Timeline for Topologies for which the ensemble of probability measures is complete
Current License: CC BY-SA 3.0
17 events
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| Apr 13, 2017 at 12:44 | history | edited | CommunityBot |
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| Dec 21, 2015 at 0:56 | comment | added | Yemon Choi | Guillaume: perhaps the following analogy would help. Consider $C[0,1]$ and a sequence $(f_n)\subset C[0,1]$ where $f_n(0)=0$, $f_n(n^{-2})=1/n$, $f_n(1)=1$ and we use piecewise-linear interpolation to define $f_n$ everywhere else. Now $C[0,1]$ is complete in the uniform norm, and $(f_n)$ is a sequence in the unit sphere of this Banach space which converges to the function $g(t)=t$ in the uniform norm. On the other hand, if we look at the Lipschitz constants of these functions $f_n$, we see that they blow up as $n\to\infty$, even though the limit function $g$ has Lipschitz constant $1$ | |
| Dec 21, 2015 at 0:44 | comment | added | Yemon Choi | @PaulSiegel I think you have your finger on the issue, but would just like to add that since not every probability distribution has finite variance, $X\to {\rm var}(X)$ isn't even well-defined -- unless you're thinking of it as taking values in the extended positive reals? | |
| Dec 20, 2015 at 23:18 | answer | added | Paul Siegel | timeline score: 1 | |
| Dec 20, 2015 at 20:20 | comment | added | Paul Siegel | I'm pretty sure that the answer to this question is "no" - there are just too many counterexamples. My guess is that the best you can hope for is convergence theorems for moments of specific classes if random variables, but I'm not sure. | |
| Dec 20, 2015 at 20:17 | comment | added | Paul Siegel | OK, I agree that your $X_n$ converge in TV to $N(0,1)$, and that the first two moments of $X_n$ converge to $1$ and $\infty$, respectively. But this is not an issue with completeness, it's an issue with continuity. I think you are really trying to ask: "Is there a topology on the space of probability distributions with the property that the functions $X \mapsto E(X)$ and $X \mapsto Var(X)$ defined in this space are continuous? | |
| Dec 20, 2015 at 19:14 | comment | added | Yoav Kallus | It seems the reason you don't like the convergence is that the moments don't converge? Maybe you should just look at convergence of moments directly? Something like in this question: mathoverflow.net/questions/102964/… | |
| Dec 20, 2015 at 19:06 | comment | added | Yoav Kallus | What do you mean by what "the limit is really"? Under total variation, as you noted, the limit is really N(0,1). | |
| Dec 20, 2015 at 18:47 | history | edited | Guillaume Dehaene | CC BY-SA 3.0 |
added 157 characters in body
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| Dec 20, 2015 at 18:45 | comment | added | Guillaume Dehaene | I have added the densities of the $X_n$ in the main post for @PaulSiegel. My problem is that the limit isn't really N(0,1) because the limit mean is 1 and the limit variance is infinite: maybe your way of phrasing it is the better way. Would it make sense to complete the space under TV for example ? | |
| Dec 17, 2015 at 16:26 | comment | added | Yoav Kallus | It seems your problem is opposite of what you state: in your example, the sequence of measures do have an actual probability measure as a measure as a limit, but you want the limit to be not a probability measure. Maybe what you want is not a topology under which the space of probability measures is complete, but rather a completion of the space under your favorite topology. | |
| Dec 17, 2015 at 16:05 | comment | added | kjetil b halvorsen | I think that maybe the concept you need is compactness not completeness. The relevant concept is often called tightness. | |
| Dec 17, 2015 at 15:51 | comment | added | Paul Siegel | Completeness isn't anything weird for measures, but the measures themselves can be weird. Could you elaborate on exactly what your $X_n$ are, perhaps by writing out their density functions? | |
| Dec 17, 2015 at 15:41 | comment | added | Guillaume Dehaene | That's interesting. The sequence of $X_n$ do converge in TV (the distance is upper-bounded: $d(X_n,X_\infty)\leq 1/n$). Is being complete weirdly defined for measures ? | |
| Dec 17, 2015 at 15:34 | history | edited | YCor | CC BY-SA 3.0 |
moved bracket from bracket to inside the post since otherwise it's too visible
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| Dec 17, 2015 at 11:42 | comment | added | Paul Siegel | The space of regular Borel probability measures on $\mathbb{R}$ is complete with respect to the total variation norm; by the Riesz representation theorem you can identify it with a closed subset of the dual of the Banach space $C_0(\mathbb{R})$, for instance. I'm a little confused by your example, but if the distributions really do converge in total variation then the limit is a regular Borel probability measure. | |
| Dec 17, 2015 at 10:37 | history | asked | Guillaume Dehaene | CC BY-SA 3.0 |