Q&A

# How to vaticinate calculating $w_{k + 1} = w_1(1 + \dfrac{1- p}{p} + \dots + [\dfrac{1- p}{p}]^k)$ separately for $p = 1 - p$ and $p \neq 1 - p$?

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Please see the orange, green, red underlines. 1. Why does $\color{limegreen}{w_{k + 1} = w_1(1 + \dfrac{1- p}{p} + \dots + p\dfrac{1- p}{p}]^k)}$ beget the two cases of $\color{red}p = 1 - p$ and $\color{red}p \neq 1 - p$?

1. Even after reading the solution, I wouldn't have prognosticated to calculate separately $\color{limegreen}{w_{k + 1} = w_1(1 + \dfrac{1- p}{p} + \dots + p\dfrac{1- p}{p})^k]}$ for the two cases $\color{red}p = 1 - p$ and $\color{red}p \neq 1 - p$. How would you divine this?

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

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