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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 ?

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Yes I know that but even knowing this result I could not show the result. –  Paul Jan 16 '13 at 12:09
    
My first suggestion is not useful. I will delete it. Let me try another approach. From where do you have this problem? Is it from a book or exercise sheet? –  user8 Jan 16 '13 at 12:30
    
From an old exam –  Paul Jan 16 '13 at 12:50
    
Cross-post on MO with answer –  Paul Feb 18 '13 at 0:55

3 Answers 3

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.

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Why would $\left ( e^{-\alpha t}\int _ 0 ^t e^{\alpha s} ~ dB_s \right)_{t\geq 0}$ be a submartingale? –  Siméon Jan 16 '13 at 15:02
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Are you sure it is ? –  Siméon Jan 16 '13 at 16:07
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Hey! I am the sceptic one here! Note $(X_t)_t$ that process and suppose it is a supermartingale with respect to its natural filtration. Then $(-X_t)_t$ is a submartingale with respect to the same filtration. Now, notice that $(X_t)_t$ and $(-X_t)_t$ have the same distribution (because of the invariances of the brownian motion). Hence $(X_t)$ would be a martingale. Your turn to convince me! –  Siméon Jan 16 '13 at 16:51
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It does not work. It is true that $e^{-\alpha t} \leq e^{-\alpha s}$ but you do not know if $\int_0^s e^{\alpha u}dB_u \geq 0$. –  Siméon Jan 16 '13 at 17:04
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Another skeptic needs a proof for the "convergence in probability hence almost sure convergence", more convincing than at present. –  Did Jan 16 '13 at 17:12

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 $$ C_t = \int_0^t e^{2\alpha s}ds = \frac{e^{2\alpha t}-1}{2\alpha}. $$

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can you give me a reference for $\textit{...has the same distribution as a Brownian motion with changed time } \{B_{C_t},t\ge 0\}$? I know that $X_t$ is normally distributed with mean zero and variance $\int_0^t e^{2\alpha s}ds$. But why is it a BM? –  user8 Jan 16 '13 at 15:04
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George Lowther's blog for instance. –  Siméon Jan 16 '13 at 15:07
    
Theorem 1.5. (Dambis / Dubins-Schwarz (1965)), that you xan find inside math.upatras.gr/dimitsana/thermo/grthermo/Files/… –  Paul Jan 16 '13 at 15:09
up vote 0 down vote accepted

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 $$ \lim_{\alpha\to\infty} \sup _ {t \in \left[0,T \right] }\left |e^{-\alpha t}\int_0 ^t e^{\alpha s} ~dB_s \right| =0\qquad\text{a.s.} $$

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@did: Sorry! I had a typo. It's $ {\left \| \dot{B^\epsilon}\right\|_\infty }/ { \alpha}$ not ${\left \| {B^\epsilon}\right\|_\infty }/{\alpha}$. This because by assumption $B^\epsilon \in \mathcal C^1$. What do you think now? –  Paul Jan 17 '13 at 12:12
    
This seems to work, I would say... (I rewrote marginally your post to simplify things, please revert to the previous version if you feel like doing so.) –  Did Jan 17 '13 at 21:52
    
@Paul: This proof seems a bit odd to me, but I do not see any big mistake. Maybe you should ask your teacher and tell us? –  Siméon Jan 17 '13 at 23:04
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Actually, I think there is a misstake. I'm not confortable with the aprotimation of $t \mapsto B_t(\omega)$ by $t \mapsto B^\epsilon (t)$ wich is not uniform in $\omega$. –  Paul Jan 17 '13 at 23:10
    
@did: thank you for rewriting! It's clear like this. –  Paul Jan 17 '13 at 23:13

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