Computing moments of exponential density f(x) = e^(-x)

[sfwweb]

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

Homework Statement

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.

Homework 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

The Attempt at a Solution

The answer is K!, but i don't know how to get there.


THANKS SO MUCH!

 

[HallsofIvy]

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]<br /> It should be easy to see that that first term is 0 at both 0 and $\infty$ and easy to do the other integral.<br /> <br /> Then the second moment is given by<br /> [tex]\int_0^\infty x^2e^{-x}dx[/itex]<br /> <br /> 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<br /> [tex]x^2e^{-x}\|_0^\infty + 2\int_0^\inty xe^{-x}dx[/itex]<br /> Again the first term is 0 and the integral is just the integral you did for the first moment!<br /> <br /> 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.[/tex][/tex][/tex]

 

[sfwweb]

Thank you so much!