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Finding E(x^k) for gamma, beta, and lognormal distributions

  1. Nov 5, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the expected value of g(X) = Xk for the
    a. gamma distribution
    b. beta distribution
    c. lognormal distribution

    2. Relevant equations

    gamma distribution: [itex] \frac{x^{\alpha -1}e^{ \frac{-x}{\beta}}}{ \Gamma (\alpha ) \beta ^{\alpha }} [/itex] with paramenter x>0

    beta distribution: [itex] \frac{ \Gamma (\alpha + \beta) x^{\alpha -1}(1-x)^{\beta -1}}{\Gamma (\alpha ) \Gamma (\beta )} [/itex] with paramenter 0<x<1

    lognormal distribution: [itex] ( \frac{1}{x \sigma \sqrt{2 \pi }}) e^{ \frac{-(lnx-\mu )^2}{2 \sigma y^2}} [/itex] with parameter x>0

    [itex] E(X) = \int_{-\infty}^{\infty} \! xf(x) \mathrm{d} x [/itex]

    3. The attempt at a solution

    Well I can set up the integrals, but that gets me nowhere.

    a. [itex] \int_0^{\infty} \! x^k \frac{x^{\alpha -1}e^{ \frac{-x}{\beta}}}{ \Gamma (\alpha ) \beta ^{\alpha }} \mathrm{d} x [/itex]

    b. [itex] \int_0^1 \! x^k \frac{ \Gamma (\alpha + \beta) x^{\alpha -1}(1-x)^{\beta -1}}{\Gamma (\alpha ) \Gamma (\beta )} \mathrm{d} x [/itex]

    c. [itex] \int_0^{\infty} \! x^k ( \frac{1}{x \sigma \sqrt{2 \pi }}) e^{ \frac{-(lnx-\mu )^2}{2 \sigma y^2}} \mathrm{d} x [/itex]
     
  2. jcsd
  3. Nov 5, 2011 #2

    micromass

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    Let's do this one first. You will have to use that

    [tex] \int_0^{\infty} \! \frac{x^{\alpha -1}e^{ \frac{-x}{\beta}}}{ \Gamma (\alpha ) \beta ^{\alpha }} \mathrm{d} x =1[/tex]

    For some alpha and beta. So what you do is

    [tex] \int_0^{\infty} \! x^k \frac{x^{\alpha -1}e^{ \frac{-x}{\beta}}}{ \Gamma (\alpha ) \beta ^{\alpha }} \mathrm{d} x = \int_0^{\infty} \! \frac{x^{k+\alpha -1}e^{ \frac{-x}{\beta}}}{ \Gamma (\alpha ) \beta ^{\alpha }} \mathrm{d} x[/tex]

    So your alpha in the above integtral will be k+alpha. Now try to adjust the integral in such a way so you can use the above integral. That is, try to have [itex]\Gamma(\alpha+k)[/itex]and [itex]\beta^{\alpha+k}[/itex] in the denominator.
     
  4. Nov 5, 2011 #3
    Okay, thanks. I'll try that.
     
  5. Nov 5, 2011 #4

    I like Serena

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    Hi ArcanaGold!

    In addition to what micro said, you may want to use the property of the gamma function that:
    [tex]\Gamma(z+1)=z \Gamma(z)[/tex]
     
  6. Nov 5, 2011 #5

    Ray Vickson

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    For problem (c), change variables to y = ln(x); that will give you something of the form E[exp(k*Y)] for normally-distributed Y. Do you recognize that quantity? (Hint: I'll bet you have seen it before, or it is in your textbook or course notes. If not, it should be.)

    RGV
     
  7. Nov 6, 2011 #6
    Okay, I got the first one, and according to the book I have the correct solution. Thanks a lot guys :)

    For the second one, (the second and third answers are not provided) I got :

    [tex] \frac{ \Gamma (\alpha ) \Gamma (\alpha + k)}{\Gamma (\alpha + \beta + k) \Gamma (\alpha + \beta )} [/tex]

    I used the same technique, I think that is right.

    Then I get stuck on the last one. Since it's x-1 instead of a variable, I'm not sure what substitution to make here.
     
  8. Nov 6, 2011 #7

    micromass

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    For the lognormal, first make the substitution t=log(x)...
     
  9. Nov 6, 2011 #8
    [tex]\int_0^{\infty} \! x^k ( \frac{1}{ \sigma \sqrt{2 \pi }}) e^{ \frac{-(t-\mu )^2}{2 \sigma y^2}} \mathrm{d} x [/tex]

    where t= ln(x), dt = 1/x

    How on earth am I going to integrate x^k?
     
  10. Nov 6, 2011 #9
    should I let some other variable, say u, =x^k?
    Can you do that, make two substitutions in one integral?
     
  11. Nov 6, 2011 #10

    Ray Vickson

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    My previous post told you exactly what you have to do. Of course, x = exp(t).

    RGV
     
  12. Nov 6, 2011 #11
    [tex] ( \frac{1}{ \sigma \sqrt{2 \pi }}) \int_a^{b} \! e^{tk} e^{ \frac{-(t-\mu )^2}{2 \sigma ^2}} \mathrm{d} t [/tex]

    the (new) current integral
     
    Last edited: Nov 6, 2011
  13. Nov 6, 2011 #12

    I like Serena

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    To get started on the exponent...

    [tex]tk - {(t-\mu)^2 \over 2 \sigma^2}[/tex][tex]
    = {2tk\sigma^2 - (t-\mu)^2 \over 2 \sigma^2}[/tex][tex]
    = {2tk\sigma^2 - t^2 + 2t\mu - \mu^2 \over 2 \sigma^2}[/tex][tex]
    = {- (t^2 - 2t(k\sigma^2 - \mu) + (k\sigma^2 - \mu)^2) + (k\sigma^2 - \mu)^2 - \mu^2 \over 2 \sigma^2}[/tex][tex]
    = {- (t - (k\sigma^2 - \mu))^2 + (k\sigma^2 - \mu)^2 - \mu^2 \over 2 \sigma^2}[/tex]

    Now substitute [itex]u=t - (k\sigma^2 - \mu)[/itex].
     
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