Consider a random walk $X\_t = \sum\_{s=1}^t D\_s$ with i.i.d. increments $D\_t \in \mathbb{R}^n$,such that $X$ is a martingale $\mathbb{E}[D\_t]=\vec{0} \in \mathbb{R}^n$, the support of $D\_t$ is bounded, and $D\_{t,i}$ has strictly positive variance for all $i \in \{1,\ldots,n\}$.


Is it true that the probability that the random walk eventually pointwise falls below some threshold $k < 0$ equals $1$, i.e.
$$
 \mathbb{P}[ \exists t \text{ such that } X\_{t,i} \leq k \text{ for all } i \in \{1,\ldots,n\} ] = 1 \,.
$$
for all $k<0$?


Also posted here the question [here](https://math.stackexchange.com/questions/3541918/multidimensional-random-walk-falling-pointwise-below-some-threeshold)



Multidimensional random walk falling pointwise below some threshold
An extended comment, answering the question asked by fedja in one of the comments. *(Edited: In order to simplify the example I chose to work in dimension two, but this example requires dimension at least three.)*


My bet was incorrect: I am very surprised to find that the answer is negative! In order to construct a counterexample (in dimension 3, for simplicity), consider two independent random walks: $A\_n$ the simple random walk in $\mathbb{Z}$, and $B\_n$ a random walk with symmetric $\alpha$-stable distribution in $\mathbb{R}^2$ for some $\alpha \in (1, 2)$. Fix $\epsilon > 0$. We have $$|B\_n| \geqslant n^{1/\alpha - \epsilon}$$ for all $n$ large enough (see Corollary 2 in [Takeuchi, doi:10.2969/jmsj/01620109](https://doi.org/10.2969/jmsj/01620109)). Furthermore, by LIL, $$A\_n \ge -n^{1/2 + \epsilon}$$ for all $n$ large enough. Choosing $\epsilon > 0$ small enough, we find that for an arbitrary constant $p > 0$, $$p A\_n + |B\_n| \geqslant 0$$ for all $n$ large enough. In other words, with probability one the random walk $(A\_n, B\_n)$ visits the cone $\{(a, b) \in \mathbb{R} \times \mathbb{R}^2 : p a + |b| < 0\}$ finitely many times.


It remains to choose $p$ and rotate $(A\_n, B\_n)$ appropriately, so that the above cone fits into the negative octant. To be specific, we set $$X\_n = A\_n \vec{v}\_1 + B\_{n,1} \vec{v}\_2 + B\_{n,2} \vec{v}\_3, $$ where $\vec{v}\_1 = \tfrac{1}{\sqrt{3}} (1, 1, 1)$ and otherwise $\vec{v}\_1, \vec{v}\_2, \vec{v}\_3$ are arbitrary orthonormal vectors in $\mathbb{R}^3$. Then, by an easy calculation, $$ \max\{X\_{n,1}, X\_{n,2}, X\_{n,3}\} \geqslant \tfrac{1}{\sqrt{3}} A\_n + \tfrac{1}{\sqrt{6}} |B\_n| \ge 0 $$
for all $n$ large enough (by choosing $p = \sqrt{2}$ above). This means that there is $k < 0$ such that with positive probability we have $X\_{n,j} \geqslant k$ for every $j = 1, 2, 3$ and every $n = 0, 1, \ldots$


