Question

Can someone tell me how the RMSD (root mean square devition) differs from the standard deviation (which I am familiar with; the formula's look so alike...)

Answer 1

RMSD is the sqrt of the mean of the sum of the squares
\[\sqrt{\frac{x_1^2+x_2^2+x_3^2+...+x_n^2}{n}}\]

standard deviation is the sqrt of variance
\[\sqrt{\frac{(x_1-\mu)^2+(x_2-\mu)^2+(x_3-\mu)^2+...+(x_n-\mu)^2}{n}}\]

Answer 2

If i did it correctly the RMSD of these numbers: 3, 4, 5, 1, 1, 2 is 3.06? while the mean is 2.67 and the standard deviation is 1.33. What does the RMSD tell me?

Answer 3

not too sure what it tells us, other than its some other attempt to define spread or density of the data set

Answer 4

maybe it can be used to relate the deviation between individual values in stead of relating them all the the mean. So if you pick two numbers from my sequence they will be 3.06 seperated from eachother.. that number seems rather large

Answer 5

the standard deviation relates the data set to a normal distribution with a mean of zero. If you were to put the mean into the set, and subtract the mean from all the data points .... youd simply be zeroing out the mean. And of course using smaller values in the operation.

the RMS doesnt move the set and therefore uses larger values in the process.

the spread itself remains the same for each data set .... my lack of experience as a whole with the RMS stuff doesnt allow me to see what, if any, significance it would play tho

Answer 6

im also coming across differening material as to how to calculate the RMS

the one I presented above was the Wolframs version of it; but i see others of it that suggest that the sample mean, instead of the population mean, is used in the operation :/

Answer 7

statisticians huh, the one i am supposed to understand is used in molecular modeling and the formula is:

Answer 8

k is the position of a particle

Answer 9

no sorry it is just a particle, not a position

Answer 10

that looks like the mean of the squares, of the differences between all the pairs of observations

Answer 11

the root of those squares

Answer 12

http://books.google.com/books?id=tPw6b1VXfkoC&pg=PA52&dq=difference+between+root+mean+square+deviation+and+standard+deviation&hl=en&sa=X&ei=uhjPUcfVNJTe8wSb9YHQCA&ved=0CDIQ6AEwAQ#v=onepage&q=difference%20between%20root%20mean%20square%20deviation%20and%20standard%20deviation&f=false

Answer 13

its like an unscaled standard deviation

Answer 14

3-{3, 4, 5, 1, 1, 2): {0,1,2,2,2,1}
4-{3, 4, 5, 1, 1, 2): {1,0,1,3,3,2}
5-{3, 4, 5, 1, 1, 2): {2,1,0,4,4,3}
1-{3, 4, 5, 1, 1, 2): {2,3,4,0,0,1}
1-{3, 4, 5, 1, 1, 2): {2,3,4,0,0,1}
2-{3, 4, 5, 1, 1, 2): {1,2,3,1,1,0}

{0,1,2,2,2,1}
sum: 0 1 4 4 4 1 = 14

{1,0,1,3,3,2}
1 0 1 9 9 4 = 24

{2,1,0,4,4,3}
4 1 0 16 16 9 = 46

{2,3,4,0,0,1}
{2,3,4,0,0,1}
4 9 16 0 0 1 = 30
30

{1,2,3,1,1,0}
1 4 9 1 1 0 = 16
--------
160 , for a sum ?

Answer 15

whats the dividor in your set up, all the squares? or the number of the original set?

Answer 16

it is multiplied by the reciproke of n

Answer 17

but also you have to root it first
right?

Answer 18

yes, if my text is good, then sqrt(160/6^2); or about 2.11

Answer 19

which is smaller than the standard deviation.. but it seems about right i think

Answer 20

i think this might be correct

Answer 21

theres a setup on one of the pages i see:\[2\sigma^2=\frac{\sum(X_i-X_j)^2}{N^2}\]
\[\sigma^2=\frac{\sum(X_i-X_j)^2}{2N^2}\]
\[\sigma=\sqrt{\frac{\sum(X_i-X_j)^2}{2N^2}}\]

Answer 22

and it gives the average difference between two molecules from the model at a certain timepoint if i'm not mistaking

Answer 23

lol, that last one didnt seem to go over to well withthe wolf

Answer 24

but yes, the average distance between all the observed pairs

Answer 25

so in my case i would have 6 numbers so 7 molecules.. you mean wolfram mathematica? :P

Answer 26

yes, the wolf :)

Answer 27

=D its stilla bit hazy but i think i can work with this. sorry for taking so much of your time and thnx!

Answer 28

it was fun :) good luck and all