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#1
04-20-2018, 03:52 PM
 Handynasty CAS SOA Join Date: Jan 2018 College: University of Virginia Posts: 4
Help with Gamma Random Variable Problem

Hi,

I run into this problem when I'm studying:

Suppose the continuous random variable X has the following pdf:

f(x) = (1/16)*(x^2)*e^(-x/2) if x > 0
0 otherwise

Find E(X^3).

Clearly the λ = 1/2 and α = 3, but I don't know how to solve it other than going through the complicated integration. Is there any good way to solve it?
#2
04-20-2018, 03:58 PM
 Abraham Weishaus Member SOA AAA Join Date: Oct 2001 Posts: 7,155

Recognize the integrand as a gamma integrand with lambda=1/2 and alpha=6, figure out what the correct constant for that gamma is, and compare with the current constant (1/16).
#3
04-20-2018, 04:25 PM
 Handynasty CAS SOA Join Date: Jan 2018 College: University of Virginia Posts: 4

I'm sorry Abraham, I don't quite follow.

I thought the lambda=1/2 and alpha=3. But even though, I still have to go through the tough integration by parts.
#4
04-20-2018, 05:58 PM
 Abraham Weishaus Member SOA AAA Join Date: Oct 2001 Posts: 7,155

You want to calculate the third moment. That means integrating x^3 f(x). Notice that x^3, when multiplied by x^2, becomes x^5. So you're integrating x^5 e^{-x/2} times some constant. That looks like a gamma density function, doesn't it? And gamma density functions got to integrate to 1. So just figure out what the constant in the density function is and voila - the integral of x^5 e^{-x/2} got to be the reciprocal of that constant!
#5
04-20-2018, 06:20 PM
 Michael Mastroianni SOA Join Date: Jan 2018 Posts: 8

Following Abraham’s method:

$\int_{0}^{\infty} x^5 \cdot e^{-x/2} \, dx= 2^{6}\cdot\Gamma(6)=2^{6}\cdot 5!$

Hence the answer would be $\frac{2^6\cdot 5!}{2^4}=4\cdot 5!=480$
__________________
Michael Mastroianni, ASA
Video Course for Exam 1/P: www.probabilityexam.com
#6
04-24-2018, 01:13 PM
 Handynasty CAS SOA Join Date: Jan 2018 College: University of Virginia Posts: 4

Thank you all!

 Tags gamma random variable, help!, p-exam