# The convergence of the integral of square of Brownian motion

Recently, I came across the convergence for the integral of Brownian motions in continuous time case.

How to prove the following almost sure convergence? \begin{eqnarray*} \frac{B_{T}\int_{0}^{T}B_{t}dt}{T\int_{0}^{T}(B_{t})^{2}dt} &\overset{a.s.}{\rightarrow }&0\,, \end{eqnarray*} where $B$ is a Brownian motion and $\overset{a.s.}{\rightarrow }$ denote convergence almost surely.

I'm thinking this is really obvious, But I cannot get my head round it. I thought about using the iterated logarithm for Brownian motion ($\limsup_{t\to \infty}\frac{B_t}{\sqrt{2t\log\log(t)}}=1$).

I will deal with $\int_{0}^{1}(B_{u})^{2}du$ as follows:

1) Since $\Bbb E\left[\exp(-u\int_0^t B^2_s\,ds)\right]={1\over\sqrt{\cosh(t\sqrt{2u})}}.$

2) We can see that \begin{eqnarray*} \mathbb{E}\left[ \int_0^t {B_s}^2 ds\right] &=&\mathbb{E}\left[\left( \int_0^t B_s dB_s\right)^2\right]=\mathbb{E}\left[\left( \frac{1}{2} B_t^2-\frac{1}{2}t\right)^2\right]\\ &=&\mathbb{E}\left[ \frac{1}{4} B_t^4+\frac{1}{4}t^2-\frac{1}{2}B_t^2t\right]= \frac{3}{4}t^2+\frac{1}{4}t^2-\frac{1}{2}t^2=\frac{t^2}{2} \end{eqnarray*}

\begin{eqnarray*} \mathrm{Var}\left[ \int_0^t {B_s}^2 ds\right] &=& \mathbb{E}\left[\left( \int_0^t {B_s}^2 ds\right)^2\right] - \left(\mathbb{E}\left[ \int_0^t {B_s}^2 ds\right]\right)^2\\ &=&\mathbb{E}\left[\left( \int_0^t {B_s}^2 ds\right)^2\right] - \frac{t^4}{4}=\frac{1}{36}\mathbb{E}\left[\left(B_t^4-4\int_0^t B_t^3dB_t\right)^2\right]-\frac{t^4}{4}\,. \end{eqnarray*}

Now, using the scaling properties, for any $\varepsilon>0$, we have \begin{eqnarray*} \frac{B_{T}\int_{0}^{T}B_{t}dt}{T\int_{0}^{T}(B_{t})^{2}dt}&\overset{d}{=}& \frac{T^{\frac{3}{2}}B_{T}\int_{0}^{1}B_{u}du}{{T}^{3}% \int_{0}^{1}(B_{u})^{2}du}=\frac{B_{T}\int_{0}^{1}B_{u}du}{T^{\frac{3}{2}} \int_{0}^{1}(B_{u})^{2}du}\leq \frac{T^{\frac{1}{2}+\varepsilon}\int_{0}^{1}B_{u}du}{T^{\frac{3}{2}} \int_{0}^{1}(B_{u})^{2}du}\\ &=&\frac{T^{\varepsilon}\mathbb{N}(0,1)}{T \int_{0}^{1}(B_{u})^{2}du}=\frac{T^{\varepsilon}O_{p}(1)}{T \int_{0}^{1}(B_{u})^{2}du}\overset{a.s.}{\rightarrow }0\,. \end{eqnarray*} where $\overset{d}{=}$ denotes equivalence in distribution and $\mathbb{N}$ denotes Normal distribution.

Is my derivation is correct?

Any help is appreciated!

Assume for now that with probability 1, $$\limsup_{T\to \infty}\frac 1 T \int_0^T \mathbf 1[|B_t|\leq 2] dt < \frac 1 3.\tag{*}$$

Assuming (*), for all sufficiently large $T$, $$\frac 1 T\int_0^T (B_t^2-|B_t|)dt \geq \frac 2 3\cdot 2+\frac 1 3\cdot \left(-\frac 1 4\right)\geq 0$$

which implies $\int_0^T B_t^2 dt \geq |\int_0^T B_t dt|.$ You know that $B_T/T$ tends to $0$ almost surely, which gives the claimed convergence \begin{eqnarray*} \frac{B_{T}\int_{0}^{T}B_{t}dt}{T\int_{0}^{T}(B_{t})^{2}dt}\xrightarrow{a.s.}0. \end{eqnarray*}

It remains to show that (*) holds with probability $1$. This should be a complete no-brainer, but I don't know the right reference. Here are two arguments:

Proof 1: Consider the embedded continuous-time random walk on $\mathbb Z$ obtained by defining $\tau_0$ to be the maximum value $\tau_0\leq t$ such that $B_{\tau_0}$ is an integer, then letting $S_t=B_{\tau_0}$. The process $S_t$ is a continuous-time (and "non-exploding") Markov chain so its transition probabilities converge, and by the shift symmetry of $\mathbb Z$ they must converge to $0$. This means the time spent in $\{-3,-2,-1,0,1,2,3\}$ up to time $T$ is eventually at most $T/4$. If you prefer positive recurrent Markov chains, take $S_t$ modulo $100$.

Proof 2: Define variables $X_1,X_2,\dots$ by $$X_n=\frac 1 {t_0} \int_{(n-1)t_0}^{nt_0} \mathbf 1[|B_t|\leq 2]dt.$$

where $t_0$ is chosen large enough such that $\mathbf E[X_n|B_{(n-1)t_0}]<1/10$ regardless of $B_{(n-1)t_0}$ - this is possible by the absolute continuity of Brownian motion occupation measure (see Mörters and Peres's Brownian Motion, Theorem 3.25) and by the scaling property of Brownian motion. We can choose variables $Y_n$, depending only on $B_{(n-1)t_0}$, to get $\mathbf E[X_n+Y_n|B_{(n-1)t_0}]=1/10$. Then by the martingale strong law of large numbers applied to the sequence $X_n+Y_n$, $$\limsup_{n\to\infty}\frac 1 {nt_0} \int_0^{nt_0} \mathbf 1[|B_t|\leq 2] dt < 1/10$$ with probability $1$. Comparing to the left-hand-side of (*) with $n=\lfloor T/t_0\rfloor$, the difference is bounded by $t_0/T$ which is less than $1/3-1/10$ for $T\geq 10t_0$ say.

• Can I use the the scaling properties to prove the convergence? Thanks for your help. How to deal with $\frac{B_{T}^{2}}{\int_{0}^{T}(B_{t})^{2}dt}\overset{a.s.}{\rightarrow }0$. – steven Sep 21 '17 at 9:40
• @steven: if you have another question could you add it to the original post? I don't see how scaling property arguments would be enough by themselves. – Dap Sep 21 '17 at 14:25