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Let $\alpha \in (0,2)$, (or for simplicity just $\alpha \in (1,2)$) and let $X_1,X_2,\dots$ be an i.i.d collection of random variables with common distribution $$ p(x,y)= \frac{c_\alpha}{|x-y|^{1+\alpha}}1_{[x\neq 0]} $$ where $c_\alpha = \frac{1}{2\zeta(1+\alpha)}$ is the normalising constant. Now, consider the random walk $\{S_n\}$ given by $S_n = \sum_{i=1}^n X_i$ and $S_0=0$.

For a given $n\ge 1$, consider $\Lambda_n = [-n,n]$, I would like to know whether good bounds for the hitting measure defined as

$$ H_{\Lambda_n}(0,x)= P_0[S_{\tau_n}=x], $$ where $\tau_n = \inf\{n\ge 0, |S_n|\ge n\}$. I have seen such problems under the name of overshoot and undershoot bounds. But I have not found an article dealing with the setting without a drift and with heavy tail random walks. It is easy to get very bad bounds, but then they are far from matching.

From a few simulations I made, it seems like it decays as a power of $(|x|-n)$, but I couldn't figure out what was the exponent for an arbitrary $\alpha$.

Are there good bounds known for $H_{\Lambda_n}(0,x)= P_0[S_{\tau_n}=x]$?

EDIT: As mentioned in the comments bellow, one can use that for $x>n$ $$ (1) \qquad \frac{c_\alpha}{(x+2n)^{1+\alpha}} \le H_{\Lambda_n}(0,x)\le \frac{c_\alpha}{(x-n)^{1+\alpha}} $$ and therefore for $x>cn$ for some $c>1$, we have that $H_{\Lambda_n}(0,x)\asymp n^{-1-\alpha}$. So the question is really about the remaining $x \in [n+1,cn]$. As the bounds given in $(1)$ would lead to upper and lower bounds that that do not match.

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  • $\begingroup$ If you're interested in the regime $|x|\ge 2n$, then it's obvious that the decay is like $1/(|x|)^{1+\alpha}$: for any point in $\Lambda_n$, the probability of jumping to $x$ is within a constant factor of $p(0,x)$. $\endgroup$ Aug 15, 2020 at 1:06
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    $\begingroup$ So I think editing questions to change what they’re really asking is frowned upon. In this case, I don’t think the question is being changed. I’d see it as a clarification rather than a new question, but I think it is preferred to mark the edit as such (e.g. by adding a paragraph at the end labeled EDIT) in order that future readers can make sense of the comments. $\endgroup$ Aug 15, 2020 at 16:11
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    $\begingroup$ So I have been doing some back-of-the-envelope calculations (I don't think I have the technical expertise to prove this). I believe that the hitting measure looks like $1/N(|x|-N)^\alpha$ for $|x|<2N$ and $1/|x|^{1+\alpha}$ for larger $x$. My rough argument is to consider jumps of size up to $N/1000$ as "small" and anything else as large. Then the variance of a single small step is something like $N^{2-\alpha}$. Thus the standard deviation coming from the small steps is of size $N$ after $N^\alpha$ steps. $\endgroup$ Aug 19, 2020 at 7:52
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    $\begingroup$ Heuristically after this order number of steps, the position is uniformly distributed on $[-N,N]$. It is then the bigger jumps that are responsible for leaving the interval. If a jump is of size $N/L$ for some $L$, then my heuristic would say you have a $1/L$ chance of leaving the interval. Summing the contributions leads to my back-of-the-envelope answer. $\endgroup$ Aug 19, 2020 at 7:52
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    $\begingroup$ @Kernel: For the stable process, see: R. M. Blumenthal, R. K. Getoor, and D. B. Ray, On the distribution of first hits for the symmetric stable processes, Trans. Am. Math. Soc., 99 (1961), 540–554. Related references: M. Kac, Some remarks on stable processes, Publ. Inst. Stat. Univ. Paris, 6 (1957), 303–306; and: M. Riesz, Intégrales de Riemann–Liouville et potentiels, Acta Sci. Math. Szeged, 9 (1938), 1–42. $\endgroup$ Aug 30, 2020 at 9:54

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