Talk:Convergence of random variables

message: The article Lévy's convergence theorem has been proposed for deletion  because of the following concern:
:This is basically a duplicate of a part of Dominated convergence theorem, but with a wrong name: no reference has been found where Lévy is credited. See the talk page for more info.
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thread_title: [[Wikipedia:Proposed deletion|Proposed deletion]] of [[Lévy's convergence theorem]]

message: If f_n(x)\xrightarrow{a.s}f(x) and $t_n\xrightarrow{a.s}t$ as ${n \to \infty}$, is this always true that $f_n(t_n)\xrightarrow{a.s}f(t)$ as ${n \to \infty}$? 
Note that $X_n \xrightarrow{a.s} X$ means $X_n$ converges to $X$ almost surely, in other words, $P(\lim\limits_{n \to \infty}X_n=X)=1$. You read more about it here:
Can you please help me how to start solve it? Is this a property?  — Preceding unsigned comment added by Profabc (talk • contribs) 17:07, 15 May 2016 (UTC)
thread_title: Properties of almost sure convergence

message: This article does not include any advanced weak convergence results, i.e. convergence in general metric spaces (for example Donsker's theorem links to this page, yet there is no mention of the type of convergence actually happening in this theorem). This should be included.
--Steffen Grønneberg 17:27, 13 April 2006 (UTC)
thread_title: Advanced Weak Convergence Results

message: What's the view on adding a section on uniform convergence?  Or should that be treated under stochastic processes?  If so, perhaps a link to the relevant article would be worthwhile?--Steve Kroon 14:14, 14 February 2007 (UTC)
thread_title:  Uniform convergence? 

message: This article seems to take for granted the difference between converging to a function (e.g., sure convergence and almost sure convergence) and converging to a random variable (e.g., the other forms of convergence). Note that you can almost surely convergence to a function that is not a random variable (i.e., not a Borel measurable function. It would be nice if this was cleared up. This would make the definitions make more sense. --TedPavlic 18:47, 8 April 2007 (UTC)
thread_title: Convergence to random variables

message: All of the convergence ideas are in terms of SEQUENCES. That is, they all involve countable index sets. Convergence is defined for uncountable index sets in general. These definitions should be adjusted so that they refer to any general random process, regardless of the countability of its index set. --TedPavlic 18:47, 8 April 2007 (UTC)
thread_title: Convegence of NETS and General Stochastic Processes

message: The article definitions of convergence don't account for usage of convergence for two sequences of random variables, X_n and Y_n, so that X_n \xrightarrow{\mathcal D} Y_n, when neither limit is sensible as such. This is usually used for various ways of stating that the sum of suitable random variables converges to a normal distribution - as used in the article under "implications": "If Sn is a sum of n real independent random variables: S_n = X_1+\cdots+X_n then Sn converges almost surely if and only if Sn converges in probability." It's also used in the central limit theorem article: "the convolution of a number of density functions tends to the normal density as the number of density functions increases without bound". It's fairly commonplace outside WP, too, so explaining it somewhere would be nice. My explanation above was maybe a bit overly general as I haven't seen this used for anything that rescaling won't fix (i.e. using mean instead of sum), but I think such statements are somewhat informal explanations of "convergence in shape of distribution". -- Coffee2theorems (talk) 18:26, 10 November 2008 (UTC)
thread_title:  Convergence to moving targets 

message: I’m removing the statement that the following d(·,·) metrizes convergence in probability: 
: 
    d(X, Y) := \operatorname{E} \big[ \min \{|X-Y|, 1\} \big],
  
since the provided reference (; chapter 2) does not give any rationale for why is this indeed a metric and why is it equivalent to convergence in probability.
I replace the old metric with the Ky Fan metric (see the article), since the referenced text does prove the theorem that this metric metrizes convergence in probability. ... stpasha » talk » 20:42, 14 September 2009 (UTC)
thread_title:  Metrics 

message: I removed this example from the "convergence almost surely" section:
The primary reason is that f(i) is not a valid limiting variable, and thus we cannot say that Ui converges f(i). Of course one can claim that what is understood here is that Ui − f(i) converges a.s. to zero, but this is equivalent to saying that Xi converges a.s. to zero, which makes this example way too similar to the example 1. Besides, the statement “history bears out that all businesses sooner or later fold up” is dubious as well: there is at least one business, the Catholic church, which survived for over 1500 years already and doesn't show any signs of giving up.  … stpasha »  07:44, 21 November 2009 (UTC)
thread_title:  Convergence almost surely: Example 3 

