How to show $2$ random variables are not independent. Let $X$ be uniform on $(0,1)$ and $Y:=3X+1$.
How that $X$ and $Y$ are not independent.
Clearly they are not independent since $Y$ is a function of $X$ but I don't know how to prov this formally. 
My notes give two ways: show that there are two event $\{X \leq s\}, \{Y \leq t\}$ that are not independent events or to show that the joint distribution function is not the product of the individual distribution functions. I don't really understand to be honest.
 A: If they are independent, then $$P(X \leq s, Y \leq t) = P(X \leq s) P(Y \leq t) \forall s,t.$$ To prove that they are not, we have to find $s,t$ such that the equality doesn't hold. 
\begin{align}P(X \leq s, Y \leq t )&= P\left(X \leq s, X \leq \frac{t-1}{3}\right) \\
&=P\left(X \leq \min \left(s, \frac{t-1}{3} \right) \right)\end{align}
In particular, let's pick $s=\frac12$ and $t=2$.
$$P( X \leq s, Y \leq t)=\frac13$$
but
$$P(X \leq s) P(Y \leq t)=P(X \leq s) P \left( X \leq \frac{t-1}{3}\right)=\frac{1}{2}\frac{1}{3} \neq P(X\leq s, Y \leq t)$$
A: Compute their covariance. Unless I'm mistaken, it's .25. If $X$ and $Y$ were independent it would be $0$.
A: What is the value of $P\{X \leq \frac 12\}$? What is the value of $P\{Y \leq 2\frac 12\}$? What is the value of $P\{X\leq \frac 12,  Y \leq 2\frac 12\} = P\left(\{X\leq \frac 12\}\cap \{Y \leq 2\frac 12\}\right)$? Does it equal $P\{X \leq \frac 12\}\cdot P\{Y \leq 2\frac 12\}$? Why, or why not? Are $\{X \leq \frac 12\}$ and $\{Y \leq 2\frac 12\}$ independent events? Why or why not?  Does $F_{X,Y}\left(\frac 12, 2\frac 12\right)$ equal $F_{X}\left(\frac 12\right)F_{Y}\left(2\frac 12\right)$? Why or why not?
