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Correlation: is this correct?

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Nuchto
Nuchto el 8 de Mzo. de 2013
ans ans ans

Respuestas (1)

George Papazafeiropoulos
George Papazafeiropoulos el 8 de Mzo. de 2013
Dear Nuchto,
See this:
x=rand(4,30); y=10*x+5; corr=sum(zscore(x',1).*zscore(y',1))'/size(x,2)
this is different from your version, where zscore scales X using the sample standard deviation, with n - 1 in the denominator of the standard deviation formula. zscore(X,0) is the same as zscore(X). In the above version zscore(X,1) scales X using the population standard deviation,with n in the denominator of standard deviation formula. This is the right way since the correlation coefficient r is given by [Σ(xi*yi)-(Σxi)*(Σyi)/N]/{[Σxi^2-(Σxi)^2/N]*[Σyi^2-(Σyi)^2/N]}^(1/2) and after some mathematical operations gives the above formula.
Greetings,
George Papazafeiropoulos
  3 comentarios
George Papazafeiropoulos
George Papazafeiropoulos el 8 de Mzo. de 2013
Alternatively, look at the function corrcoef(X). Good luck!
George Papazafeiropoulos
George Papazafeiropoulos
George Papazafeiropoulos el 8 de Mzo. de 2013
Because in the denomimators of the original formula N appears, not N-1.

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