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Is it possible to approximate any cdf by a continuous function? I was thinking that the difficulty might lie within cdf that have, for example, a dense set of discontinuities...

xyz
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1 Answers1

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Yes. There are at most a countably infinite number of discontinuities. You might, for example, replace each point of positive probability with a narrow uniform distribution, leading to

$$F_k(x)= k \int\limits_{z=x}^{x+\frac1k} F(z)\, dz$$ which is continuous and weakly increasing from $0$ to $1$ for any given $k>0$ and converges pointwise to $F(x)$ as $k$ increases.

To prove this, we can say, since $F(x)$ is a weakly increasing CDF, $F(x) \le F(z) \le F\left(x+\tfrac1k\right)$ for $z \in \left[x,x+ \frac1k\right]$, giving

$$F(x)=k \int\limits_{z=x}^{x+\frac1k} F(x)\, dz \le k \int\limits_{z=x}^{x+\frac1k} F(z)\, dz \le k \int\limits_{z=x}^{x+\frac1k} F\left(x+\tfrac1k\right)\, dz = F\left(x+\tfrac1k\right)$$

i.e. $F(x) \le F_k(x) \le F\left(x+\tfrac1k\right)$ and squeezing the continuous CDF $F_k(x)$,

then, since $F(x)$ is a right-continuous CDF meaning $\lim\limits_{k \to \infty} F\left(x+\tfrac1k\right) = F(x)$, we get the pointwise convergence of a continuous CDF to an arbitrary CDF $$\lim\limits_{k \to \infty} F_k\left(x\right) = F(x).$$

Henry
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  • If I understand correctly, the convergence is not exactly for every $x$ though (this is possible for the non-dense case, by using interpolation), but for all $x$ that are continuity points for $F$, because you are using the FTC right? Thus all we lose is a set of measure zero, and in particular convergence in distribution is preserved. So technically the answer would be no, but yes if we are fine modulo negligible (=Leb measure zero) sets. – xyz Oct 16 '24 at 15:39
  • @xyz I managed to persuade myself that it does converge to $F(x)$ everywhere, since a CDF is càdlàg and never exceeds $1$, while thinking about distributions such as that in a previous question, and that if this does not work somewhere for a dense discrete distribution then it would not work somewhere for a continuous distribution either. I may be wrong, and would be interested in any counterexample you can find. – Henry Oct 16 '24 at 18:01
  • I see the following issue: suppose F has a discontinuity point at a point $q$. For a cdf we can still have the situation that left limit is < right limit. Then the limit of $F_k(q)$ as $k$ grows is only guaranteed to be between $F(q^-)$ and $F(q^+)$. I think a function that presents a dense set of this situation is $F(x)=\sum_{n: q_n\le x}p_n$ for $x\in[0,1]$ and $q_n$ enumeration of the rationals in $[0,1]$, $p_n$ such that $\sum p_n =1$. Would your argument by contradiction work in this case? – xyz Oct 17 '24 at 15:09
  • In particular, in such situations, is $F_k(q)$ even guaranteed to converge at all, as $k$ grows? – xyz Oct 17 '24 at 15:40
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    @xyz: I do not see that. I think my $F_k(q)\ge F(q)$ for all $k$ since $F(q) \le F(z)$ for all $z \ge q$, while $F_k(z) \le F(q+\frac1k)$ for all $z \le q+\frac1k$. So $F(q^-)$ does not matter and we are only interested in $F(q^+) = F(q)$ by the right-continuity of all CDFs. In other words $F(q) \le F_k(q) \le F(q+\frac1k)$ so you do have a guarantee of $F_k(q)$ converging. I have edited my answer to provide a proof of pointwise convergence. – Henry Oct 17 '24 at 15:52
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    Thank you, I see where my confusion was, somehow I kept using the left limit instead of the point itself in the lower limit of the integral, but your definition only looks to the right so exploits the right-continuity of the cdf, thus not requiring the FTC to show convergence. – xyz Oct 17 '24 at 16:24