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Finding the mean of the gamma distribution

  1. Jan 22, 2012 #1
    Finding the mean of the gamma distribution ("reparameterise")

    1. The problem statement, all variables and given/known data
    Find the mean of the gamma distribution.
    [tex]f_{X}(x) = \frac{e^{-kx}x^{r-1}k^{r}}{(r-1)!} x>0, r ε N*, k>0[/tex]


    2. Relevant equations
    Interestingly this is the alternative form of the gamma distribution


    3. The attempt at a solution

    I didn't really know where to begin, so peeked at the answers. Apparently I have to reparameterise r→r+1. Can someone explain to me what this means? How does it differ from a substitution? Is there a way to solve it without reparameterising and how do I spot when to reparameterise?

    Thanks
    Thomas
     
    Last edited: Jan 22, 2012
  2. jcsd
  3. Jan 23, 2012 #2
    anybody?
     
  4. Jan 23, 2012 #3

    LCKurtz

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    Re: Finding the mean of the gamma distribution ("reparameterise")

    So$$
    E(X) = \int_0^\infty xf_X(x)\,dx=\int_0^\infty \frac{e^{-kx}x^{r}k^{r}}{(r-1)!}\, dx$$Try a change of variable ##u=kx## and remember the definition of ##\Gamma(r)## and that ##\Gamma(r)=(r-1)!##.
     
  5. Jan 23, 2012 #4
    Re: Finding the mean of the gamma distribution ("reparameterise")

    Here's a hint for an easy no-calculus solution. By definition, the mean is the integral of

    [tex]x f_{X}(x) = \frac{e^{-kx}x^{r}k^{r}}{(r-1)!} = \frac{e^{-kx}x^{r}k^{r+1}}{r!} \cdot \frac{r}{k} [/tex]

    Hmm... isn't that the pdf of a gamma distribution with parameters k and r+1, times r/k?

    And doesn't a pdf integrate to 1?
     
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