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Suppose we are given an $L^2(\mathcal{D})$-valued Brownian motion $W_t$ defined by $$W_t:=\sum_{k=1}^{\infty}\sqrt{\sigma_k}W_t^k\phi_k(x),$$ where $\mathcal{D}$ is bounded domain in $\mathbb{R}^d$, $\{\phi_k(x)\}$ forms the complete orthogonal basis of $L^2(\mathcal{D})$, $\{W_t^k\}_{k\in \mathbb{N}^{+}}$ are mutually independent one-dimensional standard BM, and $\sigma_k$ satisfies $$\sum_{k=1}^{\infty}\sigma_k<\infty.$$

Is it possible that

$$\lim_{k\to \infty}\sup_{t\geq 0}(-\Lambda t+\sqrt{\sigma_k}W^k_t)=0,\ \mathbb{P}-a.s.,$$

where $\Lambda$ is an given positive number?

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    $\begingroup$ the sup gives independent exponential r.v.'s with known parameters $\endgroup$ – mike Jan 9 '14 at 17:29
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I have not done the computation, but shouldn't you be able to compute $P(\sup_{t\geq 0}(-\Lambda t+\sqrt{\sigma_k}W_t^k)\geq a)$ for each fixed $a$ and $k$, then apply Borel-Cantelli?

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  • $\begingroup$ That contributes to convergence in probability only but not P-a.s. $\endgroup$ – Yue Jan 12 '14 at 22:14
  • $\begingroup$ But Borel cantelli should tell you that the probability that $\sup_{t\geq 0}(-\Lambda t+\sqrt{\sigma_k}W_t^k)\geq a$ happen for infinitely many $k$ is 0, which is just a.s. convergence, right? Isn't your $W^k_t$ independent? $\endgroup$ – Thomson Tong Jan 15 '14 at 23:34
  • $\begingroup$ Borel Cantelli give you the probability of $\limsup$ rather than $\lim$ to be zero, which implies that you could only find a subsequence which converges to zero $\mathbb{P}-a.s.$.... $\endgroup$ – Yue Jan 16 '14 at 13:03
  • $\begingroup$ $\mathbb{P}$-a.s. Convergence of the subsequence is not what I want for... $\endgroup$ – Yue Jan 16 '14 at 13:04

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