Coupling continuous random variables Let $X,Y$ be random variables with densities $f_X(x)=x^{-2}1_{[1,\infty)}$ and $f_Y(x)=3x^{-4}1_{[1,\infty)}$.

How do I couple these such that $X\leq Y$ with probability 1?

Attempt:
I know that they intersect at $x=\sqrt{3}$. Then $f_Y(x)>f_X(x)$ for $x<\sqrt{3}$ and $f_Y(x)<f_X(x)$ for $x>\sqrt{3}$.
Now we want to find random variables $X'$ and $Y'$ with joint probability function $P'$ such that its marginals return $X$ and $Y$ and $X'\leq Y'$.
How do I do this?
 A: If such $X$ and $Y$ would exist, then from $X\le Y$ one would get:
$$\infty = E[X]\le E[Y]=\frac{3}{2}$$
However, if you change the hypothesis from $X\le Y$ to $X\ge Y$, then taking $X=Y^3$ would work ($X\ge Y$ with probability $1$, and $f_Y(y)=f_X(y^3)\cdot 3x^2=3/x^4$)
A: A standard way to couple two random variables such as these is to couple them both with the uniform probability on $[0,1]$, using cumulative distribution functions, and then eliminate the middle probability.
In this case, the cumulative distribution functions are $F_X(x) = 1 - 1/x$ and $F_Y(x) = 1 - 1/x^3$ (both supported on $[1,\infty)$). Note that their inverse functions $F_X^{-1}(y) = 1/(1-y)$ and $F_Y^{-1}(y) = 1/(1-y)^{1/3}$ are both increasing functions from $[0,1)$ to $[1,\infty)$.
Let $W$ be a uniform random variable on $[0,1]$, define $X' = F_X^{-1}(W)$ and $Y' = F_Y^{-1}(W)$, and define $Z'=(X',Y')$. The definitions of $X'$ and $Y'$ ensure that they have the same distributions as $X$ and $Y$, respectively; moreover, the fact that $F_X^{-1}(y) \ge F_Y^{-1}(y)$ ensures that $X' \ge Y'$ (all the time, not just with probability $1$).
(Note that this construction ends up giving $X' = (Y')^3$, as proposed in Momo's earlier answer.)
