Random walk on a k-dimensional grid

Consider a $$k$$-dimensional grid with integer points and let's begin our random walk at the origin $$(0,0,\dots,0)$$. Every step you have to move on each cartesian axis in the following way:

$$x_1$$ axis: one step forward with probability $$\frac{1}{3}$$ , one step backward with probability $$\frac{1}{3}$$ , remain where you were with probability $$\frac{1}{3}$$.

$$\vdots$$

$$x_k$$ axis : one step forward with probability $$\frac{1}{3}$$ , one step backward with probability $$\frac{1}{3}$$ , remain where you were with probability $$\frac{1}{3}$$.

So we can define the random discrete vector $$U_i=(X_{i1},\dots,X_{ik})$$ that register the random actions taken on each axis at the $$i$$-th step of our random walk. The marginal density function of the random variable $$X_{ij}$$ is:

$$f_{X_{ij}}(t) = \begin{cases} \frac{1}{3} \mbox{ if t\in\{-1,0,1\}} \\ 0 \mbox{ otherwise}\end{cases}$$

We would like to compute the following probability: $$P\Biggl(\bigcup_{j=1}^{k}\Bigl\{\Bigl|\sum_{i=1}^{N}X_{ij}\Bigr|>\sqrt{N}\Bigr\}\Biggr)$$ as the number of steps $$N$$ approaches infinity. So we would like to know the probability after $$N$$ steps (where $$N$$ is a sufficiently large number) to be outside the $$k$$-dimensional cube with size $$2\sqrt{N}$$.

$$k=1$$ case We begin with the 1-dimensional case (a line). We want to compute: $$P\Bigl(\Bigl|\sum_{i=1}^{N}X_i\Bigr|>\sqrt{N}\Bigr) = P\Bigl(\sum_{i=1}^{N}X_i>\sqrt{N}\Bigr) + P\Bigl(\sum_{i=1}^{N}X_i<-\sqrt{N}\Bigr) = 2\cdot P\Bigl(\sum_{i=1}^{N}X_i>\sqrt{N}\Bigr)$$ Since the random variables $$\{X_i\}_{i=1}^{N}$$ are indipendents and equally distributed, and $$\mathbb{E}(X_i) = 0$$,$$Var(X_i) = \frac{2}{3}$$, then we can apply the central limit theorem, so that: $$P\Bigl(\sum_{i=1}^{N}X_i>\sqrt{N}\Bigr) = P\Bigl(\frac{\sum_{i=1}^{N}X_i}{\sqrt{N}\sqrt{\frac{2}{3}}}>\frac{\sqrt{N}}{\sqrt{N}\sqrt{\frac{2}{3}}}\Bigr) = 1- P\Bigl(\frac{\sum_{i=1}^{N}X_i}{\sqrt{N}\sqrt{\frac{2}{3}}}\leq\sqrt{\frac{3}{2}}\Bigr) \longrightarrow 1-\phi\Bigl(\sqrt{\frac{3}{2}}\Bigr) \approx 0.1112$$ where $$\phi(t) = \frac{1}{\sqrt{2\pi}}\int_{\infty}^{t}e^{-\frac{s^2}{2}}ds$$ is the cumulative distribution function of a standard normal variable ($$Z\sim\mathcal{N}(0,1)$$).

So the probability requested is $$2\cdot 0.1112 = 0.2224$$.

$$k$$-dimensional case

Call $$P\Bigl(\Bigl|\sum_{i=1}^{N}X_{ij}\Bigr|>\sqrt{N}\Bigr) = p = 0.2224$$ which is the same for all axes ($$p$$ is indipendent from the index $$j$$).

$$P\Biggl(\bigcup_{j=1}^{k}\Bigl\{\Bigl|\sum_{i=1}^{N}X_{ij}\Bigr|>\sqrt{N}\Bigr\}\Biggr) = \sum_{j=1}^{k}p - \sum_{j

Where we have applied the fact that the random variables $$\Bigl|\sum_{i=1}^{N}X_{i1}\Bigr|,\dots,\Bigl|\sum_{i=1}^{N}X_{ik}\Bigr|$$ are indipendent, so for example in the second term:

$$P\Bigl(\Bigl\{\Bigl|\sum_{i=1}^{N}X_{ij}\Bigr|>\sqrt{N}\Bigr\}\cap\Bigl\{\Bigl|\sum_{i=1}^{N}X_{im}\Bigr|>\sqrt{N}\Bigr\}\Bigr) = P\Bigl(\Bigl\{\Bigl|\sum_{i=1}^{N}X_{ij}\Bigr|>\sqrt{N}\Bigr\}\Bigr)P\Bigl(\Bigl\{\Bigl|\sum_{i=1}^{N}X_{im}\Bigr|>\sqrt{N}\Bigr\}\Bigr) = p^{2}.$$ Limit case: $$k\to\infty$$ We can now easily compute the following probability: $$P\Biggl(\bigcup_{j=1}^{\infty}\Bigl\{\Bigl|\sum_{i=1}^{N}X_{ij}\Bigr|>\sqrt{N}\Bigr\}\Biggr) = \lim_{k\to\infty} 1-(1-p)^{k} = 1$$ since $$p=0.2224\in(0,1) \implies (1-p)\in(0,1)$$.

So the probability of escaping from a $$k$$-dimensional grid of size $$2\sqrt{N}$$ as $$N\to\infty$$ and $$k\to\infty$$ approaches the value 1!

My question: Do you think that this process is correct? I first applied the CLT and then used the indipendence of these random variables (since each step our "point" has to make a move across all the axes, and the movement made on the axis $$x_i$$ does not influence the movement made on the axis $$x_j$$).

Table of values:

$$k=1 \implies P = 0.2224$$;

$$k=2 \implies P = 0.3978$$;

$$k=3 \implies P = 0.5327$$;

$$\vdots$$

\begin{align} P\Biggl(\bigcup_{j=1}^{k}\Bigl\{\Bigl|\sum_{i=1}^{N}X_{ij}\Bigr|>\sqrt{N}\Bigr\}\Biggr)& = 1-P\Biggl(\bigcap_{j=1}^{k}\overline{\Bigl\{\Bigl|\sum_{i=1}^{N}X_{ij}\Bigr|>\sqrt{N}\Bigr\}}\Biggr)\\ & = 1-P\Biggl(\bigcap_{j=1}^{k}\Bigl\{\Bigl|\sum_{i=1}^{N}X_{ij}\Bigr|\color{red}{\le}\sqrt{N}\Bigr\}\Biggr)\\ & = 1-\prod_{j}^k \left(P\Biggl(\Bigl|\sum_{i=1}^{N}X_{ij}\Bigr|\le\sqrt{N}\Bigr\}\right) \\ & = 1-(1-p)^n \end{align}