Let $ X \sim U([-1,1])$ and $ Y = {X^2} $. Are $X$ and $Y$ independent? Why?
So, here I tried to show the independence by checking whether $F_{(X,Y)}(x,y)=F(x) \cdot F(y)$.
And I wrote $$F_{(X,Y)}(x,y)=P(X\le x,Y\le x)=P(X\le x,X^{2}\le x)=P(X\le x,-\sqrt{x}\le X\le \sqrt{x})=P(-\sqrt{x}\le X\le \sqrt{x}) $$
Now, I just need to know am i doing since all right? If not what is wrong, what will you suggest to do to show the independence or dependence.
$\mathbb{P}(X\leq x, Y\leq y)=\mathbb{P}(X\leq x, X^2 \leq y)=\mathbb{P}(X\leq x, -\sqrt{y}\leq X \leq \sqrt{y})$
If $x^2>y$, then we get that $\mathbb{P}(X\leq x, Y\leq y)=\mathbb{P}(-\sqrt{y}\leq X \leq \sqrt{y})=\sqrt{y}$
If $x^2\leq y$, then we get that $\mathbb{P}(X\leq x, Y\leq y)=\mathbb{P}(-\sqrt{y}\leq X \leq x)=\frac{x+\sqrt{y}}{2}$
On the other hand, $\mathbb{P}(X\leq x)\mathbb{P}( Y\leq y)=\mathbb{P}(X\leq x)\mathbb{P}(-\sqrt{y}\leq X\leq \sqrt{y})=\mathbb{P}(X\leq x)\sqrt{y}$
If $x>0$ then $\mathbb{P}(X\leq x)=\frac{1+x}{2} \implies \mathbb{P}(X\leq x)\mathbb{P}( Y\leq y)=\frac{1+x}{2}\sqrt{y}$
If $x\leq 0$ then $\mathbb{P}(X\leq x)=\frac{1-x}{2} \implies \mathbb{P}(X\leq x)\mathbb{P}( Y\leq y)=\frac{1-x}{2}\sqrt{y}$