Let $U,V$ be random variables on $\mathbb{N}_0$ with pmf's
\begin{equation} f_U(x) = \frac{1}{2} \mathbb{1}_{\{0,1\}}(x), f_V(x) = \frac{1}{3} \mathbb{1}_{\{0,1,2\}}(x) \end{equation}
Give a coupling of U and V under which $\{U \geq V\}$ with probability 1.
I really don't see how, as the marginals of the coupling have to be equal to the pmf's of $U$ and $V$, and $0 = \mathbb{P}(U = 2) < \mathbb{P}(V = 2) = \frac{1}{3}$.
Help?
$U \geq V$ with probability $1$ implies that $P\{U = 0, V = 1\} =P\{U = 0, V = 2\} = P\{U = 1, V = 2\} = 0$. Hence $P\{V = 2\} = 0$, which contradicts the given marginal distributions. So there is no joint distribution that satisfies the given criteria.