Equality in distribution of random variables and their probabilities

Question

Suppose we have two sequences of random variables $\{X_n(\theta)\}_{n\ge 1}$ and $\{Y_n(\theta)\}_{n\ge 1}$ and they are equal in distribution for all $n \ge 1$ and $\theta \in \Theta$ where $\Theta$ is, say, compact. Let us further assume that $$\tag{1}\label{eq1}\mathbb{P}(\sup_{\theta\in\Theta} \vert Y_n(\theta) \vert > \epsilon) \to 0.$$ Can we claim that $$\tag{2}\label{eq2}\mathbb{P}(\sup_{\theta\in\Theta} \vert X_n(\theta) \vert > \epsilon) \to 0.$$ My intuition is telling me yes, because if $X_n(\theta)$ and $Y_n(\theta)$ are equal in distribution, then $$\mathbb{P}(\{\omega \in \Omega: \sup_{\theta \in \Theta} \vert X_n(\theta; \omega) \vert > \epsilon\}) = \mathbb{P}(\{\omega \in \Omega: \sup_{\theta \in \Theta} \vert Y_n(\theta; \omega) \vert > \epsilon\})\to 0.$$ In fact, $\Theta$ needs not be compact. I want to claim \eqref{eq2} from \eqref{eq1} (provided that $X_n(\theta)$ is equal to $Y_n(\theta)$ in distribution for all $n\ge 1$, $\theta \in \Theta$). However, I am not sure if I can do that. If I can't, is there a counter example? If I can, is a proof required or is the implication simple enough that I can do without one?

Answer 1

I don't think this is correct with just the assumption that $X_n(\theta)$ and $Y_n(\theta)$ have the same distribution for all $n,\theta$. Consider $\Theta = \mathbb{N}$, $\Omega = [0,1]$ with Lebesgue measure, $Y_n(k; \omega) = 1_{[0,\frac 1n]}(\omega)$ for all $k \in \Theta$. Let $X_n(k;\omega) = 1_{\left[\frac{k-1}{n},\frac kn\right]}(\omega)$ for $k \le n$ and $X_n(k;\omega) = 1_{[0,\frac 1n]}(\omega)$ for $k > n$. Then, for all $n \ge 1$ and $\theta \in \Theta$ we have \begin{align*} \mathbb{P}(Y_n(\theta) = 1) = \mathbb{P}(X_n(\theta) = 1) = \frac 1n \end{align*} so $X_n(\theta)$ and $Y_n(\theta)$ have the same distribution. For any $\varepsilon \in (0,1)$, we have $\mathbb{P}\left(\sup_\theta Y_n(\theta) > \varepsilon \right) = \frac 1n \rightarrow 0$. However, for all $\omega \in \Omega$ we have $\sup_{\theta}X(\theta;\omega) = 1$ so $\mathbb{P}\left(\sup_\theta X_n(\theta) > \varepsilon \right) = 1$, which clearly does not converge to $0$.

If we wanted to make $\Theta$ compact, I think we could apply the same example to the set $\hat \Theta := \{0,1,\frac 12, \frac 13, ... \}$.

To rule out something like this, you would need to say something about the joint (across $\theta$) distributions of $X_n(\theta)$ and $Y_n(\theta)$.