Consider two sequences of random variables. At each point in the sequence $X_n \sim F_n$ and $Y_n \sim G_n$, and let $F_n(t)$ and $G_n(t)$ denote their respect CDFs. The distributions $(F_n, G_n)$ are not guaranteed to be continous.

Suppose that both random variables converge to the same limiting distribution. Formally, as $n \to \infty$, $X_n \to^d Z \sim H$ and $Y_n \to^d Z \sim H$, where $H(t)$ is a strictly increasing, continuous function.

Can we show that $G_n(X_n)$ converges to a uniform? If not, can we find a counter-example?


1 Answer 1


Since $G_n(r) \leqslant 1$ for all $r \in \mathbb{R}$, this function is "dominated" by the unit function over its entire domain. This allows us to apply the dominated convergence theorem to move the limit inside the probability operator. Taking any value $0 \leqslant t \leqslant 1$ we then have:

$$\begin{align} \lim_{n \rightarrow \infty} \mathbb{P} [ G_n(X_n) \leqslant t ] &= \mathbb{P} \bigg[ \lim_{n \rightarrow \infty} G_n(X_n) \leqslant t \bigg] \\[6pt] &= \mathbb{P} \bigg[ H(Z) \leqslant t \bigg] \\[6pt] &= t, \\[6pt] \end{align}$$

which proves pointwise convergence to the uniform.

  • $\begingroup$ Thanks! This is an interesting idea. How do you prove $\lim_{n \to \infty} G_n(X_n) \to H(Z)$? $\endgroup$ Commented Jul 24, 2022 at 22:35
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    $\begingroup$ That convergence is not true (since both sides are random variables) but they converge in distribution so the probabilities pertaining to those events are the same. $\endgroup$
    – Ben
    Commented Jul 25, 2022 at 11:46
  • $\begingroup$ Is it necessary for $G_n$ to converge uniformly to $H$? $\endgroup$ Commented Jul 25, 2022 at 14:53
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    $\begingroup$ I believe that pointwise convergence implies uniform convergence for CDFs (see e.g., here). $\endgroup$
    – Ben
    Commented Jul 26, 2022 at 8:39
  • $\begingroup$ That's a good point, thanks! $\endgroup$ Commented Aug 2, 2022 at 17:55

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