Let X a random variable with probability mass function:

$$f_X(x)= \frac{1}{4} I_{\{-2,-1,1,2\}} (x)$$

and let $Y:= X^2$, Proof that the Corr(X,Y)=0 and yet $X$ and $Y$ are not independent.

What I have:

$y=g(x)=x^2 \Rightarrow x=g^{-1}(y)=\sqrt{y}$

$A_x=\{-2,-1,1,2\}, B_y=\{1,4\}$

And I think, $f_Y(y)=\frac{1}{4} I_{\{1,4\}} (y)$

But that doesn't make sense, also when I try to calculate the Expected value, I can't because of the support of the function.

  • $\begingroup$ Sorry but none of this is necessary since the correlation you are after only involves $E(XY)=E(X^3)$, $E(X)$ and $E(Y)=E(X^2)$, and since, by definition, for every function $g$, $$E(g(X))=\frac14(g(-2)+g(-1)+g(1)+g(2))$$ $\endgroup$
    – Did
    Dec 2 '18 at 23:00
  • $\begingroup$ thanks a lot, that was super simple. $\endgroup$
    – pin_r
    Dec 2 '18 at 23:33
  • $\begingroup$ You have earned a marginal density function of $Y$ from that of $X$. But in general, to evaluate $E[XY]$, you need the joint distribution of $(X,Y)$. $\endgroup$ Dec 2 '18 at 23:45

Use the idea in Did's comment to get the following: $EX=0, EY=2.5,EXY=0$ so $EXY-EXEY=0$. Also, $P\{Y=1,X=2\}=P\{X=1\}=\frac 1 4, P\{Y=1\}P\{X=1\}=\frac 1 2 \frac 1 4$ which shows that $X$ and $Y$ are not independent.


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.