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Verify and Explain Binomial R.V. Identities

  1. Apr 13, 2012 #1
    If X and Y are binomial random variables with respective parameters (n,p) and (n,1-p), verify and explain the following identities:
    a.) P{X<=i}= P{Y>=n-i};
    b.) P{X=k}= P{Y=n-k}

    Relevant Equations:
    P{X=i}=nCi *p^(i) *(1-p)^(n-i), where nCi is the combination of "i" picks given "n".

    Distribution function
    ##p\left\{ x\leq i\right\} =\sum _{k=0}^{i}\left( n_{k}\right) p^{k}\left( 1-p\right) ^{n-k}
    ##, where n_k is ## (_{k}^{n})
    ##
    Solution:
    I don't know, how to start the problem. Please, help.
     
  2. jcsd
  3. Apr 13, 2012 #2

    mathman

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    If you examine the terms of the binomial you will see that the terms of (n,p) and (n,1-p) are the same except they are in opposite order from each other..
     
  4. Apr 13, 2012 #3
    What do you mean by opposite order?
     
  5. Apr 14, 2012 #4
    I know that * p and 1-p * are complement of each other, but how does this help me?
     
  6. Apr 14, 2012 #5

    mathman

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    Term of index k for (n,p) (a) = term of index n-k for (n,1-p) (b). As a result the problems you are addressing involve terms from (n,p) being equal to the corresponding (opposite order) terms in (n,1-p).

    (a) {n!/[k!(n-k)!]}pk(1-p)n-k

    (b) {n!/[(n-k)!k!]}(1-p)n-kpk

    Notice that (a) and (b) are exactly the same.
     
  7. Apr 15, 2012 #6
    From what I posted originally:
    If X and Y are binomial random variables with respective parameters (n,p) and (n,1-p), verify and explain the following identities:
    a.) P{X<=i}= P{Y>=n-i};
    b.) P{X=k}= P{Y=n-k}

    I have proven part b.) and yes it's the same.
    How do I prove part a.)? Particularly, P{Y>=n-i}.

    I know that
    P{Y>=n-i}=1-P{Y=0}-... up until Y=**A number lesser than (n-i)-1**

    Please, help.
     
  8. Apr 15, 2012 #7

    mathman

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    For part a, the X probability is gotten by adding all the terms from the beginning to k, while the Y probability is gotten by adding all the terms from n-k to the end. Since these terms match term by term, the sums match.
     
  9. Apr 15, 2012 #8
    What do you mean by " the end"?
    I know that P{Y>=n-k} has to be described as
    1-P{Y=0}-...-P{Y=n-k-1}

    Could you develop further on what you said?
     
  10. Apr 16, 2012 #9

    mathman

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    P(Y≥n-k) = P(Y=n-k) + P(Y=n-k+1) + ... + P(Y=n).

    P(Y=n-k) = P(X=k), P(Y=n-k+1) = P(X=k-1), ...., P(Y=n) = P(X=0).

    Therefore P(Y≥n-k)=P(X≤k).
     
    Last edited: Apr 16, 2012
  11. Apr 17, 2012 #10
    I understand:
    But, I do not understand this part:
    I know by one of the identities that P(Y=n-k) = P(X=k), but I don't know from where do you get the rest.
     
  12. Apr 17, 2012 #11

    chiro

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    Use the property of the combination. In other words what are the properties of combinations in pascal's triangle? (What is nCk vs nC[n-k] and similar properties in relation to your question)?
     
  13. Apr 18, 2012 #12

    mathman

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    Replace k, term by term, by k-1, k-2, etc. until you get to 0.
     
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