Probability of convergence of $\int_{1}^{\infty} \frac{B_s}{s^{3/2}} \, \mathrm d s$

Question

I'm trying to find out whether the integral $$\int_{1}^{\infty} \frac{B_s}{s^{3/2}} \, \mathrm d s,$$ where $B_s$ is a standard Brownian motion, converges with non-zero probability/almost surely as an improper integral. In the Lebesgue sense, this definitely diverges with nonzero probability (and I'd guess almost surely), because $\Bbb E[|B_t|] = \sqrt{\frac{2t}{\pi}}$. In the improper/Riemann-sense, it's not quite clear to me though, since I'd think one gets something akin of an alternating sequence, and since the law of the iterated logarithm makes $\frac{B_s}{s^{1/2+\varepsilon}}$ vanish.

Thanks!

Edit: With a substitution $t = \frac{1}{s}$ I get $$\int_{1}^{\infty} \frac{B_s}{s^{3/2}} \, \mathrm d s = \int_0^1 \frac{1}{t^{3/2}} t B_{1/t} \, \mathrm d t \overset{d}{=} \int_0^1 \frac{1}{t^{3/2}}\tilde B_t \, \mathrm dt,$$ where $\tilde B_t$ is a Brownian motion as well, by the time inversion identity. This looks more like a divergent integral to me, but this must have infinitely many zeros as well in the interval $[0,1]$ so I can't conclude that this is not convergent alternating series.

Edit2: The convergence set of the integral $\int_0^1 \frac{1}{t^{3/2}}\tilde B_t \, \mathrm dt$ should be measurable with respect to the germ-sigma-algebra $\mathcal F_0^+$, I think, and thus have measure $0$ or $1$ by Blumenthal's 0-1-law.

Answer 1

Denote \begin{gather*} I(t)=\int_1^t \frac{B_s}{s^{3/2}}\,\mathrm{d}s,\\ C=\Big\{ \omega: -\infty<\varliminf_{t\to\infty} I(t)=\varlimsup_{t\to\infty} I(t) <\infty \Big\}, \tag{1} \end{gather*} The $ \mathsf{P}(C)=0 $ will be proved as follows.

Let \begin{equation*} Y_n=I(4^{n+1})-I(4^n)=\int_{4^n}^{4^{n+1}} \frac{B_s}{s^{3/2}}\,\mathrm{d}s, \quad n\ge 0, \end{equation*} then $ \{Y_n, n\ge0 \}$ are Gaussian distributed with \begin{equation*} \mathsf{E}[Y_n]=0,\qquad \mathsf{var}[Y_n] =\log4-1 \stackrel{\triangle}{=}\sigma_0^2 , \qquad n\ge 0. \end{equation*} Taking $ a=\sigma_0\sqrt{\pi/8}(>0) $, then \begin{align*} \mathsf{P}(|Y_n|<a)&=\sqrt{\dfrac{2}{\pi}}\int_{0}^{a/{\sigma_0}}e^{-x^2/2}\,\mathrm{d}x \le \sqrt{\dfrac{2}{\pi}}\frac{a}{\sigma_0}=\frac12,\\ \mathsf{P}(|Y_n|\ge a)&=\frac12.\qquad n\ge 0 \end{align*}
Furthermore, \begin{align*} \mathsf{P}(\varlimsup_{n\to\infty} |Y_n|\ge a) &\ge \mathsf{P}( \{\omega:|Y_n|\ge a\} \; \text{i.o.})\\ & \ge \varlimsup_{n\to\infty} \mathsf{P}(|Y_n|\ge a)\qquad \text{(by Fatou Lemma)} \\ &\ge \frac12. \tag{2} \end{align*} Meanwhile, \begin{align*} C&\subset \Big\{ \omega: -\infty<\varliminf_{n\to\infty} I(4^n)=\varlimsup_{n\to\infty} I(4^n) <\infty \Big\} \subset \Big\{ \omega: \lim_{n\to\infty} |Y_n|=0 \Big\} .\\ C^c& \supset \Big\{ \omega: \varlimsup_{n\to\infty} |Y_n|\ge a \Big\} \end{align*} Furthermore, \begin{align*} \mathsf{P}(C^c)&\ge \mathsf{P}(\varlimsup_{n\to\infty} |Y_n|\ge a )\ge \frac12>0,\\ \mathsf{P}(C)& <\frac12<1. \end{align*} Now, $ \mathsf{P}(C)=0 $ holds from 0-1 law(In Edit 2 above).

Answer 2

Well this is not a full proof but more the sketch of it and as the calculations are needed to be checked as I often make mistakes on this part but if I am right the integral is not defined as a random variable in the same way as there is no proper limit to a Brownian motion as $t$ tends to infinity.

So using Integration by parts from 1 to t we get : $$\int_1^t B_s/s^{3/2}ds=2(B_1-B_t / \sqrt{t}) + 2\int_1^t 1/s^{1/2}dB_s$$ Let's note $Y_t=2\int_1^t 1/s^{1/2}dB_s$ the term $Z_t=B_t / \sqrt{t}$ is equal in law to a centered standard gaussian $N(0,1)$ and the stochastic integral term for finite time it is a true martingale as its quadratic variations equals $4\int_1^t 1/s ~ds=4ln(t)$ tends to infinity. But it's a Wiener integral so its law is also gaussian with mean 0 and variance $4ln(t)$. The last thing to check is the correlation between $Z_t$ and $Y_t$,so $E[(B_t/\sqrt{t}-B_1)*2\int_1^t 1/s^{1/2}dB_s]=E[\int_1^t dB_s*\int_1^t 1/s^{1/2}dB_s]*2/\sqrt{t}= E[\int_1^t 1/s^{1/2}ds]*2/\sqrt{t}=\int_1^t 1/s^{1/2}ds*2/\sqrt{t}=(1-1/\sqrt{t})$ which tends to 1 as t goes to infinity. I haven't checked thoroughly but the integral is certainly gaussian with constant mean $B_1$ and variance with a leading term that grows as $\ln(t)$ (so the covariance does't arm the leading term but this can be checked). So the limit of the integral doesn't exist, and as usual with this kind of stuff the lim inf and sup each must exist and tend respectively to $-\infty$ and $+\infty$ in my opinion.