Correlation of X/Z with Y vs X with YZ

Question

Let’s say that $X$, $Y$ and $Z$ have some fat tailed distribution, $X$ and $Y$ are stock returns, i.e., the sample set can have both positive and negative values. While $Z$ is some random variable such that it has a positive skew but $Z$ always assumes positive values. Then would it be fair to assume: $$Cor(X/Z, Y) = Cor(X, YZ)$$

Using actual data one of them is positive while other is negative. How is that possible?

Edit / Extension

Both the answers by @Tim and @SextusEmpiricus are good validation of my empirical observation above. However, I do not fully appreciate it yet.

Say I fit following two linear regression models:

$$X = \beta_1 * Y * Z$$

$$X / Z = \beta_2 * Y$$

Let’s say second model, i.e., $\beta_2$ is a better model with higher $R^2$ then I can get better estimate of $\hat{X}$ using $Z * \beta_2 * Y$ than say $\beta_1 * Y * Z$?

Answer 1

Why would that hold? You state nothing about the relation between $X$ and $Z$ and between $Y$ and $Z$. The correlation would influence the results. To convince yourself, create a synthetic dataset where the variables are positively or negatively correlated and check what happens. TL;DR the equality does not hold, the sign of the correlation can change.

> set.seed(42)
> N <- 100
> X <- rnorm(N)
> Y <- rnorm(N)
> Z <- rnorm(N)
> cor(cbind(X,Y,Z))
##             X          Y           Z
## X  1.00000000 0.03127984 -0.14477734
## Y  0.03127984 1.00000000  0.07122699
## Z -0.14477734 0.07122699  1.00000000
> cor(X/Z, Y)
[1] 0.05534386
> cor(X, Y*Z)
[1] -0.202128

Notice that even if you increase the $N$ to something large (like 100k samples), the correlations between the variables would be close to zeros (they are independent, so the correlation in the smaller sample was by chance), but still the equality won't hold.

Answer 2

Using actual data one of them is positive while other is negative. How is that possible?

This is not an example with fat tailed distributions, but nevertheless it will provide an intuition how this can happen.

I started generating variables $U = X/\sqrt(Z), V = Y\cdot \sqrt(Z)$ according to a joint distribution which has both an upward and a downward trend, and I define a value $Z$ that correlated with $U$. I made this such that the upward trend occurs for small values of $Z$ and the downward trend occurs for large values of $Z$. Then depending on whether we multiply or divide by $\sqrt{Z}$ we increase/decrease the two different regions.

R-code:

set.seed(1)
n = 10^4
u = runif(n)                 # x/sqrt(z)
v = u(1-u) + runif(n,0,0.1) # ysqrt(z)
z = 0.5+u + runif(n,-0.1,0.1)
x = u*sqrt(z)
y = v/sqrt(z)
plot(x/sqrt(z),y*sqrt(z),
     main = paste0("correlation = ",round(cor(u,v),2)), 
     ylim = c(0,0.4), pch = 20, cex = 0.5, col = rgb(0,0,0,0.02))
plot(x/z,y,
     main = paste0("correlation = ",round(cor(x/z,y),2)), 
     ylim = c(0,0.4), pch = 20, cex = 0.5, col = rgb(0,0,0,0.02))
plot(x,yz,
     main = paste0("correlation = ",round(cor(x,yz),2)),
     ylim = c(0,0.4), pch = 20, cex = 0.5, col = rgb(0,0,0,0.02))

---

Say I fit following two linear regression models:

$$X = \beta_1 * Y * Z$$

$$X / Z = \beta_2 * Y$$

Does this mean $b_1\neq b_2$

With that view you might think that the regression model should not change if you divide both sides with some variable and that $b_1$ and $b_2$ should be the same, but what you are changing with division/multiplication is not just the model for the mean, and it is also the residuals that changes.

Similar concepts from other questions are

• When you fit $Y = a_1 + b_1 X$ or $X = a_2 + b_2 Y$ then you do not simply get $b_1 = 1/b_2$. See: Effect of switching response and explanatory variable in simple linear regression

• Linearized regressions like $Y = a_1 \cdot \exp(b_2 X) + \epsilon$ and $\log(Y) = \log(a_1) + b_2 X+ \epsilon$ are not the same. See: Differences between approaches to exponential regression

The multiplication/division in the above example is effectively changing the weights of the data points in the regression.

With the above example the following two regressions give the same result with a slope of 0.03989

intercept = rep(1,n)
intercept2 = intercept/z
y2 = y*z
x2 = x/z
lm(y2 ~ intercept  + x  + 0)
lm(y  ~ intercept2 + x2 + 0, weights = z^2)

Note that in this case we also need to change the intercept and not just the x variable.