# Effect of log transform on Pearson correlation

Is there a generic answer as to how the log transform affects the Pearson correlation between two variables, i.e., is $\rho(\log(x),\log(y)) < \rho(x,y)$ always or the other way around and why?

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I don't have proof, but it can be both. – Julius Sep 9 '12 at 0:36
I think there may be a way to generalize, something to do with the skew of x relative to y. I obviously am not sure which is why I posted. – ganesha Sep 9 '12 at 0:44
My general tack when I ask myself this kind of question, Is to do some rapid numerical simulation in R. Make a habit of it, it will teach you a lot! – kjetil b halvorsen Sep 10 '12 at 19:49

## 1 Answer

Perhaps something can be said using other statistics, but there isn't a universal relationship of the form you specify in the question.

Suppose the joint distribution of $(x,y)$ is uniformly distributed on $\lbrace(0.1,1),(1,0.1),(.0001,.0001) \rbrace$. $\rho(x,y) = -0.335$, but $\rho(\log(x),\log(y)) = +0.885$.

Suppose $(x,y)$ is uniformly distributed on $\lbrace(.0001,1),(1,.0001),(10,10) \rbrace.$ Then $\rho(x,y) = +.984$, but $\rho(\log(x),\log(y)) = -.143$.

Variations on these examples can make one correlation close to $+1$ while the other is close to $-1$. The difference between $0.1$ vs. $0.0001$ might not make a large difference when you calculate $\rho(x,y)$, but it changes $\log x$ and perhaps $\rho(\log(x),\log(y))$ by a lot.

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