Let $$X$$, $$Y$$ and $$Z$$ be random variables. Suppose that both $$X$$ and $$Y$$ are positively correlated with $$Z$$. Are $$X$$ and $$Y$$ positively correlated?

The answer to this question is “not necessarily.” To see why, let $$\rho\in[-1,1]$$ be a constant, and define the random variables $$X=\rho Z+W \tag{1}$$ and $$Y=\rho Z-W \tag{2}$$ with $$Z\sim N(0,1)$$ and $$W\sim N(0,1-\rho^2)$$.1 Then $$W$$, $$X$$, $$Y$$ and $$Z$$ have zero means, while $$X$$, $$Y$$ and $$Z$$ have unit variances. It follows that \begin{align} \newcommand{\E}{\mathrm{E}} \newcommand{\Corr}{\mathrm{Corr}} \newcommand{\Cov}{\mathrm{Cov}} \newcommand{\Var}{\mathrm{Var}} \Corr(X,Y) &= \frac{\Cov(X,Y)}{\sqrt{\Var(X)}\sqrt{\Var(Y)}} \\ &= \Cov(X,Y) \\ &= \E[XY]-\E[X]\E[Y] \\ &= \E[XY], \end{align} and similarly $$\Corr(X,Z)=\E[XZ]$$ and $$\Corr(Y,Z)=\E[YZ]$$. Now \begin{align} \E[XZ] &= \E[(\rho Z+W)Z] \\ &= \rho\E[Z^2]+\E[WZ] \\ &= \rho\Var(Z)+\rho\E[Z]^2+\Cov(W,Z)+\E[W]\E[Y] \\ &= \rho \end{align} because $$W$$ and $$Z$$ are independent. A similar argument yields $$\E[YZ]=\rho$$. Finally, substituting $$(1)$$ into $$(2)$$ so as to eliminate $$W$$ gives $$Y=2\rho Z-X,$$ from which we obtain \begin{align} \Corr(X,Y) &= \E[XY] \\ &= \E[X(2\rho Z-X)] \\ &= 2\rho\E[XZ]-\E[X^2] \\ &= 2\rho\E[XZ]-\Var(X)+\E[X]^2 \\ &= 2\rho^2-1. \end{align} Thus, if $$\rho\in(0,1/\sqrt{2})$$ then $$X$$ and $$Y$$ share a negative correlation even though both are correlated positively with $$Z$$. Intuitively, if $$\rho$$ is sufficiently small then the negative correlation between the error terms $$W$$ and $$-W$$ dominates the positive correlations between $$X$$ and $$Z$$, and $$Y$$ and $$Z$$.

1. Here $$N(\mu,\sigma^2)$$ denotes the normal distribution with mean $$\mu$$ and variance $$\sigma^2$$. ↩︎