4
$\begingroup$

I wonder whether a $d$-dimensional random walk $S_n$, generated by the infinite i.i.d. copies of X given by:

$X=e_1=(1, 0, 0, ..., 0)$ (with probability $p_1$)

$X=e_2=(0, 1, 0, ..., 0)$ (with prob $p_2$)

. . .

$X=e_d $ (with prob $p_d$)

$X=0$ (with prob $p_0=1-p_1-p_2...-p_d$),

visits any infinite cylinder of radius $\sqrt{d}$ and parallel to the vector $\overrightarrow{p}:=(p_1, p_2,..., p_d)$ infinitely often a.s. ? More precisely, if $L$ is a line parallel to the vector $\overrightarrow{p}$ and $A:=\{ y\in \mathbb{N}^d : |y-L|\leq \sqrt{d} \}$, can I say $\mathbb{P}(S_n \in A, \; \; i.o.)=1$?

If $d=1$ this holds by law of iterated logarithm.

$\endgroup$

1 Answer 1

5
$\begingroup$

Well, for $d\geq 2$, the projection of $S_n$ onto a hyperplane orthogonal to $\vec{p}$ is a zero-mean $(d-1)$-dimensional random walk with bounded jumps. Therefore, the answer to your question is ''yes'' for $d\leq 3$ and ''no'' for $d\geq 4$. (That fact about zero-mean random walks is well known, but see e.g. Theorem 1.5.2 of the book "Non-homogeneous random walks" by Menshikov-Popov-Wade if you need a reference.)

P.S. I guess that it is implicitly assumed that all $p_i$s are strictly positive --- otherwise you can just forget about some of the coordinates and effectively reduce the dimension.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.