Suppose $\{X_n\}_n$ and $\{Y_n\}_n$ are sequences of (say, continuous) random variables with corresponding CDFs $\{F_n\}_n$ and $\{G_n\}_n$. Suppose that $$|X_n - Y_n| \xrightarrow{a.s} 0~.$$ Is it clear (without additional assumptions) that $$|F_n(t) - G_n(t)| \rightarrow 0~,$$ for all $t$? Working through the "standard" proof for the case where $Y_n = Y$ makes it seem that we might require equicontinuity of the families $\{F_n\}_n$ and $\{G_n\}_n$, for each $t$, but I was wondering if such an assumption is necessary for the result to go through.


1 Answer 1


It appears that the CDFs don't have to get close to each other at each point.

Define $X_n \sim N(1/n,e^{-n})$ and $Y_n = -X_n$. Then $|X_n - Y_n| \xrightarrow{a.s.} 0$, but $F_n(0) \to 0$ and $G_n(0) \to 1$. I think that you do need equicontinuity of the families, which is (I think?) equivalent to the densities being uniformly bounded (since we're assuming continuity).

  • $\begingroup$ Yes, thanks for the response! I guess a uniform bound on the densities is a natural way to think about this as well, although I suspect that this is a stronger condition?: We could (for example) impose that the CDFs belong to a Holder continuous family, examples of which include functions with unbounded derivatives. $\endgroup$
    – mam2017
    Jul 21, 2017 at 19:20
  • $\begingroup$ @mam2017, yeah I think you're correct I suspect it is a stronger condition as well. $\endgroup$
    – Marcus M
    Jul 21, 2017 at 19:25

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