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{\bf Question}

In a three-dimensional walk a particle starts at an origin in
three-space which can be thought of as the centre of a $2\times
2\times 2$ cube. At each step the particle moves north or south
(each with probability $\frac{1}{2}$), east or west (each with
probability $\frac{1}{2}$) and up or down (each with probability
$\frac{1}{2}$).  Thus in one step the particle moves to one of the
eight corners of the cube.  If the walk continues forever show
that the particle is not certain to return to the origin.

\vspace{.25in}

{\bf Answer}

The reasoning is very similar to the 1-dimensional case. To return
to 0 the particle must do so as a result of a simultaneous return
in in respect of movement all three directions.

$\ds P({\rm Particle\ is\ at\ 0\ after\ } k\ {\rm steps}) =
\left\{
\begin{array}{ll} 0 & {\rm if\ } k{\rm\ is\ odd} \\ \left[ \left(
\begin{array}{c} 2n \\ n \end{array} \right) \left( \ds\frac{1}{2}
\right)^{2n} \right]^3 & {\rm if\ }k = 2n \end{array} \right.$

\medskip

The expected number of returns to the origin is \begin{eqnarray*}
E & = & \sum_{k = 1}^\infty E(R_k) \cdot R_k =
\left\{\begin{array}{ll}  1 & {\rm if\ particle\ at\ 0\ after\ }
k\ {\rm steps} \\ 0 & {\rm otherwise} \end{array} \right. \\ & = &
\sum_{k=1} ^\infty P(R_k = 1) \\ & = & \sum_{n=1}^\infty  \left[
\left( \begin{array}{c} 2n \\ n \end{array} \right) \left(
\frac{1}{2} \right)^{2n} \right]^3   \\ & \approx &
\sum_{n=1}^\infty \left( \frac{1}{n \pi} \right) ^{\frac{3}{2}}
\end{eqnarray*}

using Stirling's approximation. The series converges, so $E<
\infty$. Standard theory gives $\ds P = 1 - \frac{1}{1+E}$ so $P
< 1$




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