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

  • Thread starter ArcanaNoir
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  • #1
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Homework Statement



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

Homework 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]

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]
 

Answers and Replies

  • #2
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Homework Statement



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

Homework 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]

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]
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.
 
  • #3
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4
Okay, thanks. I'll try that.
 
  • #4
I like Serena
Homework Helper
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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]
 
  • #5
Ray Vickson
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Okay, thanks. I'll try that.
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
 
  • #6
768
4
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.
 
  • #7
22,089
3,296
For the lognormal, first make the substitution t=log(x)...
 
  • #8
768
4
[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?
 
  • #9
768
4
should I let some other variable, say u, =x^k?
Can you do that, make two substitutions in one integral?
 
  • #10
Ray Vickson
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[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?
My previous post told you exactly what you have to do. Of course, x = exp(t).

RGV
 
  • #11
768
4
[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:
  • #12
I like Serena
Homework Helper
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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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