# Transformation of a discret random variable, with different support size.

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.

• 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 $g$, $$E(g(X))=\frac14(g(-2)+g(-1)+g(1)+g(2))$$
– Did
Dec 2 '18 at 23:00
• thanks a lot, that was super simple. Dec 2 '18 at 23:33
• 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 $(X,Y)$. 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.