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Math Help - Stirling's formula manipulation

  1. #1
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    Stirling's formula manipulation

    Hi, I'm stuck on part of this question involving Stirling's formula. I have to show that for p+q=1 (they're probabilities),
    \sum_{i=1}^\infty {2n \choose n}p^n q^n = \infty if and only if p=q= \frac{1}{2}
    by using Stirling's approximation, which is n! \approx \sqrt{2\pi n}\, \left(\frac{n}{e}\right)^{n} , for large n.

    I know that I should be aiming to show that <br />
{2n \choose n}p^n q^n \approx \frac{(4pq)^n}{\sqrt{\pi n}}

    I've tried rearranging it but I can't seem to get rid of the terms involving e (among others).


    Also, anyone who can solve this is a clever clogs, so maybe they can shed some light on this one, which seems to have defeated the SOS Math community. S.O.S. Mathematics CyberBoard :: View topic - Predictors after a simple linear regression is done

    Thanks VERY much
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by mongrel73 View Post
    Hi, I'm stuck on part of this question involving Stirling's formula. I have to show that for p+q=1 (they're probabilities),
    \sum_{i=1}^\infty {2n \choose n}p^n q^n = \infty if and only if p=q= \frac{1}{2}
    by using Stirling's approximation, which is n! \approx \sqrt{2\pi n}\, \left(\frac{n}{e}\right)^{n} , for large n.

    I know that I should be aiming to show that <br />
{2n \choose n}p^n q^n \approx \frac{(4pq)^n}{\sqrt{\pi n}}

    I've tried rearranging it but I can't seem to get rid of the terms involving e (among others).


    Also, anyone who can solve this is a clever clogs, so maybe they can shed some light on this one, which seems to have defeated the SOS Math community. S.O.S. Mathematics CyberBoard :: View topic - Predictors after a simple linear regression is done

    Thanks VERY much
    I don't understand where your problem is

    {2n \choose n}=\frac{(2n)!}{n!n!}=\frac{(2n)!}{(n!)^2}

    Using Stirling's formula :

    (2n)! \sim \sqrt{4\pi n} \cdot \left(\frac{2n}{e}\right)^{2n}=2\sqrt{\pi n} \cdot 2^{2n} \cdot n^{2n} \cdot e^{-2n}

    (n!)^2 \sim \left[\sqrt{2\pi n} \cdot \left(\frac ne\right)^n\right]^2=2 \pi n \cdot \left(\frac ne\right)^{2n}=2\pi n \cdot n^{2n} \cdot e^{-2n}


    Hence :
    {2n \choose n}=\frac{{\color{red}2}\sqrt{\pi n} \cdot 2^{2n} \cdot {\color{red}n^{2n}} \cdot {\color{red}e^{-2n}}}{{\color{red}2}\pi n \cdot {\color{red}n^{2n}} \cdot {\color{red}e^{-2n}}}=\frac{2^{2n}}{\sqrt{\pi n}}


    And finally :

    {2n \choose n} p^n (1-p)^n \sim \frac{(4p(1-p))^n}{\sqrt{\pi n}}


    Now, p(1-p)=\frac 14 \Leftrightarrow p=\frac 12. Otherwise, p(1-p)<\frac 14, for any p \neq \frac 12 (and in [0,1] of course). This is VERY easy to prove, by taking the derivative of f ~:~ x \mapsto x(1-x).

    If p=\frac 12, then the series would be equivalent to \sum_{n\geq 1} \frac{1}{\sqrt{n}}, which is a divergent Riemann series.



    If p \neq \frac 12, then note that \frac{1}{\sqrt{n}} \leq 1, for any n \geq 1

    Thus \frac{(4p(1-p))^n}{\sqrt{\pi n}} \leq \frac{(4p(1-p))^n}{\sqrt{\pi}}
    Since p(1-p)<\frac 14, 4p(1-p)<1 and therefore, \sum_{n \geq 1} \frac{(4p(1-p))^n}{\sqrt{\pi n}} \leq \frac{1}{\sqrt{\pi}} \sum_{n \geq 1} (4p(1-p))^n is a convergent series.
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  3. #3
    MHF Contributor matheagle's Avatar
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    The only tricky part of this was trying to figure out what that subscript i meant.
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  4. #4
    Moo
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    Quote Originally Posted by matheagle View Post
    The only tricky part of this was trying to figure out what that subscript i meant.
    I guess you replied to the wrong thread...
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  5. #5
    MHF Contributor matheagle's Avatar
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    NOPE, look at the original sum, instead of using n, the OP used i.
    Last edited by matheagle; April 17th 2009 at 07:36 AM.
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  6. #6
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    Quote Originally Posted by matheagle View Post
    NOPE, look at the original sum, instead of using n, the imposter used i.
    Yes, oops. I also can't work out what my problem was. Thanks anyway though...
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