Suppose that random variables X and Y are positively correlated, with mean zero and variance 1. Suppose Z is a positive random variable. Can it be said that X/Z and Y/Z are also positively correlated?

If no, are there some simple conditions that would make this true?

  • $\begingroup$ If you're looking at correlation of ratios X/Z and Y/Z it may be worth learning about the phenomenon of spurious correlation $\endgroup$
    – Glen_b
    Commented Aug 23, 2016 at 23:52

2 Answers 2


No, that actually does not always follow. Suppose X, Y, and Z are discrete with joint distribution proportional to this:

X  | Y  | Z    | Prob times 16
-1 | -1 | 1    | 1
-1 |  1 | 1    | 3
 1 | -1 | 1    | 3
 1 |  1 | 1    | 1
-1 | -1 | 1000 | 4
-1 |  1 | 1000 | 0
 1 | -1 | 1000 | 0
 1 |  1 | 1000 | 4

X/Z and Y/Z have approximately the (negative) correlation structure shown in the top four rows, whereas X and Y are modestly correlated, matching on average 10 times out of 16. I know you asked for X and Y to have unit variance, but that can be accomplished by rescaling them, which does not alter the correlation.

If X and Y are correlated conditional on Z for every value of Z, that should be good enough. Also, if X and Y are independent of Z, that might help, but I am not sure.

UPDATE: It is sufficient to add "Z is independent of (X, Y)". Covariance is greater than 0 if and only if correlation is zero, so I'll prove it using covariances. This proof easily extends to the case where Z and (X,Y) are not independent, but X and Y are correlated conditional on Z for every value of Z.

$$Cov[\frac{X}{Z}, \frac{Y}{Z}|Z] = Cov[X,Y|Z]/Z^2 = Cov[X,Y]/Z^2 > 0$$.

  • $\begingroup$ Another part of the question is if there are some simple conditions that make it with. I'm thinking can it still work if Z is uncorrelated from X and Y? $\endgroup$
    – SMeznaric
    Commented Aug 23, 2016 at 19:38
  • 1
    $\begingroup$ I think Z is uncorrelated with X (ignoring Y) and with Y (ignoring X) in my counterexample, but I added a bit about independence. $\endgroup$ Commented Aug 23, 2016 at 19:39
  • 2
    $\begingroup$ +1 A more general construction--perhaps providing more insight, as well as giving a more detailed solution--is to choose any positive number $\alpha$ and define $Z=1/\alpha$ when $X$ and $Y$ have the same sign and $Z=1$ otherwise. Provided there is nonzero chance that $X$ and $Y$ have the same sign and nonzero chance they have a different sign, when $\alpha \gt 1$ this increases the correlation of $(X/Z, Y/Z)$ and as $\alpha\to 0$ it eventually makes it negative. There will be a unique value of $\alpha$ making $(X/Z,Y/Z)$ uncorrelated. $\endgroup$
    – whuber
    Commented Aug 23, 2016 at 19:53
  • $\begingroup$ So if Z is uncorrelated with X*Y then it should work in general (not just for discrete +/-1 random variables), correct? $\endgroup$
    – SMeznaric
    Commented Aug 23, 2016 at 20:01
  • $\begingroup$ Eric's calculation at the end assumed nothing about the variables except that the covariances exist, which is implied by your assumption of a positive correlation, and that $1/Z^2$ has finite expectation. $\endgroup$
    – whuber
    Commented Aug 23, 2016 at 20:05

A sufficient condition for this to be true is (1) that $Z$ is independent of both $X$ and $Y$, (2) that $\text{E}[1/Z]$ exists and (3) that $0<\text{E}[1/Z^2]<\infty$.

Since $X$ and $Y$ have mean 0, we know $\text{Cov}[X,Y]=\text{E}[XY]>0$. The covariance between $X/Z$ and $Y/Z$ is then $$ \text{Cov}[\frac{X}{Z},\frac{Y}{Z}]=\text{E}[\frac{XY}{Z^2}] - \text{E}[\frac{X}{Z}]\,\text{E}[\frac{Y}{Z}] = \text{E}[XY]\,\text{E}[\frac{1}{Z^2}] - \text{E}[X]\,\text{E}[Y]\,\text{E}[\frac{1}{Z}]^2 $$ Because of (2), the second term vanishes so $$ \text{Cov}[\frac{X}{Z},\frac{Y}{Z}]=\text{E}[XY]\,\text{E}[\frac{1}{Z^2}] $$ The first factor on the right is positive and (3) holds for the second factor, so we can conclude that the covariance is positive.


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