message: The example for the archer is very good, however it is wrong !
The sequence converges in probability, but saying that it doesnt converge almost surely is false, it all depends on how fast the probability of not hitting the bullseye decreases !
For exemple, if the probability of hitting the bullseye for the n-th shot is (1 − 2−n), then Borel-Cantelli assures you that it also converges almost surely.
However, if the hit probabilty is (1 − 1/n) then it doesn't converge almost surely.  —Preceding unsigned comment added by 193.49.124.107 (talk) 09:40, 4 March 2010 (UTC)
replies: text: The example is correct, only underspecified. Although your remark is quite valid in the regards that if the probabilities were equal (1 − 2−n) then the sequence would have converged almost surely as well as in probability, this only demonstrates that in reality those probabilities cannot converge to zero so rapidly. The example appeals to the everyday's experience of a lay-reader. We know that no archer is perfect. However if the probabilities were (1 − 2−n) then we would be observing that at least some of the archers in the world have became perfect and never err anymore. Since this doesn’t happen, we can conclude that convergence almost surely doesn’t happen.  // stpasha »  00:47, 27 March 2010 (UTC)

text: Great they are, but for the wrong question.  I believe the example and the two comments above are confusing the term "infinitely often" with "almost sure convergence".  An event (or rather a sequence of events) is said to occur "infinitely often" if no matter how far down the sequence we go the event will at some point occur again.  For example if the archers chance of missing is 1/n on his nth shot then although he is continually improving he will always have another miss to come at some point in the future.  However this article is about modes of convergence, and the archer example doesn't say anything on this.  Modes of convergence is to do with how the functions(random variables) map from the probability space to the values in the range of the function.  To see the difference between "convergence almost surely" and "convergence in probability".  Take your probability space to be the uniform measure on the [0,1] interval.  Then take a function like f_n(x) = 1 if x is between [k/2^m,(k+1)/2^m] and 0 otherwise, where n= 2^m + k and k goes from 0 to 2^m -1.  That is the sequence for (m,k) goes (0,0),(1,0),(1,1),(2,0),(2,1),(2,2),(2,3). This function converges in probability to f(x) =  0 but does not converge almost surely.  This is because the "bump" is always moving across the probability space.  So the probability P(f_n ≠ 0) = 1/2^m.  However for any particular x in [0,1] f_n(x) does not converge to 0 (there is always a 1 in there from time to time). So the set of x in [0,1] where convergence holds is the empty set so we do not have almost sure convergence.  Some graphics of the sequence of functions I've described would make for a good example. --Stryac (talk) 16:38, 1 July 2010 (UTC)
thread_title:  Archer example 

message: On the "requests" page there has been a request for an article on "stochasic equicontinuity", referencing   as a possible start. This also considers "uniform convergence in probability". One or both things might reasonably be dealt with by expanding this article, rather than having separate articles. 14:34, 14 September 2010 (UTC)
replies: text: I believe this article is not the right place. Stochastic equicontinuity is the extension of the equicontinuity property to random variables (and probably should be discussed there). Stochastic equicontinuity is not a mode of convergence, but rather a technical condition needed to ensure that pointwise convergence in probability transforms into the uniform convergence in probability.  // stpasha »  16:27, 14 September 2010 (UTC)
thread_title:  Possible expansion 

message: I think it would be helpful if there was a diagram showing the relationship between the different types of convergence. I have taken a stab at making one:
http://upload.wikimedia.org/wikipedia/commons/4/49/Notions-of-convergence-in-probability-theory.jpg
I'm happy to share the photoshop and mathematica file I used to make it with anyone who would like to improve on it. I think something like this diagram would be helpful in the article for people to quickly assimilate the information.  —Preceding unsigned comment added by Emptiless (talk • contribs) 17:20, 8 October 2010 (UTC)
thread_title:  Diagram of the relationship between convergence types 

