A doubt on $d$-dimensional random walk

Question

Consider a $d$-dimensional random walk with equal probabilities in each of the $d$-directions (so, $p(v_i,v_j)=\frac{1}{d(v_i)}=\frac{1}{2d}$ here. Now, suppose the walker takes $2n$ steps. Now I have read in a book that "for large $n$" on average $\frac{2n}{d}$ steps will be taken in each of the $d$-dimension. I couldn't understand what is large $n$ doing here ?

Answer 1

Look specifically at Borel's law of large numbers. It essentially tells us that given probability $p=\dfrac{1}{2d}$ for each of $2d$ directions, as the number of steps increases, the chances of one direction having significantly more steps than another becomes smaller and smaller. This isn't true for only a small number of steps, however, because there are still reasonable odds of us having streaks of steps in some directions while possibly having no steps at all in others. It wouldn't make sense in this case to say the number of steps in each direction was approximately equal.

Two relevant quotes from the Wikipedia article:

"There is no principle that a small number of observations will coincide with the expected value or that a streak of one value will immediately be 'balanced' by the others." That is to say, there is no principle that says we should expect to have $\approx\dfrac{n}{d}$ steps in a given direction when $n$ is small.

"This theorem[BLoLN] makes rigorous the intuitive notion of probability as the long-run relative frequency of an event's occurrence." So as $n\to\infty$, the proportion of steps taken in a particular direction will be $\dfrac{1}{2d}$.