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Let $\{X_n\}_{n=1}^{\infty}$ be a sequence of random variables such that, for some random variable $X$,

$$|X_n-X| = O(a_n),$$

almost surely, for some sequence of real numbers $\{a_n\}_{n=1}^{\infty}$ converging to zero. That is, $X_n$ converges to $X$ almost surely, at rate $a_n$.

Of course, it follows that $X_n$ converges to $X$ in distribution. My question is whether the cumulative distribution function (cdf) $F_n$ of $X_n$ also converges uniformly to the cdf $F$ of $X$, at rate $a_n$. Specifically, does it hold that

$$\sup_{x \in \mathbb R} |F_n(x) - F(x)| = O(a_n)?$$

This seems true to me, but I do not have a proof. Thank you in advance for any references or suggestions on proving this.

Addendum: You may assume that the $F_n$ and $F$ are absolutely continuous, to avoid issues regarding discontinuity points in the last display above.

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The discontinuous case is false, and I think with a little hand-waving I'd say the continuous case is false also. The issue is that $|X_n - X|$ is in "units" of $X$, whereas $|F_n(x) - F(x)|$ is in "units" of probability, so you cannot bound both with the same series.

Anyway, counterexample: all the r.v.s are constants and $X = 1, X_n = 1 + b^n$ for some $0 < b < 1$.

Clearly $|X_n - X| = b^n$ surely (not just almost surely).

However, for any finite $n$, there exists $x = 1 + b^n/2$ s.t. $F_n(x) = 0, F(x) = 1$ so the sup $= 1$.

You see what I mean when I wrote about the "different units"? Another way to think about this is if you plot $F$ and all the $F_n$ on the same graph, with $x \in (-\infty, \infty)$ as the horizontal axis and $[0,1]$ as the vertical axis, then $|X_n - X|$ is a proxy for the horizontal separation but $|F_n - F|$ is the vertical separation. My counterexample is just a series of step functions which get closer and closer horizontally but the vertical distance is always $1$ at the max point.

This counter-example is discontinuous but if you want, just use small uniform r.v. to change steps into ramps and you can prove things rigorously, e.g. $X_n = Unif(1, 1+b^n), F_n(1) = 0, F(1) = 1, \sup = 1$, etc.

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  • $\begingroup$ Agreed with your discontinuous example. But in your last paragraph, the limiting cdf $F$ remains discontinuous, so I don't think this is a counter-example to my claim for the continuous case. I do see what you mean about the difference in units. My intuition was that controlling $|X_n-X|$ is typically much stronger than controlling $|F_n-F|$, but I suppose it might not be obvious that the rates are the same. $\endgroup$
    – atzol
    Mar 9, 2019 at 19:31
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    $\begingroup$ Haha, you're right! I made all the $F_n$ continuous but forgot about $F$ itself! Lemme see if this can be fixed. You might have a point re: the everybody continuous case... $\endgroup$
    – antkam
    Mar 9, 2019 at 19:47

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