How do you prove this?
Suppose converges to in distribution and converges in probability to . Show that converges to in distribution.
Thanks again.
Hello,
Convergence in probability implies convergence in distribution. From here, your problem is obvious
I can show you the proof... But it's pretty tedious I can link you to the French Wikipedia where there is a proof... Convergence de variables aléatoires - Wikipédia
where there is
(Xn converges to X in probability Xn converges to X in distribution)Théorème — Xn converge vers X en probabilité Xn converge vers X en loi.
Then prove that if Xn converges to X in distribution, and Yn converges to 0 in distribution, then Xn+Yn converges to X in distribution.
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I'll try to get you started for this problem, by using a similar method to the theorem's proof. You'll finish it because it looks very difficult to write down this in latex !
in distribution means that , for any x where is continuous. F denotes the cumulative distribution function.
This means that
This can be translated as :
in probability means that .
This can be translated as :
Now you want to show that
Write that , for any given
Since (the comma represents an intersection of events), we can say that
Similarly (this will need a little bit of thinking),
Hence we get :
Gaaah... At least this can get you started... I'm really struggling in thinking on a computer and have no much time left I don't even know if this is useful, because you can use that theorem above...
The steps after that are roughly : get an inequality in the form
And then make similar steps to get an inequality in the form
And this would mean that , which is what you want.
It may become simpler, but I would not say obvious... ?
Your proof is nice (there are just typos with the inclusion of sets, they are in the wrong direction). There are numerous equivalent ways to define convergence in distribution, and each one gives a proof. The proof I knew uses characteristic functions (there is an important theorem by Lévy saying that convergence in distribution to is equivalent to pointwise convergence of the characteristic functions to ). It is shorter but less elementary due to this theorem.
Yes indeed.... careless typos ! (which have been corrected now, thanks to Jhevon)
I realise that in the steps that would follow what I did, there would be some problems because of absolute values... I don't know if it is possible to solve them, but it makes the proof less simple
Maybe the best way is to use this theorem that the convergence in probability implies the convergence in distribution, which lets us say that Yn converges to 0 in distribution. And then prove that Xn+Yn converges to X in distribution.
Get yourself inspired from what was done above If you can't, well, just post your working here and we'll help you
There is not really a problem ; for the lower bound, you can write
Then for both the lower and the upper bound, let and use the fact that is continuous at (we prove convergence only at those points ) to choose such that . In addition, for large enough, is close from (it requires that is continuous at to say that, so we should choose accordingly; there only countably many bad choices so it's OK), and is small. That should do.
Hello again,
So we dealt with that generalized problem (when , any constant) in today's lecture..
It is known as Slutsky's theorem
We've written a proof, but it is around 1 page and a half long... And I couldn't find a proper one in the first page of google (maybe I didn't look at it very well...)
What Laurent said in #5 is exactly in the proof we've written and it looks like that what I did in #2 was not that incorrect...
If you want what we've done, you can take an appointment with me so that I copy it in latex for you lol