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I am trying to investigate the correlation values. A big problem happens. The method I used is in corpcor package in R. I have a column as missing data. I firstly calculate the correlation matrix.

cluster1corMatrix<-cor(cluster1Analytes,method="spearman")

Then I get a correlation matrix. when I did partial correlations using cor2pcor functions as follows, I got a big problem.

## 80th column and 80th row are all NAs.
cluster1corMatrix<-cluster1corMatrix[-80,]
cluster1corMatrix<-cluster1corMatrix[,-80]
### the above delete the missing value column
cluster1PcorMatrix<-cor2pcor(cluster1corMatrix)

Most of the correlation become negative. More importantly, some correlations, for instance, 0.9 correlation becomes -0.91 in partial correlation.

The method I used is in corpcor package in R (https://cran.r-project.org/web/packages/corpcor/index.html).

Can anyone comment on that? Is it reasonable to change correlation of 0.9 to -0.9 in partial correlation?

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    $\begingroup$ Correlations are not "being changed". Why would you expect that the partial correlations would be similar to the ordinary correlations? $\endgroup$
    – Glen_b
    Aug 9, 2014 at 7:34
  • $\begingroup$ I expected the partial correlation could "modify" the correlation, for instance, change 0.9 in correlation to 0.5, but it changed 0.9 to -0.9 is a completely another way around. Something are very positively correlated become very negatively correlated. $\endgroup$
    – GeekCat
    Aug 9, 2014 at 13:45

1 Answer 1

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Suppose that you have $X_1=Y+\epsilon_1$ and $X_2=Y+\epsilon_2$. Thus, let's say you want to measure $Y$, but you only can do it with error through $X_1$ and $X_2$. Let's suppose $\epsilon_1$ and $\epsilon_2$ are negatively correlated with each other and have common varioance $\sigma^2{\epsilon}$. The covariance between $X_1$ and $X_2$ is given by: $E(X_1X_2)-E(X_1)E(X_2)=E(Y^2)+E(\epsilon_1\epsilon_2)-E^2(Y)=Var(Y)+Corr(\epsilon_1\epsilon_2)*Var(\epsilon).$

The variance of $X_1$ and $X_2$ is equal to $Var(Y)+Var(\epsilon)$ (due to independence between $Y$ and both $\epsilon_1$ and $\epsilon_2$). Then, since $Corr(X_1,X_2)=\frac{Cov(X_1,X_2)}{SE(X_1)SE(X_2)}$, it is: $\rho_{X_1X_2}=\frac{\sigma^2_Y+\rho_{\epsilon_1\epsilon_2}\sigma^2_{\epsilon}}{\sigma^2_{Y}+\sigma^2_{\epsilon}}$. As for partial correlation, since it is equal to the correlation among regressions on the variable we are controlling for, it is: $\rho_{X_1X_2|Y}=\rho_{\epsilon_1\epsilon_2}$. Suppose that $\rho_{\epsilon_1\epsilon_2}=-0.91$,$\sigma^2_{\epsilon}=1$ and $\sigma^2_{Y}=18.1$. Then you have a partial correlation between $X_1$ and $X_2$ of $-0.91$, while a correlation of $\frac{18.1-0.91}{19.1}$, that is equal to $0.90$, as in your case. Thus, if you have two variables $X_1$ and $X_2$ whose correlation is a combination of a positive and a negative part, and another variable $(Y)$ explaining all the positive part of the correlation between $X_1$ and $X_2$, then the residual correlation between them is negative.

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