Entering edit mode
8.4 years ago
asifnax
▴
10
Hi,
Please see the equation below:
VJi = max1<=s<=e<=J |sigma e j=s(rij)| /sqrt(e-s +1)
It is from this paper:
http://bmcbioinformatics.biomedcentral.com/articles/10.1186/1471-2105-10-156
Equation (3).
Its based on the sum of partial residuals but clarification would be welcomed
Thanks
This questions is not directly related to bioinformatics (unless you provide additional context in your post). It will most likely be closed unless you do so soon.
Equation in original question.
which part of the expression don't you follow - how to compute it, or what it represents?
Hi,
What it represents i.e. how are residuals being summed together.
Thanks
A meaningful title would help too.
Could the admin the change the title to "Algorithm on how to calculate the sum of partial residulas ?"
Can anyone explain the 'max' bit ?
For a given value of i and J, you compute the fraction for each pair of values e,s such that 1 <= s <= e <= J. Having done this, you have K possible values, say (one for each pair e,s) and you set the value of v ^{(J)} _{i} to the largest of the K values
This is quite standard mathematical notation
So what are 'e' and 's' exactly ?
Also do you have links that explain this notion?
They are numbers between 1 and J, where J is the number of probes on the array. The paper is quite explicit: e takes every value from 1 to J, and for a given e, s takes every value from 1 to e.
You really have to get to grips with scientific / mathematical notation. Go to the library and get a maths book out
Thanks, I understand - you add two residuals every time from consecutive probes 1 ,1+2 ,2, 2+3 , 3, 3+4 ... and from them the highest is chosen as the MUF score.
Not quite. Suppose as an example, J=10, s = 2, e = 9. Then 1 <= s <= e <= J. and you have to add up r[i, j] for j=s..e. That is you do r[i, 2] + r[i, 3] + ... + r[i, 9], take the abs of that and divide it by sqrt(8). [this process is repeated for e = 1..10, s = e..10] Have you come across sigma notation before?
Yes, sigma is the summation, here it is used to sum residuals from e to s, i think.
The different s and e permutations is the most hardest part of the equation but I think you have cleared it up.
I think its called a selection algorithm.