## One dimensional lattice random walks with absorption at a point/on a half line.(English)Zbl 1234.60051

Let $$Y_{n}$$, $$n\geq 1$$, be independent, identically distributed, $${\mathbb Z}$$-valued random variables with $$\operatorname{E}Y= 0$$, $$\sigma^{2}=\operatorname{E}(Y^{2})<\infty$$ ($$Y$$ having the same law as $$Y_1$$), $$S_{n}=x+Y_{1}+\dots +Y_{n}$$, $q^{n}(x,y)=\operatorname{P}(S_{n}=y,S_{i}\neq 0,i=1,\dots ,n-1),$ $$a_{0}>1$$, $$p^{n}(x)=\operatorname{P}(Y_{1}+\dots +Y_{n}=x)$$, $$d_{0}$$ be the period, $$a(x)=\sum_{n\geq 0}(p^{n}(x)-p^{n}(-x))$$, $$a^{\ast }(x)= 1_{\{0\}}(x)+a(x)$$, $$g_{n}(u)=(2\pi n\sigma^{2})^{-1/2}\operatorname{E}^{- u^{2}/2\sigma^{2}n},$$ $$T=\inf\{n \geq 1,S_{n}\leq 0\}$$, $q_{(-\infty ,0]}^{n}(x,y)= \operatorname{P}(S_{n}=y,n<T),$ $$f_{+}$$, $$f_{-}$$ be positive harmonic on $$x>0$$, for $$S_{n}$$, $$-S_{n}$$ respectively, absorbed on $$(-\infty ,0]$$, and satisfying $$\lim_{x\to \infty }x^{-1}f_{\pm }(x)=1$$, $$h(n,y)=\operatorname{P}(S_{T}=y,T=n)$$, $$H(y)=2\operatorname{E}(f_{-}(y-Y);Y<y)/\sigma^{2}$$.
The first result (all are uniformly in $$x$$,$$y$$) is, for $$o(n^{1/2})=\min(|x|,|y|)$$, $$\max(|x|,|y|)<a_{0}n^{1/2}$$ we have $q^{n}(x,y)=(\sigma^{2}n)^{-1}(\sigma^{4}a^{\ast }(x)a(-y)+xy)p^{n}(y-x)+ o(n^{-3/2}|y|\max(|x|,1)),$ for $$a_{0}^{-1}n^{1/2}<|x|,|y|<a_{0}n^{1/2}$$, $q^{n}(x,y)=o(n^{-1/2})$ plus, if $$xy>0$$, $$d_{0}1_{{\mathbb C}\{0\}}(p^{n}(y-x))(g_{n}(y-x)-g_{n}(y+x))$$ and, for $$0<\min(|x|,|y|)<n^{-1/2}<\max(|x|,|y|)$$ and $$\operatorname{E}(|Y|^{2+\delta })<\infty$$ for some $$\delta \geq 0$$, $q^{n}(x,y)=O\left(\frac{\min(|x|,|y|)}{\max(|x|,|y|)}g_{4n}(\max(|x|,|y|))\right)+ o\left(\frac{\min(|x|,|y|)}{\max(|x|,|y|)^{2+\delta}}\right).$ In the case $$\operatorname{E}(|Y|^{3};Y<0)<\infty$$, for $$y<0<x$$, $$\max(|x|,|y|)\leq a_{0}n^{1/2}$$, $$\min(|x|,|y|)\to \infty$$, we have $q^{n}(x,y)=C^{+}\frac{|x|+|y|}{\sigma^{2}n}p^{n}(y-x)+o(n^{-3/2}\max(|x|,|y|))$ where $$C^{+}=\lim_{x\to \infty }(\sigma^{2}a(x)-x)$$.
The third is, for $$0<x,y\leq a_{0}n^{1/2}$$, $$xy/n\to 0$$, $q_{(-\infty ,0]}^{n}(x,y)= \frac{2f_{+}(x)f_{-}(y)}{\sigma^{2}n}p^{n}(y-x)(1+o(1)).$ The fourth result, appearing as an important step, is, if $$\operatorname{E}(|Y|^{2+\delta };Y<0)<\infty$$, $$d_{0}=1$$, when $$y\leq 0<x<a_{0}n^{1/2}$$, $h(n,y)=n^{-1}f_{+}(x)g_{n}(x)H(y)(1+o(1))+x\alpha_{n}(x,y)n^{-3/2}$ with $$\alpha_{n}(x,y)=o(\max(|y|,n^{1/2})^{-1-\delta })$$, $$\sum_{y\leq 0}|\alpha_{n}(x,y)|=o(n^{-\delta /2})$$, $$\sum_{y\leq 0}|\alpha_{n}(x,y)|y|^{\delta }=o(1)$$, and, when $$x\geq n^{1/2}$$, $$y\leq 0$$, $h(n,y)\leq C(n^{-1/2}g_{4n}(x)+o(x^{-(2 +\delta )}))H(y)+Cn^{-1/2}\operatorname{P}(Y<y-(x/2)).$ The fifth is $\lim_{n}\sum_{x\geq 1}\sum_{y\leq -1}q^{n}(x,y)=C^{+}/2.$ Different corollaries are also stated. The case when the absorption in $$0$$ takes place only with a certain constant probability is considered.
In the last paragraph the author proves that, if $$\operatorname{E}(|Y|^{2+\delta })<\infty$$, then $a(x)=(2\pi )^{-1}\int_{-\pi}^{\pi}Re(\frac{1-e^{ixr}}{1-\operatorname{E}(e^{irY})})\,dr$ and obtains estimates for $$\sigma^{2}a(x)-|x|$$ when $$|x|\to \infty$$.

### MSC:

 60G50 Sums of independent random variables; random walks 60J45 Probabilistic potential theory 60G51 Processes with independent increments; Lévy processes
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### References:

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