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Binomial Distribution Probability

  1. Sep 29, 2009 #1
    Let X be a Binomial B([tex]\frac{1}{2}[/tex],n), where n=2m.

    Let a(m,k) = [tex]\frac{4^m}{(\stackrel{2m}{m})}P(X = m + k)[/tex].

    Show that [tex]lim_{m->\infty}(a(m,k))^2 = e^{-k^2}[/tex].

    So far, I've found that P(X = m+k) = [tex](\stackrel{2m}{m+k}) \frac{1}{4^m}[/tex]

    Then, a(m,k)=[tex]\frac{m!m!}{(m+k)!(m-k)!}.[/tex]

    But I have no idea how to show that the limit of [tex]a^2[/tex] will be equal to [tex]e^{-k^2}[/tex].

    I think my work up to that point seems okay. I got things to cancel out, so that is usually a good sign. Any hints? Thank you!
     
  2. jcsd
  3. Sep 29, 2009 #2
    I got what you got, and this is not a good sign. If we are correct, then look at the k=3 case.

    a(m,3) simplifies to [tex]\frac{m(m-1)(m-2)}{(m+3)(m+2)(m+1)}[/tex] and the limit is 1 as m approaches infinity.
     
  4. Sep 29, 2009 #3
    Well, we are looking for the limit of [tex]a^2[/tex]. Does that make a difference?

    Also, you said you got the same thing as me, but I'm a little confused how what you wrote (even though I know it's for the specific k=3 case) is the same as what I got. I'll look at it again to see if I can see the connection, but maybe what I got for a was wrong?
     
  5. Sep 30, 2009 #4
    [tex]a(m,3)=\frac{m!m!}{(m+3)!(m-3)!}=\frac{m!\cdot m(m-1)(m-2)(m-3)!}{(m+3)(m+2)(m+1)m!\cdot (m-3)!}
    =\frac{m(m-1)(m-2)}{(m+3)(m+2)(m+1)}[/tex]

    and

    [tex]\lim_{m\to\infty}a(m,3)^2=\left(\lim_{m\to\infty}a(m,3)\right)^2=1^2=1[/tex]



    I think you should check if the problem is stated correctly. I get the same thing as you did for a(m,k) every time I check.
     
  6. Sep 30, 2009 #5
    Yep, that's definitely what the problem says so, I'm not sure.

    What you've shown definitely seems right and I'm assuming it works for other cases when k is something else, as well. I'll check it out and ask my teacher about it if need be. Thanks!
     
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