correlation: the difference of two correlations is positive, but the correlation of the difference is negative? There are 3 random variables, $X_1$, $X_2$ and $Y$. We know $$corr(X_2, Y)>corr(X_1, Y)>0$$, but $$corr(X_2 - X_1, Y)<0$$
In other words, $X_2$ is more positively correlated to $Y$ than $X_1$, but the difference ($X_2-X_1$) is actually negatively correlated with $Y$.
Is this possible? and can you construct such random variables?
 A: It is possible.  Simple algebra shows that
\begin{align}
\rho_{X_2 - X_1, Y} = \frac{\sigma_{X_2}\rho_{X_2, Y} - \sigma_{X_1}\rho_{X_1, Y}}{\sigma_{X_2 - X_1}}.
\end{align}
So although $\rho_{X_2, Y} > \rho_{X_1, Y}$, the sign can be flipped if $\sigma_{X_1}   \gg \sigma_{X_2}$.  Following this observation, a valid counterexample can be easily constructed as $(X_1, X_2, Y) \sim N_3(0, \Sigma)$, where
\begin{align}
\Sigma = \begin{bmatrix}
100 &  0 &   2 \\
  0 &  1 & 0.3 \\
  2 & 0.3 &  1
\end{bmatrix}.
\end{align}
In this case, $0 < \rho_{X_1, Y} = 0.2 < \rho_{X_2, Y} = 0.3$, but $\sigma_{X_2} = 1 < \sigma_{X_1} = 10$.  In addition, the joint distribution of $(X_2 - X_1, Y)$ is $N_2(0, \Sigma')$, where
\begin{align}
\Sigma' = \begin{bmatrix}
101 & -1.7 \\
-1.7 & 1
\end{bmatrix}.
\end{align}
Hence $\rho_{X_2 - X_1, Y} = \frac{-1.7}{\sqrt{101 \times 1}} < 0$.
A: A small example in R:
X1 <- c(0, 1, 3)
X2 <- c(0, 1, 2)
Y  <- c(0, 1, 2)

cor(X1, Y)
# 0.9819805  # sqrt(27/28)
cor(X2, Y)
# 1
cor(X2-X1,Y)
# -0.8660254 # -sqrt(3/4)

It would not be possible with covariance, as cov(X2-X1, Y) = cov(X2, Y) - cov(X1, Y)
