# Scaled integrated Brownian motion has limit

Let $B$ be a standard Brownian motion and put $$X(t)=\frac{1}{\sqrt{t}}\int_{0}^{t}f(B(s))ds,$$ where $f \in L_1(\mathbb{R}^{1})$ and $\int f(x)dx=1$. Show that $$\lim_{t \rightarrow \infty} EX(t) =\sqrt{2/\pi}$$ and $$\lim_{t \rightarrow \infty} E[X(t)^2]=1.$$

I was thinking to try writing them as Riemann sums, but it seems that it does not work out well. Any hints or help would be appreciated.

## 1 Answer

Let

$$X_t := \frac{1}{\sqrt{t}} \int_0^t f(B_s) \, ds.$$

Using that $B_s \sim N(0,s)$, $s \geq 0$, and applying Fubini's theorem yields

\begin{align*} \mathbb{E}(X_t) &= \frac{1}{\sqrt{t}} \int_0^t \mathbb{E}(f(B_s)) \, ds \\ &= \frac{1}{\sqrt{t}} \int_{\mathbb{R}} \int_0^t f(x) \frac{1}{\sqrt{2\pi s}} \exp \left(- \frac{x^2}{2s} \right) \, ds \, dx. \end{align*}

Now we substitute $r := \frac{\sqrt{s}}{\sqrt{t}}$ and obtain

$$\mathbb{E}(X_t) = \frac{2}{\sqrt{2\pi}} \int_{\mathbb{R}} \left[ \int_0^1 \exp \left( - \frac{x^2}{2r^2 t} \right) \, dr \right] f(x) \, dx.$$

From the dominated convergence theorem, it follows that

$$\lim_{t \to \infty} \mathbb{E}(X_t) = \sqrt{\frac{2}{\pi}} \int f(x) \, dx = \sqrt{\frac{2}{\pi}}.$$

For the second equation, write

$$\left| \int_0^t f(B_r) \, dr \right|^2 = \int_0^t \int_0^t f(B_r) f(B_s) \, dr \, ds,$$

and use that $(B_r,B_s)$ is Gaussian with mean vector $(0,0)$ and covariance matrix

$$\begin{pmatrix} r & r \\ r & s \end{pmatrix}$$

for $r \leq s$. By choosing a suitable substitution, we obtain - as in the first part - an integral which can be calculated explicitely.

• Thank you very much for the hint. For the second one I obtain something like $$\frac{1}{t}\int_{\mathbb{R}}\int_{\mathbb{R}}(\int_{0}^{t}\int_{0}^{t} \frac{1}{2\pi\sqrt{|rs-r^2|}}exp(-\frac{x^2r+y^2s}{2})ds dr)f(x)f(y)dxdy$$ . From here it doesn't seem very clear for me what kind of substitutions I need to make, the ones that you did will not work as we have that factor $|r-s|$. Any hint on what substitutions I could make? I would be grateful. – Frank Zermelo Jun 24 '14 at 8:26
• @FrankZermelo No, that's not the joint density of the Brownian motion. The exponential term should read $$\exp \left( - \frac{(y-x)^2}{2(s-r)} - \frac{x^2}{2r} \right)$$ instead. Then use the substitution $$u := \sqrt{\frac{r}{t}} \qquad v := \sqrt{\frac{s-r}{t}}.$$ – saz Jun 24 '14 at 9:09