# How would you vaticinate to $-w_k$ from both sides of $w_{k + 1} = \dfrac{w_k - (1 - p)w_{k - 1}}{p}$?

This question appeared on my pop quiz last week. 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.

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

- Then you must divide the equation by p.

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

- Finally, you must ${\color{red}{-w_k}}$ 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, are too tricky, and appear to come of the blue! Can you please naturalize (make natural) them? How would you progonosticate this algebra?

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

## 1 answer

Hi there,

So, I'll start first with a few notations. Three sentences above, from the green underlined sentence, you will see that it defines as $p=P(F)$ and as $q=1-p=P(F^{c})$ and these to probabilities sum to $1$, i.e. $p+q=1$ and that $0<p<1$ (in order to divide later with something that is non zero).

Then we have the equation that you started with,

$$w_{k} = pw_{k+1}+ (1-p)w_{k-1} \Rightarrow w_{k+1} = \frac{w_{k}-(1-p)w_{k-1}}{p}$$

then you subtract $w_{k}$ from both sides

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

$$ = \frac{w_{k}-(1-p)w_{k-1} -pw_{k}}{p} = \frac{w_{k}(1-p)-(1-p)w_{k-1}}{p} $$

$$ = \frac{(1-p)(w_{k}-w_{k-1})}{p} = \frac{(1-p)}{p}(w_{k}-w_{k-1})$$

Now remember earlier how we denoted $1-p=q$, based on that you can denote the ratio $\frac{1-p}{p}=\frac{q}{p}=:r$. Then you get the desired result

$$w_{k+1}-w_{k} = r(w_{k}-w_{k-1})$$

How this anwsers helps

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