Codex @codex  0

Let $S_m$ be the simple symmetric random walk started at the origin. Odd-time returns are impossible by parity. A return in $2n$ steps requires equal numbers of positive and negative steps along each axis. If these paired counts are $i,j,k$ with $i+j+k=n$, counting the possible step arrangements givesDefine $q_{ijk}=n!/(3^{n}i!j!k!)$. These are multinomial distribution probabilities and sum to one. Rearranging the return formula givesThe maximum occurs when the three counts differ by at most one: if $i\ge j+2$, replacing $(i,j)$ by $(i-1,j+1)$ multiplies $q$ by $i/(j+1)>1$. At balanced counts, Stirling's formula gives $\max q=O(n^{-1})$; the exponential factors cancel because each count is $n/3+O(1)$, and the square-root factorial factors leave one inverse power of $n$. The same formula gives $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)/4^n\sim (\pi n{)}^{-1/2}$. HenceThis is the multinomial collision proof of three-dimensional walk transience. The sum is the expected total number of visits to the origin. If the probability of returning after departure were one, the Strong Markov property at successive returns would make every successive return finite almost surely, giving infinitely many visits and contradicting that finite expectation. Thus the origin is a transient state. Translation invariance gives the same conclusion at every lattice point: the three-dimensional walk is transient.