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Gamma Distribution (Statistics)

  1. Apr 15, 2008 #1
    A random variable X is gamma distributed with a=3, b=2 (alpha/beta). Determine the probability that 1[tex]\leq[/tex]X[tex]\leq[/tex]2

    The first thing I did was plug a & b into the gamma distribution formula 1/(b^a [tex]\Gamma[/tex]a * x^(a-1) * e^(-x/b)

    Which I ended up with

    1/16(x^2 * e^(-x/2)

    Which I then though I'd take the integral from 1 to 2

    [tex]\int[/tex] 1/16(x^2 * e^(-x/2) dx = 1/16(-2x^2 * e^-.5 - 8x * e^-.5 - 16e^-.5) & then I evaluated that from 1 to 2. Which gave .06...

    The answer in the book gives (b-a)^4/80

    So where am I going wrong & am I just approaching this problem in totally the wrong way?
     
  2. jcsd
  3. Apr 16, 2008 #2
    Try this for Gamma density.

    [tex]f(x)= \frac {\beta^\alpha}{\Gamma(\alpha)}x^{\alpha -1} e^{-\beta x}[/tex]

    The books answer would be negative ???
     
    Last edited: Apr 16, 2008
  4. Apr 17, 2008 #3
    The book answer wouldn't be negative, it'd be ((2-3)^4)/80.
     
  5. Apr 17, 2008 #4
    Also tried the formula, it does not work, it gives a probability over 11, which clearly can't be. The only difference between the formula I posted & yours is you have e^-bx & I had e^(-x/b).
     
  6. Apr 17, 2008 #5

    exk

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    I am assuming montoyas7940 looked up the gamma pdf on wikipedia (as I did) and I think that parametrization is correct for your application.

    I get:

    [itex]\int_{1}^{2} 4x^{2}e^{-2x}dx[/itex]

    That doesn't evaluate to 1/80.

    Try this maybe: P(1<X<2) = P(X<2)+P(X<1)-1
     
  7. Apr 17, 2008 #6
    Oh, I see. I assumed an extra set of parentheses in the books answer. I used the gamma density from "Probability for Risk Management".
     
  8. Apr 17, 2008 #7
    Unless I fat fingered a key on my calculator,

    [itex]\int_{1}^{2} 4x^{2}e^{-2x}dx[/itex]

    is not .6 either.
     
  9. Apr 17, 2008 #8
    I read the answer book wrong... sorry. :) Its NOT (b-a)^4/80.

    The answer was approx. .0659.

    Someone asked about the fact Probability for Risk Management gave a different answer today, but I'm not sure what they eventually decided. Using the formula you give I'm still getting 11.14, but I do recall the professor saying Probability for Risk Managements define their alpha as our beta... what I mean is its flip-flopped from what our notation is in the text we use. Anyway, thanks for your time & the help.
     
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