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Math Help - Simple Binominal Distribution Question

  1. #1
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    Simple Binominal Distribution Question

    Hi all,

    Simple question that I've been staring at for too long to think about sensibly.

    Question is:

    Given X \sim Bi(n, \theta ), Y \sim Bi(n, 1- \theta ), prove that P(X\leq x)=P(Y\geq n-x).

    Now obviously can easily prove using formula for binomial distribution that P(X=x)=P(Y=n-x), and on up the sequence to P(X=0)=P(Y=n). But I don't know how to prove in general that the sums of the two series are such that P(X\leq x)=P(Y\geq n-x).

    If I write the two sequences as sums I get,

    P(X\leq x)=\sum ^x_{x=0} \left(\begin{array}{cc}n\\x\end{array}\right) \theta^x (1- \theta)^{n-x}

    P(Y\geq n-x)=\sum ^n_{y=n-x} \left(\begin{array}{cc}n\\y\end{array}\right) \theta^{n-y} (1- \theta)^{y}

    Am I on the right track...?
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  2. #2
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    Re: Simple Binominal Distribution Question

    Quote Originally Posted by TimM View Post
    Hi all,

    Simple question that I've been staring at for too long to think about sensibly.

    Question is:

    Given X \sim Bi(n, \theta ), Y \sim Bi(n, 1- \theta ), prove that P(X\leq x)=P(Y\geq n-x).

    Now obviously can easily prove using formula for binomial distribution that P(X=x)=P(Y=n-x), and on up the sequence to P(X=0)=P(Y=n). But I don't know how to prove in general that the sums of the two series are such that P(X\leq x)=P(Y\geq n-x).

    If I write the two sequences as sums I get,

    P(X\leq x)=\sum ^x_{x=0} \left(\begin{array}{cc}n\\x\end{array}\right) \theta^x (1- \theta)^{n-x}

    P(Y\geq n-x)=\sum ^n_{y=n-x} \left(\begin{array}{cc}n\\y\end{array}\right) \theta^{n-y} (1- \theta)^{y}
    In the first summation do not us x twice.
    P(X\leq x)=\sum\limits_{y = 0}^x {\binom{n}{y}\theta^y(1-\theta)^{n-y}}

    Now note that \sum\limits_{y = 0}^n {\binom{n}{y}\theta^y(1-\theta)^{n-y}}=1
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  3. #3
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    Re: Simple Binominal Distribution Question

    Plato: thanks.

    I understand that \sum^{n}_{x=0}p(x)=1.

    I also get that the sum of the two sequences must equal 1 because, due to their respective distributions, the two events, P(X\leq x) and P(Y\geq n-x) partition the sample space, so that P(X\leq x)+P(Y\geq n-x)=1.

    Since X is the probability of success and Y is the probability of failure of the same experiment, the probability that there are x successes in n Bernouli trials must equal the probability that there are n-x failures in the same n Bernouli trials.

    Probably I'm being a bit dense, but I don't understanding how to prove it.

    [I am a bit confused by the notation as well. If I change the notation per your suggestion, then aren't I expressing the sum P(X=0)+P(X=1)+...+P(X=y)? Does it make sense to write that?]
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  4. #4
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    Re: Simple Binominal Distribution Question

    Quote Originally Posted by TimM View Post
    I understand that \sum^{n}_{x=0}p(x)=1.

    I also get that the sum of the two sequences must equal 1 because, due to their respective distributions, the two events, P(X\leq x) and P(Y\geq n-x) partition the sample space, so that P(X\leq x)+P(Y\geq n-x)=1.
    I don't think you can show that.
    You do want P(X\leq x)=P(Y\geq n-x) do you not
    \sum\limits_{k = 0}^x \binom{n}{k}\theta^k(1-\theta)^{n-k}=\sum\limits_{k = n-x}^n \binom{n}{k}(1-\theta)^k \cdot\theta^{n-k}
    Now there will be some overlapping there.
    You know that \binom{N}{k}=\binom{N}{N-k}

    This is a post script.
    You have two distinct cases.
    If x<n-x the proof is straight forward.
    In the first sum when k=3 and k=N-3 the second sum then the two corresponding terms are equal.

    However, if N-x\le x you must take special care.
    You have overlapping terms is each sum.
    Last edited by Plato; September 7th 2011 at 04:09 PM.
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  5. #5
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    Re: Simple Binominal Distribution Question

    Plato: Thanks again for your help with this. IN work now, but will have a look at this and try to post my answer tonight.
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