Stochastic Processes: Find the function f(n) , n = 0,1,2,... that satisfies f(0) = 0, f(n) = 1/3f(n-1) + 1/3f(n+1) +1/3f(n+2), n >= 1

Question

Find the function $$f(n),\ n \in \mathbb{N},$$ that satisfies $$f(0) = 0,$$

$$f(n) = \dfrac{f(n-1)}{3} + \dfrac{f(n+1)}{3} +\dfrac{f(n+2)}{3},$$ and $$\lim_{n\rightarrow +\infty} f(n) = 1.$$

Answer 1

Rewrite the recursion to get the form $$f(k)=3f(k-2)-f(k-3)-f(k-1)$$

The cubic $x^3=3x-1-x^2$ has roots $1,-1\pm \sqrt 2$ so the general solution is $$f(n)=A+B(-1+\sqrt 2)^n+C(-1-\sqrt 2)^n$$

For large $n$, the $(-1-\sqrt 2)^n$ term oscillates to $\pm \infty$ and the $(-1+\sqrt 2)^n$ term goes to $0$, so the limit condition tells us that $C=0$ and $A=1$.

Finally, setting $n=0$ tells us that $0=A+B$ so $B=-1$ and the solution is $$f(n)=1-\left(-1+\sqrt 2\right)^n$$