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Show that Y ~ Gamma()

  1. Oct 6, 2015 #1
    1. The problem statement, all variables and given/known data

    Suppose X ~ Gamma(α, scale = β and that Y = kX with k > 0 a constant. Show that Y ~ Gamma(α, scale = kβ).

    2. Relevant equations

    Gamma distribution, etc.

    3. The attempt at a solution

    [itex] λ = \frac{1}{β}

    [/itex]

    [itex] f(x) = \frac{λ^{α}}{Γ(α)}x^{α-1}e^{-λx}

    = {(\frac{1}{β})^{α}}{\frac{1}{Γ(α)}}{(\frac{y}{k})}^{α-1}e^{-λ{(\frac{y}{k})}}

    = k{(\frac{1}{kβ})^{α}}{\frac{1}{Γ(α)}}{({y})}^{α-1}e^{-{(\frac{y}{βk})}}

    [/itex]

    What's with the extra k?
     
  2. jcsd
  3. Oct 6, 2015 #2

    Ray Vickson

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    It comes from ##dy = k dx##, so that ##\int f_X(x) \, dx = \int f_X(y/x) \, dy/x##.
    Thus, ##f_Y(y) = (1/k) f_X(y/k) ##.
    First, tell us what YOU think.
     
  4. Oct 6, 2015 #3
    Eh. It's related to the scale factor and property of the gamma function. In the first line, shouldn't it be y/k and dy/k?
     
    Last edited: Oct 6, 2015
  5. Oct 6, 2015 #4

    Ray Vickson

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    Yes.

    Unfortunately, when I was composing the response, I first gave that detailed answer, then changed my mind and went instead with the "non-answer" that would leave the work to you. I thought I had deleted the detailed answer, but the PF editor is tricky: if one is not careful, material one thinks has been deleted turns out to not have been. That is what happened here! I told you more than I wanted to.
     
  6. Oct 6, 2015 #5
    Hm. Are they fixing the bug? Is my addition below correct?

    dy=kdx, so that ∫ fX(x)dx = ∫ fX(y/k)dy/k = ∫ fY(y)dy.

    I assume that this is the CDF: FY(y)=(1/k)FX(y/k).[/sub]
     
    Last edited: Oct 6, 2015
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