message: The examples provided in the "almost sure convergence" section are also examples of "sure convergence".  I would move them to "sure convergence" examples.  The main difference is that "almost sure convergence" allows for violations on a set of measure 0, and these examples don't highlight this.  I would suggest putting an example that involves a set of measure 0.  Off the top of my head, maybe something like this: consider a random variable x drawn uniformly on (0,1).  Let C_n be a sequence that converges to the Cantor set.  Then what is the probability that x is in C_n?  Or, to make a random variable, the random variable y_n is 1 if x_n is in C_n, and 0 else.  Then because the Cantor set has zero measure, y_n converges to 1 almost surely.   Sorry this example isn't very precise -- maybe there's a much better one from standard textbooks.  But my point is, I think the example should involve a zero-measure set. Lavaka (talk) 19:27, 2 December 2010 (UTC)
thread_title:  Almost sure convergence vs. sure convergence 

message: for example Q_n(Y_n,theta) converges to Q(theta) almost surely uninformly in theta. How is this defined exactly??? Jackzhp (talk) 21:21, 13 January 2011 (UTC)
thread_title: Uniformly in some parameter

message: The animation shows that for n=3,4,.. the distribution is smooth - makes no sense to me.  — Preceding unsigned comment added by 193.219.42.53 (talk) 14:23, 10 February 2012 (UTC)
thread_title: Graphic example of convergence on the right

message: I am removing the subsection on convergence in rth-mean because it replicates (badly) the section it belongs to! Also the example is kind of silly IMO, but if someone feels like it is useful then put it back by all means. 189.166.57.181 (talk) 16:44, 10 September 2012 (UTC)
thread_title: Convergence in mean

message: The examples seem to me to give the impression that in order to converge almost surely, the sequence must (with probability 1) become constant.  But that's not right.  It's just saying that with probability 1 the sequence converges -- in other words, you can pick a large enough N to ensure that X_n is no more than eps away from X, for any particular choice of eps.  To compare to the archer example: the reason it converges in probability but doesn't converge almost surely is because we can pick some definition of a "bad miss", and the archer will still miss by that much every once in a while, but those misses become rarer and rarer, such that we can make them arbitrarily rare by taking the number of previous shots to be large enough.  But we can't take the number of previous shots large enough to ensure he will never have a bad miss again.  The probability of a bad miss goes to zero, but doesn't reach zero in some finite number of shots.  In contrast, if for each definition of a "bad miss" we could pick some number of previous shots after which he would never have a bad miss again (i.e., the probability of another bad miss is zero), then it would converge almost surely.  But that doesn't mean he never misses by *any* amount from then on.  Rather, after some number of shots he will no longer have a "bad miss", after some greater number of shots he will no longer have a "somewhat bad miss", after some even greater number of shots he will no longer have a "just slightly bad miss", etc. - Tim314 (talk) 16:41, 25 June 2015 (UTC)
replies: text: Convergence almost surely does not imply that the limit is constant, however, it _does_ mean that it's marginal distribution is eventually constant (except a zero-measure set). My example would be the following: suppose our sequence is the following one: X_1 \sim N(0,1) , and the rest X_i , are equal to X_1. This sequence converges almost surely (I think, even surely), but each X_i's distribution is not constant, far from it. However, the marginal distribution given a concrete realisation of X_1 is, indeed, singular.  — Preceding unsigned comment added by Lockywolf (talk • contribs) 04:13, 30 July 2023 (UTC)
thread_title:  Converges Almost Surely doesn't mean eventually constant 

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thread_title:  External links modified 

message: The article affirms that the fact that convergence amost everywhere implies convergence in measure is a (seemingly straightforward ) consequence of Fatou's Lemma. I know that this implication is true, what I would like to know is how to prove it by using Fatou's Lemma, and not the usual argument of the continuity of the probability measure. Brunosmaniotto (talk) 21:08, 1 June 2018 (UTC)
thread_title:  Convergence almost everywhere implies convergence in measure via Fatou's Lemma 

message: About the implications between convergences, it seems to me that a sequence which is bounded in  L^p  and converges in probability converges in  L^q  for any q<p .
This follows for example from  |X_n -X|^q  being uniformly integrable, because bounded in L^{p/q} .
It seems so useful to me that seeing it missing let me fear that I am doing a mistake. Does anyone have a reference for this?  — Preceding unsigned comment added by 134.157.64.107 (talk) 16:05, 22 July 2020 (UTC)
thread_title:  Bounded in Lp implies convergence in Lq 

message: I find it a bit misleading that the animation for convergence in distribution depicts a PDF converging to the normal; doesn't convergence in distribution entail that the CDF converges, and says nothing about the PDF? 2601:644:400:8430:6DB7:B5DC:FD18:70E (talk) 08:32, 11 May 2024 (UTC)
thread_title:  Graphic example of convergence in distribution