Homework Help: Moment of Density Problem midterm in 2 hours helpppp

1. Oct 14, 2009

sfwweb

Moment of Density Problem.... midterm in 2 hours helpppp

1. The problem statement, all variables and given/known data
If f is a nonnegative function whose integral is equal to 1, then f defines a probability density; the kth moment of this distribution is defined to be the average value of x^k with respect to this density. Compute all moments of the density defined by f(x) = e^(-x) on the positive half-line.

2. Relevant equations
\begin{displaymath}M(\theta) = E[e^{X\theta}] = \int_{-\infty}^{\infty} e^{x\theta} f(x) dx. \end{displaymath}

The kth central moment of a random variable X is given by E[(X-E[X])k

3. The attempt at a solution
The answer is K!, but i don't know how to get there.

THANKS SO MUCH!

2. Oct 14, 2009

HallsofIvy

Re: Moment of Density Problem.... midterm in 2 hours helpppp

Since the pdf is defined as $e^{-x}$ for $0\le x< \infty$, the first moment is defined as
$$\int_0^\infty xe^{-x}dx$$
. Integrate that by parts, letting u= x, $dv= e^{-x}dx$. Then du= dx, $v= -e^{-x}$ and the integral becomes
[tex]xe^{-x}\|_0^\infty+ \int_0^\infty e^{-x}dx[/itex]
It should be easy to see that that first term is 0 at both 0 and $\infty$ and easy to do the other integral.

Then the second moment is given by
[tex]\int_0^\infty x^2e^{-x}dx[/itex]

Again, do that by parts taking $u= x^2$, $dv= e^{-x}dx$ so that u= 2xdx[/itex] and $v= -e^{-x}$. Now the integral becomes
[tex]x^2e^{-x}\|_0^\infty + 2\int_0^\inty xe^{-x}dx[/itex]
Again the first term is 0 and the integral is just the integral you did for the first moment!

Try the same thing for the third and maybe fourth moments. That should tell you how to prove that the kth moment is k! using induction.

3. Oct 14, 2009

sfwweb

Re: Moment of Density Problem.... midterm in 2 hours helpppp

Thank you so much!