User:Steven J. Koch/Notebook/Kochlab/2009/06/08/Stepping statistics: Difference between revisions
(New page: ~~~~: Read a bunch of kinesin papers yesterday and the day previous. I'm not going to record these kinds of things in my notebook, but you can see what I'm reading via citeulike, del.icio...) |
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** [http://pipes.yahoo.com/pipes/pipe.info?_id=NHLMoeXk3RGVOmisbbsjiw The Yahoo Pipe]. You can use this pipe with your own citeulike account, by changing the input parameter to your citeulike feed | ** [http://pipes.yahoo.com/pipes/pipe.info?_id=NHLMoeXk3RGVOmisbbsjiw The Yahoo Pipe]. You can use this pipe with your own citeulike account, by changing the input parameter to your citeulike feed | ||
* http://delicious.com/skoch3/toread Here is my delicious bookmark tag for articles I'd like to read. As of now, I don't plan on removing them once I've read them. | * http://delicious.com/skoch3/toread Here is my delicious bookmark tag for articles I'd like to read. As of now, I don't plan on removing them once I've read them. | ||
==Stepping statistics== | |||
A particularly good article I read was the Thorn, Ubersax, and Vale 2000 paper on engineering the processivity of kinesin-1. http://www.citeulike.org/user/skoch3/article/1479768 | |||
At the bottom of page 1093, they talk about inferring the stepping versus detachment probability by counting the mean number of steps. This made me realize I hadn't thought about the math in a while, and for fun I solved the summation problem. | |||
* Probability of stepping = x | |||
* Probability of detaching = 1-x | |||
* Let N = number of steps before detaching. What is <N>, the expected number of steps, given x, the stepping probability? | |||
<math><N>=\sum_{n=0}^{\infty}{n(1-x)x^{n}}</math> <br> | |||
<math>(1-x)\sum_{n=0}^{\infty}{nx^{n}}=(1-x)\sum_{n=0}^{\infty}{ \frac{d}{dx} x^{n+1}-x^{n})}</math> <br> | |||
<math>=(1-x)\frac{d}{dx}\sum_{n=0}^{\infty}{ x^{n+1}} - (1-x)\sum_{n=0}^{\infty}{ x^{n}} = (1-x)\frac{d}{dx}\sum_{n=0}^{\infty}{ x^{n+1}} - 1</math> <br> | |||
<math>=(1-x)\frac{d}{dx}(\sum_{n=1}^{\infty}{ x^{n}} -1) - 1 = (1-x)\frac{d}{dx}(\frac{1}{1-x} -1) - 1</math> | |||
<math><N>=\frac{1}{1-x}-1</math> | |||
Latest revision as of 13:52, 9 June 2009
Steve Koch 14:43, 9 June 2009 (EDT): Read a bunch of kinesin papers yesterday and the day previous. I'm not going to record these kinds of things in my notebook, but you can see what I'm reading via citeulike, del.icio.us, and friendfeed:
- http://www.citeulike.org/user/skoch3 For the past several months, I have been using citeulike for articles I have at least skimmed. Thus, I type in some notes when I add the article.
- http://friendfeed.com/stevekoch This is my friendfeed aggregation of all my activities. My citeulike feed goes into here, and it's processed with Yahoo Pipes to add the comment as well.
- The Yahoo Pipe. You can use this pipe with your own citeulike account, by changing the input parameter to your citeulike feed
- http://delicious.com/skoch3/toread Here is my delicious bookmark tag for articles I'd like to read. As of now, I don't plan on removing them once I've read them.
Stepping statistics
A particularly good article I read was the Thorn, Ubersax, and Vale 2000 paper on engineering the processivity of kinesin-1. http://www.citeulike.org/user/skoch3/article/1479768
At the bottom of page 1093, they talk about inferring the stepping versus detachment probability by counting the mean number of steps. This made me realize I hadn't thought about the math in a while, and for fun I solved the summation problem.
- Probability of stepping = x
- Probability of detaching = 1-x
- Let N = number of steps before detaching. What is <N>, the expected number of steps, given x, the stepping probability?
[math]\displaystyle{ \lt N\gt =\sum_{n=0}^{\infty}{n(1-x)x^{n}} }[/math]
[math]\displaystyle{ (1-x)\sum_{n=0}^{\infty}{nx^{n}}=(1-x)\sum_{n=0}^{\infty}{ \frac{d}{dx} x^{n+1}-x^{n})} }[/math]
[math]\displaystyle{ =(1-x)\frac{d}{dx}\sum_{n=0}^{\infty}{ x^{n+1}} - (1-x)\sum_{n=0}^{\infty}{ x^{n}} = (1-x)\frac{d}{dx}\sum_{n=0}^{\infty}{ x^{n+1}} - 1 }[/math]
[math]\displaystyle{ =(1-x)\frac{d}{dx}(\sum_{n=1}^{\infty}{ x^{n}} -1) - 1 = (1-x)\frac{d}{dx}(\frac{1}{1-x} -1) - 1 }[/math]
[math]\displaystyle{ \lt N\gt =\frac{1}{1-x}-1 }[/math]
