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Could someone give some advice in order to study $$\underset{t\to\infty}{\operatorname{a.e.-lim}} \frac {1}{t} \int_0 ^ t W_s ~ds,$$ where $(W_t)_{t\geq 0}$ is a standard brownian motion starting at zero ?

Thank's in advance.

Paul
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  • What's the definition of $W_s$. – Mhenni Benghorbal May 27 '13 at 17:48
  • @MhenniBenghorbal: $W_s$ si a brownian motion as writen – Paul May 27 '13 at 17:57
  • what do you mean by ps-lim? – Tim May 27 '13 at 18:24
  • @Tim: Sorry! I was thinking in french so ps (presque sure) means a.e. (almost ever). Edited – Paul May 27 '13 at 18:28
  • OK, but would think that the sequence diverges for almost every path. If it converges it must be to $0$ by symmetry. It might converge in probability to some random variable. – Tim May 27 '13 at 18:39
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    For fixed $t$, the integral is normally distributed with mean zero and variance of order $t^3$, so $\frac{1}{t}$ times the integral is normal with variance of order $t$. I don't think this limit exists. – Chris Janjigian May 27 '13 at 19:19
  • What about using double-logarithm grow arguments, or passing through the Brownian motion with an argument $\frac1t$? – SBF May 28 '13 at 12:02
  • @Ilya: I've tried to use an mean value argument as yours also with $1/t$ but it doesn't work. I wonder the limit does not exist. – Paul May 28 '13 at 12:04
  • As @ChrisJanjigian indicated (did you read this comment?), the distributions diverge hence the random variables cannot converge almost surely (and a little more work show that they diverge almost surely). – Did May 28 '13 at 13:29
  • @Paul: I meant using these arguments to show divergence – SBF May 29 '13 at 08:31

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I'll turn my comment above into an answer. Call $X_t = \frac{1}{t} \int_0^t W_s ds$. We claim that the family $\{ X_t\}$ is not tight and therefore cannot converge in distribution (and hence cannot converge almost surely). This is easy to see, since $X_t$ is Gaussian with mean zero and variance of order $t$. Therefore, for any compact set $K$ $P ( X_t \in K) \to 0$ as $t \to \infty$.

As Did suggested, a little more work gets a stronger result. Clearly, for any $M$ the event $A = \{\limsup_{t \to \infty} X(t) > M \}$ is a tail event in the sigma algebra of the Brownian motion $W_s$. Therefore $P(A) \in \{0,1 \}$. An explicit computation using the normality above gives that $P(X(t) > M) \to \frac{1}{2}$ as $t \to \infty$ and (the reverse) Fatou's lemma then gives that $P(A) \geq \frac{1}{2}$, so $P(A) = 1$. Therefore $P(\{\limsup_{t \to \infty} X(t) = \infty \}) = 1$. Symmetry gives that $P(\{\liminf_{t \to \infty} X(t) = - \infty \}) = 1$. As a result, the limit almost surely does not exist.

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Chris Janjigian
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