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This question appeared on my pop quiz. I got 0%. I achieved everything until the green equation, then I didn't know how to proceed. After reading this solution, I see that you must isolate $pw_{k + 1}$ and move $w_k$ to the right.

$$\color{limegreen}{w_k = pw_{k + 1} + (1- p)w_{k - 1}} \iff pw_{k + 1} = w_k - (1 - p)w_{k - 1}$$

  1. Then you must divide the equation by p.

$$w_{k + 1} = \dfrac{w_k - (1 - p)w_{k - 1}}{p}$$

  1. Finally, you must subtract ${\color{red}{w_k}}$ from both sides!

$$w_{k + 1} -{\color{red}{w_k}} = \dfrac{1\color{red}{-p}}{p}(w_k - w_{k - 1})$$

These steps eluded me and came out of the blue! How would you prognosticate this tricky algebra?

Image alt text

Tsitsiklis, Introduction to Probability (2008 2e), p 63.

Lorenzo B.
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    The idea is to find a suitable $\lambda$ such that $w_{k+1}-\lambda w_k$ is a geometric progression., whose general expression you know already - see my answer here for more details. That's assuming you don't know (or don't want to use) the characteristic polynomial of the linear homogeneous recurrence to solve it directly. – dxiv Sep 11 '21 at 07:03
  • Because of experience I thought of subtracting $w_{k-1},$ giving $w_{k+1} -w_{k-1}=.(w_k- w_{k-1})/p,$ which looks promising . It looks even better when you substitute $f_k=w_k- w_{k-1}$ as it gives $f_k/p= w_{k+1}- w_{k-1}=( w_{k+1}-w_k)+(w_k- w_{k-1})=f_{k+1}+f_k.$ – DanielWainfleet Sep 11 '21 at 11:59

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