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Math Help - [SOLVED] A convergence proof.

  1. #1
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    [SOLVED] A convergence proof.

    How do you prove this?

    Suppose X_n converges to X in distribution and Y_n converges in probability to 0. Show that X_n + Y_n converges to X in distribution.

    Thanks again.
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by akolman View Post
    How do you prove this?

    Suppose X_n converges to X in distribution and Y_n converges in probability to 0. Show that X_n + Y_n converges to X in distribution.

    Thanks again.
    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
    Théorème — Xn converge vers X en probabilité \Rightarrow Xn converge vers X en loi.
    (Xn converges to X in probability \Rightarrow Xn converges to X in distribution)


    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.


    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    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 !

    X_n \to X in distribution means that F_{X_n}(x) \to F_X(x), for any x where F_X(x) is continuous. F denotes the cumulative distribution function.
    This means that \lim_{n \to \infty} \mathbb{P}(X_n \leqslant x)=\mathbb{P}(X \leqslant x)
    This can be translated as :
    \forall \epsilon, ~ \exists N_1,~ \forall n \geqslant N_1,~ |\mathbb{P}(X_n \leqslant x)-\mathbb{P}(X \leqslant x)|<\epsilon/2

    Y_n \to 0 in probability means that \forall \delta >0,~\lim_{n \to \infty} \mathbb{P}(|Y_n-0|> \delta)=\lim_{n \to \infty} \mathbb{P}(|Y_n|> \delta)=0.
    This can be translated as :
    \forall \epsilon,~ \exists N_2,~ \forall n \geqslant N_2,~ \mathbb{P}(|Y_n|> \delta)<\epsilon/2


    Now you want to show that \boxed{\lim_{n \to \infty} \mathbb{P}(X_n+Y_n \leqslant x)=\mathbb{P}(X \leqslant x)}

    Write that \mathbb{P}(X_n+Y_n \leqslant x)=\mathbb{P}(X_n+Y_n \leqslant x ~,~ |Y_n|>\delta )+\mathbb{P}(X_n+Y_n \leqslant x ~,~ |Y_n|<\delta ), for any given \delta >0

    Since \{|Y_n|>\delta\} \supset \{X_n+Y_n \leqslant x\ ~,~ |Y_n|>\delta \} (the comma represents an intersection of events), we can say that
    \mathbb{P}(X_n+Y_n \leqslant x ~,~ |Y_n|>\delta ) \leqslant \mathbb{P}(|Y_n|> \delta)
    Similarly (this will need a little bit of thinking), \{X_n \leqslant x+\delta\} \supset \{X_n+Y_n \leqslant x ~,~ |Y_n|<\delta \}


    Hence we get :
    \mathbb{P}(X_n+Y_n \leqslant x) \leqslant \mathbb{P}(|Y_n|> \delta)+\mathbb{P}(X_n \leqslant x+\delta)



    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 \mathbb{P}(X_n+Y_n \leqslant x) \leqslant F_X(x)+\epsilon

    And then make similar steps to get an inequality in the form \mathbb{P}(X_n+Y_n \leqslant x) \geqslant F_X(x)-\epsilon

    And this would mean that \mathbb{P}(X_n+Y_n \leqslant x) \longrightarrow F_X(x), which is what you want.
    Last edited by Jhevon; March 7th 2009 at 07:26 AM.
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  3. #3
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    Quote Originally Posted by Moo View Post
    Hello,

    Convergence in probability implies convergence in distribution. From here, your problem is obvious
    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 X is equivalent to pointwise convergence of the characteristic functions to \Phi_X). It is shorter but less elementary due to this theorem.
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  4. #4
    Moo
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    Quote Originally Posted by Laurent View Post
    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 X is equivalent to pointwise convergence of the characteristic functions to \Phi_X). 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
    Last edited by Moo; March 7th 2009 at 07:43 AM.
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  5. #5
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    Quote Originally Posted by Moo View Post
    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
    There is not really a problem ; for the lower bound, you can write

    P(X_n+Y_n\leq x)\geq P(X_n\leq x-\delta,|Y_n|\leq\delta) =P(X_n\leq x-\delta)-P(X_n\leq x-\delta, |Y_n|>\delta)\geq P(X_n\leq x-\delta)-P(|Y_n|>\delta).

    Then for both the lower and the upper bound, let \varepsilon>0 and use the fact that F_X is continuous at x (we prove convergence only at those points x) to choose \delta such that |F_X(x)-F_X(x\pm\delta)|\leq\varepsilon. In addition, for n large enough, F_{X_n}(x\pm\delta) is close from F_X(x\pm\delta) (it requires that F_X is continuous at x\pm\delta to say that, so we should choose \delta accordingly; there only countably many bad choices so it's OK), and P(|Y_n|>\delta) is small. That should do.
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  6. #6
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    Thanks a lot for the help, I think I get how to set the inequalities and take the limits.
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  7. #7
    Moo
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    Hello again,

    So we dealt with that generalized problem (when Y_n \to c, 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
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  8. #8
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    Thank you very much Moo. Actually, my professor handed me the solution for this homework problem. Your hints and suggestions were really helpful .
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