How do you prove this?

Suppose converges to in distribution and converges in probability to . Show that converges to in distribution.

Thanks again.

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- March 6th 2009, 02:08 AMakolman[SOLVED] A convergence proof.
How do you prove this?

Suppose converges to in distribution and converges in probability to . Show that converges to in distribution.

Thanks again. - March 7th 2009, 01:13 AMMoo
Hello,

Convergence in probability implies convergence in distribution. From here, your problem is obvious :D

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

Quote:

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. - March 7th 2009, 03:37 AMLaurent
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. - March 7th 2009, 07:29 AMMoo
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 :D If you can't, well, just post your working here and we'll help you (Tongueout) - March 7th 2009, 10:02 AMLaurent
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.

- March 7th 2009, 01:32 PMakolman
Thanks a lot for the help, I think I get how to set the inequalities and take the limits.

- March 11th 2009, 10:34 AMMoo
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 (Rofl) 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 (Doh) - March 11th 2009, 12:50 PMakolman
Thank you very much Moo. Actually, my professor handed me the solution for this homework problem. Your hints and suggestions were really helpful (Bow).