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Suppose I have a random sequence $X_n$ of cadlag functions on $[0,1]$ that converge weakly to $X$. In general this is meant with respect to the Skorkhod metric but suppose here I have that $X$ is continuous and $X_n$ converge weakly to $X$ with respect to the uniform metric on the space of cadlag processes $[0,1]$. (Under this metric this is a complete metric space however it is not separable.) I need that $X_n$ is well defined with respect to the uniform topology but suppose I have this as well.

Now suppose $X_n$ are piecewise constant and consider $\tilde{X}_n$, the linearized version of $X_n$. By this I mean the function I get by drawing straight lines between the jump discontinuities of $X_n$. Is it true that $\tilde{X}_n$ converges in law to $X$ again with respect to the uniform metric on continuous functions?

I'm pretty sure that this is true, and I believe I have a proof but I'd like confirmation. I thought this would be written down somewhere. I tried Billingsley, Convergence of Probability Measures, especially Chapter 3, Section 18 (The Uniform Topology), but I couldn't find it. Thanks for your help.

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  • $\begingroup$ The question is not really clear, so let's translate it to distributions. You consider functions $f:[0,1] \to \Bbb R$ that are cadlag, right? Let's say $\Omega$ is a space of all such functions with the uniform metric convergence. You have probability measures $\mu_n$ that converge weakly (w.r.t. topology of $\Omega$) to $\mu$. These $\mu_n$ are distributions of $X_n$. Now you suppose that $X_n$ is piecewise constand and $\tilde X_n$ is a linearized version of $X_n$. What does that mean. Also, your question now sounds like is it true that if I consider $\tilde X_n$, then $X_n \to X$ $\endgroup$ – Ilya Jul 23 '14 at 19:55
  • $\begingroup$ The linearised version of a piecewise function is the corresponding continuous function constructed by drawing straight lines between all the jump discontinuities. I fixed the tilde so that the question makes sense. Thank you. $\endgroup$ – Ben Jul 23 '14 at 23:07
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    $\begingroup$ So $\tilde X_n$ is not defined uniquely, you have same parameters of the linearization - correct? Depending on how you define it, perhaps the continuous mapping theorem could work. As far as I understand, $\|\tilde f - \tilde g\| \leq \|f - g\|$ where $f,g$ are piece-wise constant functions, $\tilde f,\tilde g$ are their linearizations done in a good way and $\|\cdot\|$ is the uniform sup norm. $\endgroup$ – Ilya Jul 24 '14 at 6:51
  • $\begingroup$ Nice. That's a lot cleaner than what I had. Thanks. $\endgroup$ – Ben Jul 24 '14 at 15:31
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If I understand the question correctly, the linearization map $l$ have some parameters and can be chosen to satisfy some nice properties. For example, we can define it in such a way that $$ \|l(f) - l(g)\|\leq \|f-g\| $$ where $\|f\|:=\sup_{x\in [0,1]}|f(x)|$ denotes the uniform sup-norm. Hence, $l$ is a Lipschitz map and thus continuous. The result follows from the continuous mapping theorem.

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