# Estimator converges in probability - interesting simulation wise? [closed]

I am currently reading a paper in which the authors are constructing an estimator for a cumulative distribution (like) function (some context here), without going too much into the details they have the following convergence result for the estimator $\hat N_n(t)$ of $N(t)$ $$\sqrt n(\hat{N}_n(t)-N(t))\overset{\mathcal L} \longrightarrow \mathbb G \text{ in } \ell^\infty (\mathbb R)$$ This relation does imply convergence in probability, so we have $$\hat{N}_n(t)\overset{P} \longrightarrow N_n(t)\iff \forall t\in\mathbb R,\, \varepsilon>0: \lim_{n\to \infty}P(|\hat N_n(t)-N(t)|>\varepsilon)=0$$ My question: Is it worth to try to implement this estimator and to run it for some $n$ and $t$?

I mean, I don't know anything about the convergence rate or such alike, so from my naive point of view I am not sure whether there is anything I can draw from this implementation simulation wise, even if I could implement it in an efficient way. If I would have a.s. convergence things would be different, that's clear, but here...any ideas why an implementation could be interesting?

• Without the details of the situation, and not knowing the purpose of the simulation, there's nothing anyone can (validly) say--even in the case of a.s. convergence.
– whuber
Jun 5 '17 at 21:10
• @whuber Well, I first thought I could try to approximate the distribution function by taking an explicit example - so basically just a plot of the approximation depending on the order of $n$ or investigate different classes of distributions (pretty much like in the case of an integral approximation etc.) - maybe this too naive. In the case of a.s. convergence things should be clearer, don't they? In this case the estimator should approximate the real distribution, since this resembles just the pointwise convergence. But convergence in prob. seems too weak to get anything like it. Jun 5 '17 at 21:30