$n$-th return of a random walk on $\Bbb Z^d$

Lets define $f_n = P(X_n =0 , X_k \ne 0, k< n)$ the first return distribution of the random walk $X_n$ on $\mathbb{Z}^d$, and lets go ahead and assume that $f_n \approx n^{-(1+\alpha)}$ for some $\alpha\in(0,1)$ and $\sum_n f_n =1$, so our walk is naturally recurrent.

We can now let $f_n^{*k} := \sum_{j_1+\cdots j_k =n} \prod_{i=1}^n f_{j_i}$ be the $k$-fold convolution, or more popularly, the probability of $k$-th return at step $n$.

We define $g_k(x) = \sum_{n=2k}^\infty f_n^{*k} e^{-x} \frac{x^n}{n!}$.

I would like to show that for $x^\alpha >k$ we will have $g_k(x) \approx kg(x)$

and for $x^\alpha < k$, we will have $g_k(x) \approx O(e^{-ax})$ for some $a>0$ ?

Notice that this is the $k$-fold convolution of the sojourn times of a continuous random walk with exponential jump wait times.

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The random walk is transient for $d \geq 3$, so your question makes sense only for $d=1,2$ i.e. $\sum_n f_n < 1$ in those cases – Felipe Olmos Mar 27 '12 at 23:19
you are thinking of the simple symmetric random walk, im only talking about a general random walk, which can be made recurrent in any dimension – sqz Mar 28 '12 at 1:00
Ok, I haven't got that. So the transition probabilities are different in each site of $\mathbb{Z}^d$? – Felipe Olmos Mar 28 '12 at 15:13
yes thats one way to make a walk recurrent, but the fact that im on $\mathbb{Z}^d$ is pretty irrelevant to the main question – sqz Mar 28 '12 at 19:48