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Q&A

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

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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.

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

  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 ${\color{red}{-w_k}}$ 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?

Image alt text

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

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$$(1:999)(99:1)=(99:999)=(11:111)\ne(1:111)$$ (2 comments)

1 answer

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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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How does your answer address my question? If it doesn't, please delete your answer? You expounded t... (1 comment)

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