Does no correlation but dependence imply a symmetry in the joint variable space?

Question

I was looking through the answers to this question, and all of them seem to have some form of symmetry between the variables.

I'll walk through the examples in that question so you can see what I mean.

Coin Game Example $Y = \pm X$ and $X = |Y|$ (symmetrical functions)

The $x^2$ and $x$ Example Both the uniform distribution and squaring the samples of the uniform distribution are symmetric about the y axis.

Circle Example Circles are symmetrical across both X and Y axes. (Can include many other axes as well, but I'm not 100% that if you chose different axes that you would get no correlation, so I'm just going to say X and Y.)

Car Velocity Example Similar to $x$ and $x^2$ case, $K$ is essentially $V^2$, and $V$ is essentially $\pm \sqrt{K}$. Again, both functions are symmetric.

So, does 0 correlation and dependence imply a symmetry in the joint variable space? Can anyone describe the symmetry mathematically or provide a mathematical proof that there is a symmetry in the joint space?

Answer 1

NO

Consider the following bivariate distribution, with uniform probability across the six points.

$$ \{ (1,1),(2,2),(3,3),(4,4),(5,5),(2,15) \} $$

The correlation is zero. However, $P(Y=5)=\frac{1}{6}$ while $P(Y=5\mid X=5)=1$, so the variables are dependent. At the same time, I see no obvious symmetry in the bivariate distribution when I graph it.

Proof that the correlation is zero

Correlation is zero if the covariance between the variables is zero.

$$\operatorname{Cov}(X, Y) = \mathbb E[XY] - \mathbb E[X]\mathbb E[Y]$$ $$\mathbb E[XY] = \dfrac{1}{6}\left((1\times1) + (2\times2) + (3\times3)+(4\times4) + (5\times5)+(15\times 2)\right) = \dfrac{85}{6}$$ $$\mathbb E[X] = \dfrac{1}{6}(1+ 2+ 3+ 4+ 5+ 2) = \dfrac{17}{6}$$ $$\mathbb E[Y] = \dfrac{1}{6}(1 + 2 + 3 + 4 + 5 + 15) = 5$$ $$\implies$$ $$\operatorname{Cov}(X, Y) = \dfrac{85}{6} - \left(\dfrac{17}{6} \times 5\right) = 0$$

Thus, the variables are uncorrelated.

# Code for the plot with the dotted lines

library(ggplot2)
set.seed(2022)
x <- c(1, 2, 3, 4, 5)
y <- c(1, 2, 3, 4, 5)
x <- c(x, 2)
y <- c(y, 15)
cor(x, y)
plot(x, y)
sqrt((1 - 2)^2 + (1-15)^2)
sqrt((5 - 2)^2 + (5-15)^2)
d <- data.frame(X = x, Y = y)
e <- data.frame(x = c(1, 2, 5, 1), y = c(1, 15, 5, 1))
f <- data.frame(x = c(1, 5), y = c(1, 5))
p <- ggplot(d, aes(x = X, y = Y)) + 
  geom_point() + 
  scale_x_continuous(limits = c(0, 16)) +
  scale_y_continuous(limits = c(0, 16))
p <- p + geom_line(data = e, aes(x = x, y = y),linetype = "dotted") 
p <- p + geom_line(data = f, aes(x = x, y = y),linetype = "dotted")
# p <- p + geom_line(data = g, aes(x = x, y = y),linetype = "dotted")
p# + theme_bw()

Answer 2

This example from the dinosauRus package shows that you do not need symmetry for zero correlation.

library(datasauRus)
selection = which(datasaurus_dozen$dataset == 'dino')
x = datasaurus_dozen[selection,'x']
y = datasaurus_dozen[selection,'y']
plot(x,y,pch=20)
title(paste0("correlation = ", cor(x,y)))

If you rotate this dataset then you can get the correlation equal to zero

z = cbind(x,y)
newz = z %*% expm::sqrtm(solve(cov(z)))
plot(newz, pch = 20)
title(paste0("correlation = ", cor(newz[,1],newz[,2])))

Answer 3

Yet another discrete counterexample. Let the joint pmf of $(X, Y)$ be given as follows:

\begin{align} & P(X = -2, Y = 0) = 1/6, P(X = 0, Y = 0) = 0, P(X = 1, Y = 0) = 1/3, \\ & P(X = -2, Y = 1) = 0, P(X = 0, Y = 1) = 1/2, P(X = 1, Y = 1) = 0. \end{align}

As $P(X = 1, Y = 1) = 0 \neq P(X = 1)P(Y = 1) = 1/3 \times 1/2 = 1/6$, $X$ and $Y$ are not independent.

As $P(XY = 0) = 1$ and $E(X) = 0$, $\operatorname{Cov}(X, Y) = E(XY) - E(X)E(Y) = 0 - 0 = 0$, i.e., $X$ and $Y$ are uncorrelated.

Clearly, there are no symmetry in either $x$-direction or $y$-direction. The idea of this construction is to get a "non-symmetrical" random vector such that $P(XY = 0) = 1$, which is relatively easy (by setting probabilities of pairs with non-zero product to be $0$) to achieve under the discrete case.

Answer 4

No, symmetric examples are simply easier to generate and easier to verify (if a graph is horizontally symmetric, then it's obvious that the correlation is zero, without having to do any math). There are plenty of non-symmetric examples. The residuals from a linear regression have zero correlation, so you can take any set data at all, subtract off the line of best fit, and get a zero-correlation set. Moreover, correlation is a single degree of freedom, so it's easy to make it zero. Given any set, there are points that, if added to the set, will result in zero correlation. And there is always some angle such that if the scatterplot is rotated by that much, the result will have zero correlation.