I have two variables, say A and B. Pearson's correlation between the raw data of A and B is positive (not significant). But after taking the logarithm of one variable, say correlation(A, log(B)), the sign of the correlation becomes negative. How is this possible?

  • $\begingroup$ You can answer this yourself by examining the scatterplots of the pairwise variables. $\endgroup$ – Mike Hunter Mar 26 '16 at 14:44
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    $\begingroup$ The logarithm is a monotonic transformation. I don't think this should happen. How large is the positive & negative correlation coefficients? Are they both almost 0? Can you provide a scatterplot of your data? $\endgroup$ – gung Mar 26 '16 at 14:55
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    $\begingroup$ One way this could be possible is that the positive sign is largely a side-effect of an outlier that is subdued by the transformation. Note that changes of sign sound dramatic, but a change between very small positive and very small negative could be quite unimportant $\endgroup$ – Nick Cox Mar 26 '16 at 17:48
  • $\begingroup$ @gung I've constructed a pair of random variables $A$ and $B$ in my answer that satisfy $\rho(A, B)\approx 1$ and $\rho(A, \log(B)) < 0$. The basic trick I used was assigning $B$ values very close to 0 when $A$ takes high values, meaning $log(B)$ takes values as close as I want to $-\infty$. $\endgroup$ – josliber Mar 26 '16 at 17:53
  • $\begingroup$ I saw that, @josliber, +1. But I think the data need that special setup. I think it will be more typical for a monotonic transformation to yield a correlation with the same sign. $\endgroup$ – gung Mar 26 '16 at 18:27

Base Result

Random variables $A$ and $B$ can have a positive correlation but a negative correlation when one is logged. An easy way to see this is to consider two random variables that have positive correlation but where $B$ takes values arbitrarily close to 0 when $A$ is large; since $\log(B)$ will then take values arbitrarily close to $-\infty$ as $A$ gets large, $A$ and $\log(B)$ will be negatively correlated.

As a worked example, let $A$ and $B$ be discrete random variables whose values always coincide; for constants $x, y, \epsilon, \delta > 0$, let $(A, B)$ take value $(0, x)$ with probability $\frac{1-\delta}{2}$, let $(A, B)$ take value $(1, y)$ with probability $\frac{1-\delta}{2}$, and let $(A, B)$ take value $(2, \epsilon)$ with probability $\delta$. Then arithmetic shows:

\begin{align*} \newcommand{\cov}{{\rm cov}} \cov(A, B) &= -\frac{(1-\delta)(1+3\delta)}{4}x + \frac{(1-\delta)(1-3\delta)}{4}y + \frac{3\delta-3\delta^2}{2}\epsilon\\[7pt] \cov(A, \log(B)) &= -\frac{(1-\delta)(1+3\delta)}{4}\log x + \frac{(1-\delta)(1-3\delta)}{4}\log y + \frac{3\delta-3\delta^2}{2}\log \epsilon \end{align*}

Setting $x=1, y=2, \delta=0.01, \epsilon=e^{-100}$ yields the desired result:

\begin{align*} \cov(A, B) &\approx 0.2252 \\ \cov(A, \log(B)) &\approx -2.8036 \end{align*}

Obtaining $\rho(A, B)\approx 1$ and $\rho(A, \log(B)) < 0$

We can construct a pair of random variables $A$ and $B$ with $\rho(A, B)\approx 1$ and $\rho(A, \log(B)) < 0$ using the above example with $0 < x < y$, setting $\epsilon = e^{-1/\delta^2}$ and taking the limit as $\delta\rightarrow 0$. Basically we are selecting very small $\epsilon$ and $\delta$, but $\epsilon$ is getting small much faster than $\delta$.

We have the following results from the equations above:

\begin{align*} \cov(A, B) &\xrightarrow[\delta\rightarrow 0]{} \frac{y-x}{4}\\[7pt] \cov(A, \log(B)) &= \frac{-3}{2\delta} + o(\frac{1}{\delta}) \end{align*}

As a result, we see that $\cov(A, \log(B)) < 0$ as $\delta\rightarrow 0$ and therefore $\rho(A, \log(B)) < 0$ (it approaches 0 from below with the selected $\epsilon$ and $\delta$). To calculate the correlation between $A$ and $B$, we need to compute the standard deviations of the random variables:

\begin{align*} \sigma(A) &= \sqrt{\frac{1+8\delta-9\delta^2}{4}} \\[7pt] \sigma(B) &= \sqrt{\frac{1-\delta^2}{4}(x^2+y^2) + \frac{(1-\delta)^2}{2}xy - 2(x+y)\frac{1-\delta}{2}\delta\epsilon + \epsilon^2\delta(1-\delta)} \end{align*}

From these equations, we obtain $\sigma(A) \xrightarrow[\delta\rightarrow 0]{} 0.5$, $\sigma(B) \xrightarrow[\delta\rightarrow 0]{} \frac{y-x}{2}$, and therefore $\rho(A, B) \xrightarrow[\delta\rightarrow 0]{} 1$.


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