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Assume the two random variables $X$ and $Y$. The correlation of these two variables is given by the formula:

$\rho_{X,Y} = \frac{Cov(X,Y)}{\sigma_{X}\sigma_{Y}}$

Assume now that the two random variables become $\frac{1}{X}$ and $Y$. In this case the correlation becomes:

$\rho_{\frac{1}{X},Y} = \frac{Cov(\frac{1}{X},Y)}{\sigma_{\frac{1}{X}}\sigma_{Y}}$

My question is what is the relationship between $\rho_{X,Y}$ and $\rho_{\frac{1}{X},Y}$?

Does it hold that $\rho_{X,Y}= -\rho_{\frac{1}{X},Y}$? And if yes then what's the proof?

develarist
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Whitebeard13
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    In general, it doesn't hold (this can be seen by trying a few examples that have support on more than two points). Indeed it's possible for both of them to have the same sign or even to be equal. – Glen_b Sep 16 '20 at 10:44
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    If you do not know any thing about the distribution of $X$, you can't say anything about the $Var(\frac{1}{x})$ (https://stats.stackexchange.com/q/41896/144441). – OmG Sep 16 '20 at 10:47

2 Answers2

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You can also make a simulation in R for 1000 samples and see that your relation doesn't seem to hold:

rm(list=ls())
set.seed(42)

cor_xy=c()
cor_invxy = c()
n=1000

for (i in 1:1000){
  x = rnorm(n)
  y = x+rnorm(n)/2
  cor_xy = cbind(cor_xy,cor(x,y))
  cor_invxy = cbind(cor_invxy,cor(1/x,y))
}
mean(cor_xy)
[1] 0.8941694
mean(cor_invxy)
[1] 0.02236089

By printing one sample of $X\sim N(0,1)$ and $Y = X + \frac{N(0,1)}{2}$:

$$\hat{\rho}_{X,Y} = 0.8999498$$

enter image description here

$$\hat{\rho}_{\frac{1}{X},Y} = 0.009259998$$

enter image description here

Ale
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1

For example, If $X$ be Rademacher distributed, since $X$ and $1/X$ are the same, the two correlations are even equal, i.e. $\rho_{X,Y}=\rho_{1/X,Y}$.

If $X=\{1,2\}$ uniform and $Y=X$, you'll have $\rho_{X,Y}=1=-\rho_{1/X,Y}$ as well.

If $X=Y$, it won't be hard to find an example satisfying, $1=|\rho_{XY}|>|\rho_{1/X,Y}|$. For the same case, letting $Y=1/X$ would reverse inequality.

There is no general relation.

gunes
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