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Let $\nu(x)$ be a symmetric probability measure with respect to the origin on $x\in[-1,1]$ such that $\nu(\{0\})\neq 1$.

Consider a random walk started at $S_0=0$, denoted $S_n=X_1+\dotsb+X_n$, where $X_1,X_2, \dotsc$ are i.i.d. random variables such that $X_i\sim \nu(x)$.

Fix $1\leq L<\infty$, and put $\tau=\inf\{n\geq0: S_n>L\}$ and $\hbar_{\nu,L}=\mathbb{E}(S_\tau)-L$. In other words, $\hbar_{\nu,L}$ is the mean value of exitpoint distance from $L$.

My question is how to derive the explicit formula for $\hbar_{\nu,L}$.

Maybe one can start by fixing $L = 1$ and choosing some simple $\nu(x)$, say with probability density function $\mu(x)$ given by

  1. $\mu(x)=1/2$, $x\in[-1,1]$ or

  2. $\mu(x)=\frac{2}{\pi}\sqrt{1-x^2}$, $x\in[-1,1]$.

Could you recommend some relevant papers or books for me? Anyway, any hints or help would be appreciated.

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  • $\begingroup$ Don't use math mode for formatting. For example, ** ** forces bold in Markdown, and Markdown natively supports enumerated lists; neither should be faked with an equation environment. $\endgroup$ – LSpice Dec 9 '18 at 2:43
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If we denote the density $d\mu/dx=p(x)$ (considered as a function on the whole line), for the function $f(L)=\hbar_{\nu,L}$ (also considered on the whole line) we get $f(L)=-L$ whenever $L<0$ and $$f(L)=\int_{-\infty}^\infty f(t)p(L-t) dt,\quad L\geqslant 0.$$ This is Wiener--Hopf equation for the function $f$ on $[0,\infty)$, and on first glance your problem (in full generality) does not seem to be much less general than general Wiener--Hopf equation.

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The question you ask is studied by the renewal theory (Feller, An introduction to probability theory and its applications, vol. 2, Chapter XI or Asmussen, Applied probability and Queues, Chapter V). The function $f(L)=E S_{\tau(L)} - L$ solves the so-called renewal equation (written above by Fedor Petrov), and it follows from the renewal theorem that $f(L)$ has a limit as $L \to \infty$ (Asmussen, eq. (V.6.3)). However, the value of this limit involves $f(0)= E S_{\tau(0)}$ and $E S_{\tau(0)}^2$. If your distribution is atomless, it follows from the symmetry assumption (Feller, Section XVIII.5) that $f(0)=\sqrt{E (X_1^2)/2}$, and it seems that a similar formula should be available, with an effort, for $E S_{\tau(0)}^2$.

Moreover, the distribution function $F(x, L) = P(S_{\tau(L)} - L<x)$ satisfies the renewal equation for every fixed $x\ge 0$. The renewal theory implies that there is a limit distribution for $S_\tau - L$ as $L \to \infty$ with the explicit distribution function (Feller, Sec. XI.3). The expectation of this distribution is the limit for $f(L)=E S_\tau - L$, the quantity you are up to.

Nothing specific can be said for fixed $L$ in general case. The following known relation gives a way to think about $f(L)$ for a fixed $L$: $$P(\tau(L)> n) \sim (f(L)+L) \sqrt{\frac{2}{E (X_1^2) \pi n}}, \qquad n \to \infty.$$

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