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I am deriving a formula for a volume of Wiener sausage in one dimension.

$$\mathbb{E}[\operatorname{vol}(W(t))] = 2r+\sqrt{\frac{8t}{\pi}}$$ where $W(t) = \bigcup_{s\leq t}\mathcal{B}(\epsilon(s),r)$. $\mathcal{B}$ - ball (just a segment in this case). $\epsilon(s)$ - standard Brownian motion. Vol for volume (length in 1d).

I follow the trick my lecturer used for higher dimensions. Let $\tau_A$ stand for a hitting time of $\epsilon$ for $A$:

$$\mathbb{E}[\operatorname{vol}(W(t))]=\int_{\mathbb{R}}\mathbb{P}(y \in \bigcup_{s\leq t}\mathcal{B}(\epsilon(s),r))\;dy=\int_{\mathbb{R}}(\tau_{\mathcal{B}(y,r)}\leq t))\;dy=\operatorname{vol}(\mathcal{B}(0,r))+$$ $${}+\int_{\mathbb{R}\backslash \mathcal{B}(0,r)}(\tau_{\mathcal{B}(y,r)}\leq t))\;dy=2r+\int_{-\infty}^{-r}\mathbb{P}(\tau_{y+r}\leq t)\;dy+\int_{r}^{\infty}\mathbb{P}(\tau_{y-r}\leq t)\;dy$$

I use $\mathbb{P(\tau_{|a|} \leq t)=\frac{2}{\sqrt{2\pi t}}\int_{|a|}^{\infty}}e^{-\frac{x^2}{2t}}dx$ and the latter 2 integrals evalute to $\infty$ each (after use of Fubini).

Thank you for any suggestions.

Edit: $\int_{-\infty}^{-r}\mathbb{P}(\tau_{y+r}\leq t)\;dy+\int_{r}^{\infty}\mathbb{P}(\tau_{y-r}\leq t)\;dy = 2\int_0^\infty\mathbb{P}(\tau_y\leq t)\;dy=$=$2\int_0^\infty\int_y^\infty\frac{2}{\sqrt{2\pi t}}e^{-\frac{x^2}{2t}}\;dx\;dy=\frac{4}{\sqrt{2\pi t}}\int_0^\infty\int_0^xe^{-\frac{x^2}{2t}}\;dy\;dx=\frac{4}{\sqrt{2\pi t}}\int_0^\infty xe^{-\frac{x^2}{2t}}\;dx=\sqrt{\frac{8t}{\pi}}$

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1 Answer 1

up vote 3 down vote accepted

In other words, you are after $V(t)=2r+\mathrm E(M_t-m_t)$ where $(B_t)$ is a Brownian motion starting from $B_0=0$, $M_t=\max\limits_{0\leqslant s\leqslant t}B_s$ and $m_t=\min\limits_{0\leqslant s\leqslant t}B_s$. This is also $V(t)=2r+2\mathrm E(M_t)$. The reflection principle shows that $\mathrm P(M_t\geqslant x)=2\mathrm P(B_t\geqslant x)$ for every $x\geqslant0$, hence $\mathrm E(M_t)=2\mathrm E(B_t;B_t\gt0)$. By scaling, $\mathrm E(B_t;B_t\gt0)=\sqrt{t}\mathrm E(B_1;B_1\gt0)$. Finally, $\mathrm E(B_1;B_1\gt0)=\frac1{\sqrt{2\pi}}\int\limits_0^{+\infty}x\mathrm e^{-x^2/2}\mathrm dx=\frac1{\sqrt{2\pi}}$. Putting everything together yields your formula. (But to call this a Wiener sausage is a bit deceptive.)

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This is very neat. Indeed the name is deceptive, but I don't know a better name for it. Thank you for showing me the short way. I still feel that I am so close to the answer, trying to pursue. –  Tom Artiom Fiodorov Jan 18 '12 at 18:35
    
For $r=0$, this is the range (note that $r$ is useless anyway). // Your post says that the latter 2 integrals evalute to $\infty$ each... No they do not. –  Did Jan 18 '12 at 19:24
    
I got the answer! I might post it later. r is indeed useless. Thank you. –  Tom Artiom Fiodorov Jan 18 '12 at 19:31

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