Hi every body,
I don’t understand the answer to this simple question:
Suppose X,Y,Z are independent random variables show that X/Y and Y/Z are negatively correlated. I think that their correlation is zero.
Can you help
thanks
JME
I don’t see how this is related to Mathematica but in any event I think you’d need to give more details if for no other reason that the correlation between X/Y and Y/Z might not even exist depending on the distributions being considered. Have you tried any specific examples in Mathematica so that your code could be shared?
If Y goes up, X/Y goes down and Y/Z goes up, so it makes sense that they are negatively correlated.
Agree with Jim that there doesn’t seem to be any Mathematica involved in this question. But given some simple types of distributions, it would be easy to prove the statement. Consider if we just looked at Y and 1/Y. Those are clearly negatively correlated.
Here is a visualization using simple uniform distributions.
correlation.bmp (83.5 KB)
You can look at simulations (as Eric Rimbey shows) to see that a negative relationship is certainly possible and not unlikely (as Frank Kampas describes logically). Another way is attempting to work out out the appropriate integrals if you’re feeling energetic. And one more way is to use a discrete distribution for X, Y, and Z and just enumerate all of the possibilities followed by the calculation of the correlation coefficient.
Suppose X, Y, and Z are independent of each other and uniform on the integers 1 through n. We can write a function to enumerate the possibilities for (not too large) values on n. Below is such a function.
corr[n_] := Module[{t, x, y, z, varyx, varzy, meanyx, meanzy, cov},
(* Make a table of all possible values of y/x and z/y (and the product of y/x and z/y = z/x *)
t = Flatten[Table[{y/x, z/y, z/x}, {x, 1, n}, {y, 1, n}, {z, 1, n}], 2];
(* Calculate mean, variance, and covariance *)
meanyx = Mean[t[[All, 1]]];
varyx = Mean[t[[All, 1]]^2] - meanyx^2;
meanzy = Mean[t[[All, 2]]];
varzy = Mean[t[[All, 2]]^2] - meanzy^2;
cov = Mean[t[[All, 3]]] - meanyx*meanzy;
(* Now calculate correlation coefficient *)
cov/(varyx^0.5 varzy^0.5)
]
(* Plot results for n going from 2 to 100 *)
ListPlot[Table[{i, corr[i]}, {i, 2, 100, 1}], Frame -> True,
FrameLabel -> {{"Correlation coefficient", ""}, {"n", ""}}]
So the answer from this example is “Yes, the correlation can be negative.” If you need a more global statement (such as “When the correlation exists for Y/X and Z/Y, then the correlation will negative.”), then more specifics about X, Y, and Z will be needed (assuming that statement is true).