \int _ 0 ^t e^{\alpha s} ~ dB_s \right | =0, \ \ \text{a.e.} $$

where $\left (B_s \right)_{s\geq 0}$ is a real standard brownian motion starting from zero ?

Answer 1

The set of functions of class $\mathcal C^1$ on $[0,T]$ is dense into the set of continous functions on $[0,T]$. Since $t\mapsto B_t$ is continuous we have, a.s. for every $\epsilon >0$, there exists $\left ( B_t ^\epsilon\right)_{t \in \left[0,T \right] }$ of class $\mathcal C^1$, such that

$$\sup _ {t \in \left[0,T \right] } \left | B_t-B_t^\epsilon\right | < \epsilon.$$

Since we have by Itô's lemma that

$$ \int_0 ^t e^{\alpha s} ~dB_s = e^{\alpha t}B_t - \int_0 ^t B_s \alpha e^{\alpha s}ds$$

and also by integration by parts

$$ \int_0 ^t e^{\alpha s} ~dB_s ^\epsilon= e^{\alpha t}B_t ^\epsilon - \int_0 ^t B_s ^\epsilon \alpha e^{\alpha s}ds\qquad\text{a.s.}$$

this shows that

\begin{align} \left |\int_0 ^t e^{\alpha s} ~dB_s - \int_0 ^t e^{\alpha s} ~dB_s ^\epsilon\right| &=\left | e^{\alpha t}B_t -e^{\alpha t}B_t ^\epsilon - \int_0 ^t B_s \alpha e^{\alpha s}ds + \int_0 ^t B_s ^\epsilon \alpha e^{\alpha s}ds\right| \\ &\leq e^{\alpha t}\left |B_t -B_t ^\epsilon\right| + \int_0 ^t \alpha e^{\alpha s}\left|B_t -B_t ^\epsilon\right|ds \\ &\leq \epsilon e^{\alpha t} + \epsilon \int_0 ^t \alpha e^{\alpha s} ds \leq 2\epsilon e^{\alpha t}\qquad\text{a.s.}\end{align}

so $$\left |e^{-\alpha t}\int_0 ^t e^{\alpha s} ~dB_s - e^{-\alpha t}\int_0 ^t e^{\alpha s} ~dB_s ^\epsilon\right| \leq 2\epsilon\qquad\text{a.s.} $$

Furthermore,

$$ \left| e^{-\alpha t}\int_0 ^t e^{\alpha s} ~dB_s ^\epsilon\right|\le\frac {\left \| \dot{B^\epsilon}\right\|_\infty }{\alpha} \left( 1 - e^{-\alpha t} \right)\le\frac {\left \| \dot{B^\epsilon}\right\|_\infty }{\alpha}\qquad\text{a.s.} $$ Summing these and considering the supremum over $t\in[0,T]$, one gets $$ \sup _ {t \in \left[0,T \right] }\left |e^{-\alpha t}\int_0 ^t e^{\alpha s} ~dB_s \right| \leq 2\epsilon+\frac {\left \| \dot{B^\epsilon}\right\|_\infty }{\alpha}\qquad\text{a.s.} $$ hence $$ \limsup_{\alpha\to\infty} \sup _ {t \in \left[0,T \right] }\left |e^{-\alpha t}\int_0 ^t e^{\alpha s} ~dB_s \right| \leq 2\epsilon\qquad\text{a.s.} $$ This holds for every $\epsilon\gt0$ hence $$ \lim_{\alpha\to\infty} \sup _ {t \in \left[0,T \right] }\left |e^{-\alpha t}\int_0 ^t e^{\alpha s} ~dB_s \right| =0\qquad\text{a.s.} $$

Answer 2

I believeI have found also an other solution to my own question. Could you check if it's ok, please?

Since, the process $\left ( \int _ 0 ^t e^{\alpha s} ~ dB_s \right)_{t\geq 0}$ is a Wiener integral, it's a gaussian process with law $ \mathcal N \left(0,\int _ 0 ^t e^{2\alpha s} ~ ds \right)$. Also, $\left ( e^{-\alpha t}\int _ 0 ^t e^{\alpha s} ~ dB_s \right)_{t\geq 0}$ is a supermartingale so we have for all $C>0$

\begin{align} \mathbb P \left \{ \sup _{t \in \left [ 0,T\right ] }\left |e^{-\alpha ^t} \int _ 0 ^t e^{\alpha s} ~ dB_s\right | \geq C\right\} &\leq \frac {1}{C^2} \mathbb E \left \{ \left |e^{-\alpha ^T} \int _ 0 ^T e^{\alpha s} ~ dB_s\right |^2\right \} \\&=\frac {1- e^{-\alpha T}}{2C^2 \alpha} \underset {\alpha \rightarrow \infty} {\longrightarrow} 0 \end{align}

which shows the convergence in probability. Finally we can conclude the a.e. convergence by Borel-Cantelli's theorem.

Answer 3

The following is more a comment than an answer, as I do not know if it is useful.

Let $$X_t = \int_0^{t}e^{\alpha s}dB_s$$ You can prove that the martingale $\{X_t,t\geq 0\}$ has the same distribution as a Brownian motion with changed time $\{B_{C_t},t\geq 0\}$ where $$ C_t = \int_0^t e^{2\alpha s}ds = \frac{e^{2\alpha t}-1}{2\alpha}. $$