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Let $X_1, X_2, \dots \in \mathbb{R}^d$ be random vectors, each with cdf $F_n$. Let $F$ denote the cdf of another random vector $X$. Suppose they are all continuous w.r.t. Lebesgue measure for now. I know that when $d=1$, $F_n(x) \to F(x)$ implies uniform convergence, i.e., $$\sup_{x \in \mathbb{R}} |F_n(x)-F(x)| \to 0.$$ And this link has a nice proof for it: Convergence in law implies uniform convergence of cdf's

I wonder whether this holds true in general for $d>1$. And if it's not, what would be the conditions for it to be true?

Intuitively I thought this is true and I should be able to just modify the proof for the $d=1$ case to prove for higher dimensions. However, I've been thinking about this for a while but still couldn't figure out how to set it up even for $d=2$. Any thoughts would be greatly apprecaited!

statstats
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  • What does it mean that $F_n(x) \to F(x)$ "almost surely"? These are not random variables. Same with the conclusion "with probability $1$" is meaningless as nothing is random in that statement. – Michh Dec 30 '23 at 21:42
  • You are right. For the original problem I had in mind, $Fn$ is rather $F_{\hat{\theta}_n}$, i.e., it's a parametric cdf based on estimators $\hat{\theta}(X_n)$, and hence I had the "a.s." and "w.p.1" there. I forgot to remove these when I posted the question. I'll edit the question. Thanks for catching it! – statstats Dec 30 '23 at 22:25

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