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 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
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 /> \int_0^\infty x^2e^{-x}dx[/itex]&lt;br /&gt; &lt;br /&gt; 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&lt;br /&gt; x^2e^{-x}\|_0^\infty + 2\int_0^\inty xe^{-x}dx[/itex]&amp;lt;br /&amp;gt; Again the first term is 0 and the integral is just the integral you did for the first moment!&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; 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!