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Bernoulli Trials

  1. Oct 19, 2011 #1

    jgens

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    1. The problem statement, all variables and given/known data

    Suppose [itex]0 < p < 1[/itex]. Define [itex]b(n,k) = \binom{n}{k}p^k(1-p)^k[/itex]. For what value of [itex]k[/itex] is [itex]b(n,k)[/itex] a maximum?

    2. Relevant equations

    N/A

    3. The attempt at a solution

    Is there any way to get a nice closed form solution to this problem? I've already proved that it has a maximum so there must be some j such that [itex]b(n,j)[/itex] is maximal. Then we know [itex]b(n,j+1) < b(n,j)[/itex] and [itex]b(n,j-1) < b(n,j)[/itex]. And I figure we can use these relations to figure out j or something like that. But I'm not sure if this approach will work.

    Could someone help me with this please?
     
  2. jcsd
  3. Oct 20, 2011 #2

    Ray Vickson

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    Develop an expression for r(j) = b(n,j+1)/b(n,j) [which, by the way, is also useful for computing the b(n,k) recursively]. What can you say if r(j) > 1? If r(j) < 1?

    RGV
     
  4. Oct 20, 2011 #3

    jgens

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    Well, if I let [itex]j[/itex] be such that [itex]b(n,j)[/itex] is maximal, then

    [tex]b(n,j-1) < b(n,j) \implies 1 < \frac{b(n,j)}{b(n,j-1)} \implies 1 < \frac{p}{1-p}\frac{\binom{n}{j}}{\binom{n}{j-1}}[/tex]

    [tex]b(n,j+1) < b(n,j) \implies \frac{b(n,j+1)}{b(n,j)} < 1 \implies \frac{p}{1-p} \frac{\binom{n}{j+1}}{\binom{n}{j}} < 1[/tex]

    But I'm still not sure where to go from here. Do you think you could give me another hint in the right direction?
     
  5. Oct 20, 2011 #4

    Ray Vickson

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    Sure. Work at simplifying the ratio.

    RGV
     
  6. Oct 20, 2011 #5

    jgens

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    Okay, so I think I've got it. Rather than TeX the whole thing out right now, is the answer [itex]j < p(n+1) < j+1[/itex] correct?
     
  7. Oct 20, 2011 #6

    Ray Vickson

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    OK, if you replace [itex] < [/itex] by [itex] \leq[/itex].

    RGV
     
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