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Homework Help: Statistics pmf

  1. Feb 27, 2006 #1
    Hi,
    I know this isn't hard, but I have a block on it.
    Here's the question:
    Let [tex]f(x)=1/6, x=1,2,3,4,5,6[/tex] zero elsewhere be the pmf of a distribution of the discrete type. Show that the pmf of the smallest observation of a random sample of size 5 from this distribution is:
    [tex]g_1(y_1)=\left(\frac{7-y_1}{6}\right)^5-\left(\frac{6-y_1}{6}\right)^5,y_1=1,2,3,4,5,6[/tex]

    I know I'm looking for the pmf of the minimum, which is the y_1, and I did this:
    [tex]P(Y_1=y_1)=P(Y_1\leq y_1)-P(Y_1<y_1)[/tex]
    I'm having trouble seeing why there's a 7 in the solution. I'm missing something.
    I know that wheneve I get the probabilities of each piece, they will be to the 5th power because there's 5 observations in the sample and we multiply the probabilities together. What the probabilities are is the part that has me confused.
    We've been focusing on continuous problems for so long I can't wrap my mind around this.
    Any hints or pointers will be appreciated.
     
  2. jcsd
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