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Regarding probability bound of flip coins

  1. Aug 7, 2013 #1
    Suppose you flip a fair coin 10,000 time how can you characterize the distribution of the occurrence of head?

    From the textbook, it says that P[head>n/2 + k√n] < e^(-k^2)/2, why is that and what is the derivation? What theorem is this, we had only learn Bernoulli distribution and Chebyshev so far, it seem odd that the textbook would jump to such a conclusion without rigorous proof.

    Thanks in advance
     
  2. jcsd
  3. Aug 7, 2013 #2

    mathman

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    I haven't worked on it, but it looks like an approximation for the tail of the Bernoulli distribution for a large sample.
     
  4. Aug 7, 2013 #3
    thanks for the reply, could you give me a hint for how would you go about deriving this
     
  5. Aug 8, 2013 #4

    mathman

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    The first step is to approximate the tail by an integral.
     
  6. Aug 8, 2013 #5

    Stephen Tashi

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    I haven't worked on the problem either. I agree that the result is not a simple consequence of the bernoulli distribution and the Chebyshev inequality, but you might be able to prove it from them with some work.

    The mean of N independent tosses of a fair coin (landing "0" or "1") is the sum of the mean of the results of the individual tosses and the variance is the sum of the individual variances. So you have a mean of N/2 and variance of N(1/2)(1-1/2). To prove the result from the Chebyshev inequality, you'd need to work out an inequality relating 1/x^2 and e^(-x^2).
     
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