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Prove that the function given with f(x) is a PDF

  1. May 28, 2014 #1
    Hi,
    Could someone help me prove that the function given with
    ##f(x) = \binom{x-1}{r-1} p^{r}(1-p)^{x-r}## is a probability density function, where ## x= r, r+1,..., \infty ## and ## 0<p<1 ##


    I thought to solve it somehow by using the binomial theorem, but since it's the upper part that's changing on the binomial coefficients, this proved futile.
     
    Last edited: May 28, 2014
  2. jcsd
  3. May 28, 2014 #2
    Let's start simple:

    Can you prove that the value is always larger than zero?
     
  4. May 28, 2014 #3

    Fredrik

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    Use ## instead of $ for inline LaTeX.
     
  5. May 28, 2014 #4

    pasmith

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    You need to show that [tex]\sum_{x=r}^\infty \binom{x-1}{r-1} p^r (1-p)^{x-r} = 1.[/tex] Setting [itex]q = 1- p[/itex] and [itex]m = x - r[/itex] gives [tex]
    \sum_{x=r}^\infty \binom{x-1}{r-1} p^r (1-p)^{x-r}
    = (1 - q)^r \sum_{m=0}^\infty \frac{(m + r - 1)!}{(r-1)!m!}q^m.[/tex] What does the series on the right have to equal if the left hand side is to equal 1, and can you prove that it does?
     
  6. May 28, 2014 #5

    Ray Vickson

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    For discrete random variables, I prefer to use symbols like ##j## and ##k## instead of ##x##. So, in this notation: what properties must be possessed by
    [tex]f_k = {k-1 \choose r-1} p^r (1-p)^{k-r}, \, k = r, r+1, r+2, \ldots [/tex]
    in order that it be a legitimate probability mass function of a discrete random variable? Note that ##f_k## is NOT a probability density function; it is a probability mass function. There is a world of difference!
     
  7. May 28, 2014 #6
    Hi, thanks for the reply .
    It must equal 1. Though I have no idea how to prove it.
    Thanks, I had the terminology mixed up.
    To be a probability mass function , ##f_k## has to be positive, and the sum over all ## k ## has to be 1.
    It's easy to see that ##f_k## is positive, but I cannot seem to prove the second property
     
  8. May 29, 2014 #7

    pasmith

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    Remember, you need to show that
    [tex]
    (1 -q)^r \sum_{m=0}^\infty \frac{(m + r - 1)!}{(r-1)!m!}q^m = 1
    [/tex] for every [itex]0 < q < 1[/itex]. This suggests that the series [tex]\sum_{m=0}^\infty \frac{(m + r - 1)!}{(r-1)!m!}q^m[/tex] must be the binomial expansion of ... what function of [itex]q[/itex]?
     
  9. May 29, 2014 #8
    ## \frac{1}{(1 -q)^r } ##
    But how would I prove that ?
    ---
    Edit:
    I checked a wikipedia article about binomial series, and it turns out that the series [tex]\sum_{m=0}^\infty \frac{(m + r - 1)!}{(r-1)!m!}q^m[/tex] is of this form , where ##β= r-1## and ##z=q##

    Thanks for the help :)
    Though, is there any other way to prove it, without directly using the formula ?
     
    Last edited: May 29, 2014
  10. May 29, 2014 #9

    pasmith

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    Ultimately the only way to sum power series is to relate them to power series of known functions.

    There are other ways of showing that [itex]\sum_{m=0}^\infty \frac{(m + r - 1)!}{(r-1)!m!}q^m = (1 - q)^{-r}[/itex]. For example, using the formula [tex]
    \frac{d^k}{dx^k} \left( \sum_{n=0}^\infty a_n x^n \right)
    = \sum_{n=0}^\infty a_{n+k} \frac{(n + k)!}{n!} x^n[/tex] we see that [tex]
    \sum_{m=0}^\infty \frac{(m + r - 1)!}{(r-1)!m!}q^m = \frac{1}{(r - 1)!} \frac{d^{r-1}}{dq^{r-1}} \left( \sum_{m=0}^\infty q^m \right).[/tex] [itex]\sum_{m=0}^\infty q^m[/itex] is a geometric series and is easily summed.
     
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