5
$\begingroup$

I don't really have a motivation for this - but I was thinking about this and couldn't work it out.

Suppose I have a random variables $X$ and $Y$ which are correlated. Is it possible that the partial correlation between $X$ and $X\cdot Y$ is zero after taking into account Y? In other words, would a regression of $X$ on $Y$ and $X\cdot Y$ possibly result in a zero coefficient on $X\cdot Y$?

$\endgroup$
  • $\begingroup$ what about Y = 1 / X (over a reasonably defined range) ? $\endgroup$ – Andre Holzner Aug 14 '11 at 7:10
  • $\begingroup$ @Andre, simple code in R says that $y=1/x$ is not a solution: x <- runif(100,0.5,0.7); y<-1/x; summary(lm(x~y+I(x*y)-1)) $\endgroup$ – mpiktas Aug 16 '11 at 11:10
7
$\begingroup$

Yes, that is possible. Take these data for example

      x      y       xy
  .2217  .5000    .1108
  .3048 -.9787   -.2983
-1.6445  .3512   -.5775
 -.2461 -.4866    .1197
 -.3170 -.0954    .0302
-1.1603 1.8352  -2.1294
 -.8720  .1372   -.1196
-1.7852 -.2160    .3856
 1.0100  .0165    .0166
  .3000 -.3251   -.0975

$XY$ is a product of $X$ and $Y$. Multiple regression of $X$ on $Y$ and $XY$ yields $b$ for $XY$ as 0 and $b$ for $Y$ as -.444. Constant is -.386.

Note the theoretical prerequisite for this: $bXY$ will be 0 if and only if $rX.XY$ (i.e. correlation bw $X$ and $XY$; "." here means "with") $= rX.Y * rY.XY$. Here, .280 = (-.361) * (-.776).

$\endgroup$
  • $\begingroup$ could you please add how did you get this data set. $\endgroup$ – mpiktas Aug 16 '11 at 11:08
  • $\begingroup$ @mpiktas, I generated random correlated data, then I altered some of the values until received the wanted result (b for XY = 0). As for the "theoretical prerequisite", it is deduced from the well-known formula for b in the form it exists for OLS regression. $\endgroup$ – ttnphns Aug 16 '11 at 13:01
  • 1
    $\begingroup$ if there is a code to reproduce this example, then it would be very good to include it in the text. $\endgroup$ – mpiktas Aug 16 '11 at 13:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.