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Math Help - random variable convergence, need help please

  1. #1
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    Exclamation random variable convergence, need help please

    Wn has density [1+(x/n)]/[1+(0.5n)] for 0<x<1, and 0 otherwise. The random variable W has a uniform distribution on [0,1]. Prove that {Wn} converges in distribution to W.

    I have so far that
    mWn(s)=E(e^(sWn))= integral (0 to 1): e^(sWn)*((1+(x/n)))/(1+(1/2n))dWn
    = ((1+(x/n)))/(1+(1/2n)) * integral (0 to 1): e^(sWn)dWn
    which equals (as lim n-> inf.)
    integral (0 to 1): e^(Sw)dw


    mW(s)=E(e^(Sw))=integral (0 to 1): e^(Sw)dw

    But now how do I show the convergence?

    Any help is appriciated.
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  2. #2
    MHF Contributor matheagle's Avatar
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    IT's hard to follow.
    It would be nice if you typed in tex, that's why I couldn't understand your last post

    BUT [1+(x/n)]/[1+(0.5n)] isn't a density, one of the n's are wrong.

    And forget MGFs, just get the cummulative distribution and show they converge.
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  3. #3
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    i doubled checked the question in my book, and its exactly what i typed.... maybe its some different kind of density function?

    Anyways i got the c.d.f. of both random variables to match, but is that all i do for this?
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  4. #4
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    Quote Originally Posted by Sneaky View Post
    i doubled checked the question in my book, and its exactly what i typed.... maybe its some different kind of density function?

    Anyways i got the c.d.f. of both random variables to match, but is that all i do for this?
    If I were a betting man I would put money on it actually saying the equivalent of:

    [1+(x/n)]/[1+1/(2n)] for 0<x<1, and 0 otherwise

    CB
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  5. #5
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    what?
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  6. #6
    MHF Contributor matheagle's Avatar
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    one of the n's are wrong.

    \int_0^1[1+(x/n)]dx=[1+{1\over 2n}]
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  7. #7
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    Quote Originally Posted by Sneaky View Post
    what?
    That makes it a density and consistent with what you actually use in your third line.

    CB
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  8. #8
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    Quote Originally Posted by matheagle View Post
    one of the n's are wrong.

    \int_0^1[1+(x/n)]dx=[1+{1\over 2n}]
    so if i am understanding this right, for the cdf of Wn, as n approaches infinity, the cdf will become 1. But for the cdf of W, its just

    0 if w<=0
    w 0<w<1
    1 w>= 1

    I don't fully understand why Wn converges in distribution to W.
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  9. #9
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    Quote Originally Posted by Sneaky View Post
    Wn has density [1+(x/n)]/[1+(0.5n)] for 0<x<1, and 0 otherwise. The random variable W has a uniform distribution on [0,1]. Prove that {Wn} converges in distribution to W.

    I have so far that
    mWn(s)=E(e^(sWn))= integral (0 to 1): e^(sWn)*((1+(x/n)))/(1+(1/2n))dWn
    = ((1+(x/n)))/(1+(1/2n)) * integral (0 to 1): e^(sWn)dWn
    which equals (as lim n-> inf.)
    integral (0 to 1): e^(Sw)dw


    mW(s)=E(e^(Sw))=integral (0 to 1): e^(Sw)dw

    But now how do I show the convergence?

    Any help is appriciated.
    For x\le 1 put

    \displaystyle F_n(x)=\int_{\xi=0}^x \dfrac{1+\frac{\xi}{n}}{1+\frac{1}{2n}}\ d\xi

    \displaystyle F_(x)=\int_{\xi=0}^x 1 \ d\xi

    You need to show that:

    F_n(x) \to F(x)

    CB
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  10. #10
    MHF Contributor matheagle's Avatar
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    just integrate, calc 1, and show that the limit of Fn approaches x for 0<x<1
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  11. #11
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    after integration i get

    x^2 + 2nx
    ----------- -> x
    2n +1

    I'm not sure what to do to the left hand side, since subbing n as infinity here will not work.

    EDIT: i factored out x, then the 2n+1 cancelled out, and was just left with x, so x -> x, so yeah i think i got it
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  12. #12
    MHF Contributor matheagle's Avatar
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    this is basic calc 1

     {x+{x^2\over 2n}\over 1+{1\over 2n}} =x\left({1+{x\over 2n}\over 1+{1\over 2n}}\right)\to x
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  13. #13
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    i'm not sure where you got those equations
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  14. #14
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    Quote Originally Posted by Sneaky View Post
    i'm not sure where you got those equations
    1. Quote what you are referring to, saves us guessing.

    2. Presumably you have done calculus and so are familiar with convergence of sequences?

    CB
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