Sign of correlation changes after taking log 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?
 A: 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 $\rho(A, B) \xrightarrow[\delta\rightarrow 0]{} 1$.
