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Let $X \sim U[-1,1]$ and $Y=X^2$. Show that $X$ and $Y$ aren't independent.

Of course we have $F_X(x) = \frac{x+1}{2}$ and $F_Y(y) = \frac{\sqrt{y}+1}{2}$. But how can I find $$F_{XY}(x,y) = \Pr(X \le x, Y \le y) = \Pr(X \le x , X \le \sqrt{y})$$

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"Of course we have that..." is not true. This is false for $F_Y$. – Jimmy R. Mar 16 '14 at 11:13

The formula that you give for $F_Y(y)$ is false. Actually $$\begin{align*}F_Y(y)&=P(Y\le y)=P(X^2 \le y)=^{1>y>0}P(-\sqrt{y}\le X \le \sqrt{y})=\\&=F_X(\sqrt{y})-F_X(-\sqrt{y})=\frac{-\sqrt{y}-(-1)}{1-(-1)}-\frac{\sqrt{y}-(-1)}{1-(-1)}=\frac{-\sqrt{y}+1-\sqrt{y}-1}{2}=\\&=\frac{2\sqrt{y}}{2}=\sqrt{y}\end{align*}$$ for $0<y<1$. But you do not need the above, since proving that two variables $X,Y$ are not independent is much easier than proving that they are independent. In your case, it suffices to show that that there exist sets $A, B$ such that $$P(Y\in B|X \in A )\neq P(Y \in B)$$ For example take $B=[\frac12, 1]$ and take $A=[-\frac12, \frac12]$. Then of course $$P(Y \in B)>0$$ but $$P(Y \in B| X \in A)=0$$ since given that $X \in [-\frac12, \frac12]$ you have that $$Y=X^2 \in [0, \frac14]$$ Another more elegant way to show that they are dependent is exhibited here, in the accepted answer. This is based on the fact that if two random variables $X,Y$ are independent then random variables that are functions of $X, Y$ should also be independent (this is intuitively obvious, but admitts also a strict mathematical proof). In symbols $X,Y$ independent should imply that $f(X)$ and $g(Y)$ are also independent for any choice of $f, g$. Obviously taking $f(X)=X^2$ and $g(Y)=Y$ you have that $$f(X)=X^2=Y=g(Y)$$ and therefore obviously the random variables $X, Y=X^2$ are not independent.

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For a small $\delta$, compare the probabilities $$\mathbb P\{X\in [0,\delta], Y\in [0,\delta]\}\mbox{ and }\mathbb P\{X\in [0,\delta\}\cdot \mathbb P\{Y\in [0,\delta]\}.$$

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