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
Giannis
·
2021-09-16T16:22:40Z (about 3 years ago)