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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

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    $\begingroup$ What do you mean by "$D_t$ has strictly positive variance?" Does it mean that the covariance matrix of $D_t$ is positive-definite? $\endgroup$
    – Sangchul Lee
    Commented Feb 11, 2020 at 20:23
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    $\begingroup$ By the central limit theorem and Borel-Cantelli, the probability that this happens infinitely often is positive, and by Kolmogorov's 0-1 law it is necessarily one. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Feb 11, 2020 at 23:24
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    $\begingroup$ If that means component-wise positive variance, it is not enough to capture the dimensionality of your random walk, and so, the answer may be false. For instance, consider the $n=2$ case with the step distribution $$\mathbb{P}(D_t=(1,-1))=\mathbb{P}(D_t=(-1,1))=\frac{1}{2}.$$ Then $\mathbb{E}[D_t]=\vec{0}$ and $\operatorname{Var}(D_{t,i})=1$ for each $i=1,2$, but the walk $(X_t)$ is constrained on the line $x+y=0$, which never touches the region $(-\infty,k)^2$ if $k<0$. $\endgroup$
    – Sangchul Lee
    Commented Feb 12, 2020 at 19:09
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    $\begingroup$ @MateuszKwaśnicki Indeed. But what happens if only the expectation of the increment exists? (assuming that $D$ is not contained in any hyperplane, of course). $\endgroup$
    – fedja
    Commented Feb 12, 2020 at 23:56
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    $\begingroup$ @fedja: Good question! Obviously – yes, if $n = 1$ (the random walk is then recurrent). Also – yes, if the distribution of $X_1$ is in the domain of attraction of a stable law, by exactly the same argument as in my previous comment. I bet the answer is the same in the general case, but at this moment I fail to see a proof. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Feb 13, 2020 at 9:30

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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). 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 $n = 0, 1, \ldots$

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