# Showing that two scalar random variables are independent when only their distributions are known

Suppose that $X$ and $Y$ are two scalar random variables, and we know only their distributions. Is there some general strategy to show whether or not these two random variables are independent?

In particular we have $$\mathbb{P}(X \leq x) = \begin{cases} 1 - e^{-2x}, & x \geq 0, \\ 0 , & x < 0, \end{cases}$$ and $$\mathbb{P}(Y \leq y) = \begin{cases} 1 - e^{-y}, & y \geq 0, \\ 0 , & y < 0. \end{cases}$$ I feel as if I am missing something extremely trivial. If I knew the joint density or the distribution of the random vector $(X,Y)$ I could check whether $$\mathbb{P}((X,Y) \in (-\infty, a] \times (-\infty, b]) = \mathbb{P}(X \leq a) \mathbb{P}(Y \leq b) \ .$$ However I am not given this information. Any ideas? I feel as if what I am asking is impossible.

$$\Pr(Y \leq y) = \begin{cases} 1 - e^{-y}, & y \geq 0, \\ 0 , & y < 0. \end{cases} \tag 1$$
Suppose what appears above is true and $X = Y/2$. Then it follows that $$\Pr(X\le x) = \begin{cases} 1 - e^{-2x}, & x\ge 0, \\ 0, & x<0, \end{cases} \tag 2$$ (since $X\le x$ if and only if $Y\le 2x$) and $X$ and $Y$ are then as far from independent as two random variables can be: the value of each determines the value of the other, and the correlation is exactly $1$.
But it is also possible for $(1)$ and $(2)$ to hold and $X$ and $Y$ to be independent.
You are indeed correct, in order to determine if $X$ and $Y$ is independent you need to know the simultaneous distribution. Knowing the marginal distributions is not enough.