Understanding Mean Centering in Spectroscopic Data Analysis

  • Thread starter Thread starter physical101
  • Start date Start date
  • Tags Tags
    Data Mean
Click For Summary

Homework Help Overview

The discussion revolves around the concept of mean centering in the context of spectroscopic data analysis, particularly as it relates to preparing data for Principal Component Analysis (PCA). Participants explore the implications of subtracting the mean from a dataset and question how this affects the representation of the data.

Discussion Character

  • Conceptual clarification, Assumption checking, Exploratory

Approaches and Questions Raised

  • Participants discuss the mechanics of mean centering, questioning how the operation results in a dataset where the sum of deviations equals zero. They explore whether mean centering is limited to normally distributed data and express confusion about the implications of having unequal distributions of positive and negative deviations.

Discussion Status

The discussion is ongoing, with participants providing insights into the mathematical definition of mean centering and its effects. Some participants express gratitude for the assistance received, indicating a supportive environment, though no consensus has been reached on the broader implications of mean centering.

Contextual Notes

Participants are grappling with the foundational concepts of mean centering and its application in data analysis, highlighting uncertainties about the distribution of data points and the assumptions underlying the mean centering process.

physical101
Messages
41
Reaction score
0

Homework Statement



Hi there I have been currently working on spectroscopic data and I have mean centered them all before I carry out PCA on them. The mean centering and standardisation operations are simple, just take away the mean and divide through by the standard deviation respectively. I have only today wondered however, how the mean centering operation actually works. If you take away the mean from the dataset will you not redistribute the data set so that the new dataset is representative of the distance of the original data point from the mean? How then is the data mean centered such that the addition of columns in a matrix will be O. I would of though that for this to be true then the negitive values must equal the positive values in the mean centered data. How does this happen, would this not mean that you can only use mean centered data on equally distributed data? If so what would be the point as it would be restricted to a very few cases? Please help I am really stuck
 
Physics news on Phys.org
I'm not sure this is the right forum... Anyway... yes, mean centering means that you substract the mean, so you get only the "deviations". This is needed if you're going to fit it to a standard shape, i.e.: a gaussian or lorenzian...
 
But once you have the deviations from the mean, why is the sum of their total equal to 0?
 
An example. Let's say your data are 4, 5, 6. Mean: 5. Substracting the mean: -1, 0, 1.
 
Okay but the data above is equally distributed both sides of the mean. What if you had more negitives than positives, how come this still equates to 0? So sorry to bother you, just really stuck
 
Want a proof, eih? :) [btw, no bother at all!]

The mean of [tex]x_i[/tex] with [tex]i=1\cdots N[/tex] is defined as

[tex]\bar x={1\over N} \sum x_i[/tex]

OK, now substract the mean from the data [tex]y_i=x_i-\bar x[/tex] and take the mean of these values:

[tex]\bar y_i = {1\over N} \sum (x_i - \bar x) = \bar x - \bar x = 0[/tex]

In more simple terms. Let's say you have some data and its average is 5. If you add 7 to all the values... the new average is 12, right?
 
thank you so much - i have been strugling all day with this - if i knew you id but you chocolates
 
:) I appreciate them even if they're virtual... ;)
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 11 ·
Replies
11
Views
6K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K