# Homework Help: Finding E(x^k) for gamma, beta, and lognormal distributions

1. Nov 5, 2011

### ArcanaNoir

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: $\frac{x^{\alpha -1}e^{ \frac{-x}{\beta}}}{ \Gamma (\alpha ) \beta ^{\alpha }}$ with paramenter x>0

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

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

$E(X) = \int_{-\infty}^{\infty} \! xf(x) \mathrm{d} x$

3. The attempt at a solution

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

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

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

c. $\int_0^{\infty} \! x^k ( \frac{1}{x \sigma \sqrt{2 \pi }}) e^{ \frac{-(lnx-\mu )^2}{2 \sigma y^2}} \mathrm{d} x$

2. Nov 5, 2011

### micromass

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

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

For some alpha and beta. So what you do is

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

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 $\Gamma(\alpha+k)$and $\beta^{\alpha+k}$ in the denominator.

3. Nov 5, 2011

### ArcanaNoir

Okay, thanks. I'll try that.

4. Nov 5, 2011

### I like Serena

Hi ArcanaGold!

In addition to what micro said, you may want to use the property of the gamma function that:
$$\Gamma(z+1)=z \Gamma(z)$$

5. Nov 5, 2011

### Ray Vickson

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. Nov 6, 2011

### ArcanaNoir

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 :

$$\frac{ \Gamma (\alpha ) \Gamma (\alpha + k)}{\Gamma (\alpha + \beta + k) \Gamma (\alpha + \beta )}$$

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. Nov 6, 2011

### micromass

For the lognormal, first make the substitution t=log(x)...

8. Nov 6, 2011

### ArcanaNoir

$$\int_0^{\infty} \! x^k ( \frac{1}{ \sigma \sqrt{2 \pi }}) e^{ \frac{-(t-\mu )^2}{2 \sigma y^2}} \mathrm{d} x$$

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

How on earth am I going to integrate x^k?

9. Nov 6, 2011

### ArcanaNoir

should I let some other variable, say u, =x^k?
Can you do that, make two substitutions in one integral?

10. Nov 6, 2011

### Ray Vickson

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

RGV

11. Nov 6, 2011

### ArcanaNoir

$$( \frac{1}{ \sigma \sqrt{2 \pi }}) \int_a^{b} \! e^{tk} e^{ \frac{-(t-\mu )^2}{2 \sigma ^2}} \mathrm{d} t$$

the (new) current integral

Last edited: Nov 6, 2011
12. Nov 6, 2011

### I like Serena

To get started on the exponent...

$$tk - {(t-\mu)^2 \over 2 \sigma^2}$$$$= {2tk\sigma^2 - (t-\mu)^2 \over 2 \sigma^2}$$$$= {2tk\sigma^2 - t^2 + 2t\mu - \mu^2 \over 2 \sigma^2}$$$$= {- (t^2 - 2t(k\sigma^2 - \mu) + (k\sigma^2 - \mu)^2) + (k\sigma^2 - \mu)^2 - \mu^2 \over 2 \sigma^2}$$$$= {- (t - (k\sigma^2 - \mu))^2 + (k\sigma^2 - \mu)^2 - \mu^2 \over 2 \sigma^2}$$

Now substitute $u=t - (k\sigma^2 - \mu)$.