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Math Help - Distribution problem

  1. #1
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    Distribution problem

    Let X1, X2,...,Xn be n mutually independent random variables, each of which is uniformly distributed on the integers from 1 to k. Let Y denote the minimum of the Xi's. Find the distribution of Y.
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  2. #2
    MHF Contributor matheagle's Avatar
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    The continuous case is easier, but the same logic works here.

    P(Y=1)=P({\rm at \; least \; one\;} X_i=1)=1-P({\rm all\;} X_i>1)=1-\biggl({k-1\over k}\biggr)^n

    Then it gets messy.

    P(Y=2)=P({\rm at \; least \; one\;} X_i=2\; {\rm and \; none \; are \; equal \; to \; one})

    Here you have to think about all the possible strings of length n of numbers 1 through k.

    All k^n sequences are equally likely.

    To have exactly one 2 (and no 1's) that's  n\biggl({1\over k}\biggr)\biggl({k-2\over k}\biggr)^{n-1}

    To have exactly two 2's (and no 1's) that's  {n\choose 2}\biggl({1\over k}\biggr)^2\biggl({k-2\over k}\biggr)^{n-2}

    Then exactly three 2's....

    The sum of all of these will give you P(Y=2).

    The last is easy.

    P(Y=k)=P(X_1=k, \ldots ,X_n=k) =P(X_1=k) \cdots P(X_n=k)=\biggl({1\over k}\biggr)^n
    Last edited by matheagle; August 2nd 2009 at 09:13 PM.
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  3. #3
    Moo
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    Hello,

    Here is the good ol'trick of the "cumulative" probability ! (better than baldeagle's method )

    For any j, integer between 1 and k :

    \mathbb{P}(Y\geqslant j)=\mathbb{P}(\forall i=1..n ~,~ X_i\geqslant j)=\left(\frac{k-j+1}{k}\right)^n

    Similarly, for any j, integer between 1 and n-1, we have :

    \mathbb{P}(Y\geqslant j+1)=\left(\frac{k-j}{k}\right)^n

    If j=k, this probability equals 0. So this formula works for any j, integer between 1 and k.


    Since Y has integer values, we can easily see that \mathbb{P}(Y=j)=\mathbb{P}(Y\geqslant j)-\mathbb{P}(Y\geqslant j+1)


    So finally,

    \boxed{\mathbb{P}(Y=j)=\left(\frac{k-j+1}{k}\right)^n-\left(\frac{k-j}{k}\right)^n}
    Last edited by Moo; August 3rd 2009 at 03:28 AM. Reason: problems with letters...
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