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I had a probability test and that was one of the questions:

We have an infinite grid of hexagons like this: octagons

Each edge have length 1 and all the degrees are 120°.

There's a particle in one of the vertices and each second it randomly moves to one of it's neighbours.

After n seconds what is the expected distance squared of the particle from it's starting position?

I spent a lot of time on that but I really have no idea how to even approached this question.

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  • 1
    $\begingroup$ Note: Those are hexagons, not octagons. $\endgroup$ – Mees de Vries Feb 7 at 12:34
  • $\begingroup$ @Mees de Vries sorry I meant hexagons $\endgroup$ – Tomer Wolberg Feb 7 at 12:36
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Let $(Z_n)$ be i.i.d. $\mathbb{C}$-valued random variables having the law $ \mathbb{P}(Z_n = -\omega^k) = \frac{1}{3}$ for any $ k \in \{ 0, 1, 2 \}$, where $\omega = e^{2\pi i/3}$. Using the $90^{\circ}$-rotation of the lattice in OP's figure, the $n$-th step $X_n$ of the random walk started at $0$ can be realized as

$$ X_n = \sum_{k=1}^{n} Z_1 \cdots Z_k. $$

Then the expectation of the square-distance from the starting point is

\begin{align*} \mathbb{E}[|X_n|^2] &= \mathbb{E}[X_n \overline{X_n}] \\ &= \sum_{j,k=1}^{n} \mathbb{E}[(Z_1\cdots Z_j)\overline{(Z_1 \cdots Z_k)}] \\ &= n + \sum_{1 \leq k < j \leq n} \underbrace{\mathbb{E}[Z_{k+1}\cdots Z_j]}_{=0} + \sum_{1 \leq j < k \leq n} \underbrace{\mathbb{E}[\overline{Z_{j+1}\cdots Z_k}]}_{=0} \\ & = n. \end{align*}

The following is a simulation using 2500 samples of $X_{100}$, numerically verifying the above computation.

Numerical simulation

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