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Thread: Convergence in Distribution

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    Convergence in Distribution




    If X_n is a sequence of random variables independent and X_n is Binomial(n,p) with 0<p<1 and either T_n=\frac{X_n}{n} for all n

    1) Show that \sqrt{n}(T_n-p)\rightarrow^DN(0,p(1-p))

    I know that
    \frac{\sqrt{np(1-p)}}{\sqrt{np(1-p)}}\sqrt{n}(T_n-p)=\sqrt{np(1-p)}N(0,1)=N(0,np(1-p))

    But I can not get this result at all, I've tried using characteristic function but also yielded nothing.

    Please take a look here too http://www.talkstats.com/showthread....-Limit-Theorem
    Last edited by askazy; 12-07-2014 at 08:41 AM.

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    Re: Convergence in Distribution

    I'm a little confused about the sequence of random variables

    If X_n is a sequence of independent random variables, that means that E[X_n]=E[X] or E[X_n]=\sum_{i=1}^nE[X]?

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    Re: Convergence in Distribution

    A sequence of random variables is nothing special; as you can have a sequence of numbers, a sequence of functions etc. for different mathematical objects.

    In this question you have typed that all random variables in this sequence X_n have the common distribution (and they are independent). And this is often abbreviated as i.i.d. sequence of random variables. As they have the common distribution, they share the common mean and all other moments / properties as well.

    Of course for the sequence T_n is not the same.

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    Re: Convergence in Distribution


    Quote Originally Posted by BGM View Post
    A sequence of random variables is nothing special; as you can have a sequence of numbers, a sequence of functions etc. for different mathematical objects.

    In this question you have typed that all random variables in this sequence X_n have the common distribution (and they are independent). And this is often abbreviated as i.i.d. sequence of random variables. As they have the common distribution, they share the common mean and all other moments / properties as well.

    Of course for the sequence T_n is not the same.
    Thank you, solved.

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