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Find boundary conditions that allow the recursion $f(k)=\frac{1}{2} f(k-1) + \frac{1}{2} f(k+1)$ for $-B

The boundary conditions $f(A) = 1$ and $f(-B)=0$ uniquely determine $f$, but I am not certain as to why these values give the unique solution. I've encountered this in several books and I always took it for granted, but never quite figured out the intuition behind the solution.

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    Got something from an answer below?2012-08-11

2 Answers 2

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The condition $$f(k)=\frac{1}{2} f(k-1) + \frac{1}{2} f(k+1)$$ can be rewritten as $$f(k)-f(k-1) =f(k+1)-f(k).$$ This means that $f$ is a straight line function, and so is completely determined by its values at the two boundary points.

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As @Byron explained, for every pair of conditions $(a,b)$, there exists a unique function $f$ solving the recursion and such that $f(A)=a$, $f(-B)=b$. If the question is why one should use $(a,b)=(1,0)$ in the case at hand, the answer is that $\tau=0$ when $S_0=A$ and when $S_0=-B$, hence the probability of the event $[S_\tau=A]=[S_0=A]$ is $1$ when $S_0=A$ and $0$ when $S_0=-B$.