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Comments on 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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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)
$$(1:999)(99:1)=(99:999)=(11:111)\ne(1:111)$$
Moshi‭ wrote over 3 years ago

You made a mistake in your calculations,

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

Using $(11:111)$ gives the expected result of $\frac{11}{122}\approx 0.09016$

DNB‭ wrote over 3 years ago

Thanks! Boy am I scatter brained! I already got one downvote here. If I keep getting downvotes, I'll construe them to mean that this question is too half-baked, and I'll delete