Communities

Writing
Writing
Codidact Meta
Codidact Meta
The Great Outdoors
The Great Outdoors
Photography & Video
Photography & Video
Scientific Speculation
Scientific Speculation
Cooking
Cooking
Electrical Engineering
Electrical Engineering
Judaism
Judaism
Languages & Linguistics
Languages & Linguistics
Software Development
Software Development
Mathematics
Mathematics
Christianity
Christianity
Code Golf
Code Golf
Music
Music
Physics
Physics
Linux Systems
Linux Systems
Power Users
Power Users
Tabletop RPGs
Tabletop RPGs
Community Proposals
Community Proposals
tag:snake search within a tag
answers:0 unanswered questions
user:xxxx search by author id
score:0.5 posts with 0.5+ score
"snake oil" exact phrase
votes:4 posts with 4+ votes
created:<1w created < 1 week ago
post_type:xxxx type of post
Search help
Notifications
Mark all as read See all your notifications »
Q&A

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

Post

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

+0
−1

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.

History
Why does this post require attention from curators or moderators?
You might want to add some details to your flag.
Why should this post be closed?

1 comment thread

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