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Let $(X_i)_{n\in\mathbb{N}}$ be independent with $\mathbb{P}(X_i=1)=p=1-\mathbb{P}(X_i=-1)$, $i\in\mathbb{N}$.Let $a,b\in\mathbb{N}$. For $n\in\mathbb{N}$, let $S_n=\sum_{i=1}^nX_i$ and finally $\tau$ be the first hitting time of $\{ -a,b\}$ by $(S_n)_n$.

For $p=\frac{1}{2}$, I can calculate $\mathbb{P}(S_\tau = b$) as follows:

For $x\in\mathbb{N}\cap [-a,b]$, let $\tau_x$ be the first hitting time of $\{ -a,b\}$ by $(x+S_n)_n$ and $\phi (x)=\mathbb{P}(x+S_{\tau_x}=b)$. Then we see that $\phi(-a)=0$, $\phi(b)=1$ and $$\phi (x)=\frac{1}{2}\phi (x+1) + \frac{1}{2}\phi (x-1) \quad \mbox{for }x\notin\{-a,b\}.$$ Hence, $\phi(x)=mx+n$ for some $m,n\in\mathbb{N}$ (we can use the formula above for induction) and with the two boundary conditions, we can calculate $\phi (0)$.

My Question:
I want to generalize the solution above to arbitrary $p\in (0,1)$. But I don't recognise the function $\phi$ after calculating $\phi(x)$ for $x=-a+1,\dots ,-a+5$. Does anybody recognise it or maybe even have a more elegant idea to obtain $\phi$? These are the values I calculated: \begin{align} \phi(-a)&=0\cdot \phi(-a+1)\\ \phi(-a+1)&=1\cdot\phi(-a+1)\\ \phi(-a+2)&=\frac{1}{p}\cdot\phi(-a+1)\\ \phi(-a+3)&=\frac{1}{p^2}(p^2-p+1)\cdot\phi(-a+1)\\ \phi(-a+4)&=\frac{1}{p^3}(2p^2-2p+1)\cdot\phi(-a+1)\\ \phi(-a+5)&=\frac{1}{p^4}(p^4-2p^3+4p^2-3p+1)\cdot\phi(-a+1) \end{align}

amars
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1 Answers1

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Write $q=1-p$ for brevity, and assume that $p\neq q$. Then the equation defining $\phi$ becomes $$ \phi(x)=p\phi(x+1)+q\phi(x-1)$$ with the boundary conditions $\phi(-a)=0$ and $\phi(b)=1$.

This sort of difference equation can be solved in essentially the same way that one finds solutions to linear constant-coefficient differential equations: The characteristic polynomial is $$ pt^2-t+q=p\Big(t-1\Big)\Big(t-\frac{q}{p}\Big)$$ hence $f_1(x)=1$ and $f_2(x)=(\frac{q}{p})^x$ both solve the difference equation. Therefore there are constants $c_1$ and $c_2$ such that $$ \phi(x)=c_1f_1(x)+c_2f_2(x)$$ and then the constants $c_1$ and $c_2$ can be determined from the boundary conditions.

carmichael561
  • 53,688