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Homework Help: Simplified C.D.F.

  1. Mar 16, 2009 #1
    Derive a simplified formula for the c.d.f., F(n) = (P [tex]\leq [/tex] n), using:

    [tex]\sum_{k=0}^{n}r^k=\frac{1-r^{n+1}}{1-r} [/tex]


    [tex]p(n)=\alpha (1-\alpha)^n [/tex]

    [tex]\alpha \sum_{k=0}^{n}r^k=\frac{1-(1-\alpha)^{n+1}}{1-(1-\alpha)} [/tex]

    How do I break this down to a simplified version?
     
    Last edited: Mar 17, 2009
  2. jcsd
  3. Mar 17, 2009 #2
    Alright, I may have figured this out hopefully...

    [tex]
    \alpha \sum_{k=0}^{n}r^k=\frac{1-(1-\alpha)^{n+1}}{1-(1-\alpha)}
    [/tex]

    [tex]
    \alpha \frac{1-(1-\alpha)^{n+1}}{1-(1-\alpha)}
    [/tex]


    [tex]
    \alpha \frac{-(1-\alpha)^{n+1}}{-1+\alpha}
    [/tex]

    Is this right?


    [tex]
    \frac{-(1-\alpha)^{n+1}}{-1}
    [/tex]


    [tex]
    (1-\alpha)^{n+1}
    [/tex]
     
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