Correlation after dividing with a random variable 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?
 A: 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.
$$Cov[\frac{X}{Z}, \frac{Y}{Z}|Z] = Cov[X,Y|Z]/Z^2 = Cov[X,Y]/Z^2 > 0$$.
A: 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.
