Moment of Density Problem midterm in 2 hours helpppp

[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 [itex]e^{-x}[/itex] for [itex]0\le x< \infty[/itex], the first moment is defined as
[tex]\int_0^\infty xe^{-x}dx[/tex]
. Integrate that by parts, letting u= x, [itex]dv= e^{-x}dx[/itex]. Then du= dx, [itex]v= -e^{-x}[/itex] 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 [itex]\infty[/itex] 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 [itex]u= x^2[/itex], [itex]dv= e^{-x}dx[/itex] so that u= 2xdx[/itex] and [itex]v= -e^{-x}[/itex]. 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.

 

[sfwweb]

Thank you so much!