5
$\begingroup$

Suppose $X$ and $Y$ are positive valued continuous random variables. Is it possible to find a non-linear function $f$, such that

independence between $Y$ and $X+\frac{X}{Y}$ implies independence between $Y$ and $X+f\left(\frac{X}{Y}\right)$?

I am guessing it is not possible. Trying to find a counter example, to show if the independence to be hold then $f$ must be a linear function. For showing this I am using product moment correlation, which leads covariance of the two part are same. But covariance same may not imply that the transformation is linear. Here I stuck.

$\endgroup$

closed as unclear what you're asking by whuber Dec 13 '15 at 15:31

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ What is $W$? Another random variable? A typo for $X$? It would be interesting to consider whether it is possible for $Y$ and $X+\frac XY$ to be independent random variables at all. Do you have some joint distribution for $X$ and $Y$ in mind where this condition holds? $\endgroup$ – Dilip Sarwate Dec 13 '15 at 14:56
  • $\begingroup$ Until the issues raised by @Dilip are cleared up, there is not an answerable question here. $\endgroup$ – whuber Dec 13 '15 at 15:32
  • $\begingroup$ @DilipSarwate. Thanks for pointing out the typo. $W$ is typo for $Y$. I don't have any joint distribution in my mind for this problem. Actually this question arises from a real scientific problem. $\endgroup$ – Janak Dec 30 '15 at 11:03
  • $\begingroup$ Unless you have an example of positive continuous random variables $X$ and $Y$ such that $Y$ and $X+\frac XY$ are independent random variables, there is no point in trying to prove that $$Y~\text{and}~X+\frac XY~\text{independent}~\Longrightarrow Y~\text{and}~X+f\left(\frac XY\right)~\text{independent}.$$ Starting from a false hypothesis, one can prove any statement, true or false. $\endgroup$ – Dilip Sarwate Dec 31 '15 at 4:43