# Uniform convergence rate of cumulative distribution functions

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.

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.

• 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. – atzol Mar 9 at 19:31
• 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... – antkam Mar 9 at 19:47