Let $(X_n)$ be a simple random walk on $\Bbb{Z}^d$ starting at $0$. (The dimension $d$ will vary, but I will suppress the dependence on $d$ for brevity.) I encountered a statement which claims that
For any $l = 1, 2, \cdots$, there exists a constant $c_l > 0$ (depending only on $l$) such that $$\Bbb{P}(X_{4l} = 0) \leq c_l (2d)^{-2l}.$$
It was given without any explanation, so I tried to prove it by myself and was able to check this with some dirty calculation using complex analysis(!).
But I would like to know how to prove this using probability theory or any intuition on why this should be true.
Thanks in advance!
Let $n_i,n_{-i},i=1,\dots,d$ denote a move of a (simple symmetric) RW in $i$-th and the opposite direction, respectively. Then
$$P\{X_{2n}=0\}=P\{n_1=n_{-1},\dots,n_d=n_{-d}\}$$ $$=\sum_{n_1+n_{-1}\cdots+n_d+n_{-d}=2n}1\{n_1=n_{-1},\dots,n_d=n_{-d}\}\binom{2n}{n_1,n_{-1}\dots,n_d,n_{-d}}\left(\frac{1}{d}\right)^{2n}$$
$$=(2d)^{-2n} \binom{2n}{n} \sum_{n_1+\dots+n_d=n}\binom{n}{n_1,\dots,n_d}^2$$ $$\le (2d)^{-n}2^{-n}\binom{2n}{n}\sum_{n_1+\dots+n_d=n}M_{n,d}\binom{n}{n_1,\dots,n_d} \left(\frac{1}{d}\right)^n=(2d)^{-n}2^{-n}M_{n,d}\binom{2n}{n}$$
where $M_{n,d}$ is the maximum of $\binom{n}{n_1,\dots,n_d}$ over all $n_1\dots,n_d$ s.t. $n_1+\cdots +n_d=n$ which is attained at $n_1=n_2=\cdots=n_d=n/d$.