Moment of a probability distribution

[dbb04]

when you calculate the Moment of the following equation

<br /> <br /> p(x)=\left\{\begin{array}{cc}2Axe^{-Ax^2},&amp;\mbox{ if }<br /> x\geq 0\\0, &amp; \mbox{ if } x&lt;0\end{array}\right.<br />

We get

<br /> Mn =2A \int_0^\infty x^{n+1}e^{-Ax^2}<br />

solving it by parts I am getting

<br /> Mn=(n+1)\int_0^\infty x^{n-1}e^{-Ax^2}<br />

but, apparently, the right solution is

<br /> Mn=n\int_0^\infty x^{n-1}e^{-Ax^2}<br />


What am I doing wrong? What is the proper way to solve it? Could you please do it step by step?

Thanks

 

[Tide]

It looks like you didn't differentate properly when you integrated by parts:

\frac {d x^n}{dx} = n x^{n-1}

 

[dbb04]

sorry, still not following you.
If we integrate by parts we have

<br /> Mn =2A \int_0^\infty x^{n+1}e^{-Ax^2}<br />

<br /> \int_0^\infty u\frac{dv}{dx} dx=uv-\int_0^\infty v\frac{du}{dx} dx<br />

where

<br /> u=x^{n+1} \ \ \ \ \ \frac{du}{dx}=(n+1)x^n<br />

and

<br /> \frac{dv}{dx}=e^{-Ax^2} \ \ \ \ \ v= -\frac{e^{-Ax^2}}{2xA}<br />

so

<br /> Mn=2A \ \{-x^{n+1} \ \frac{e^{-Ax^2}}{2xA} \ \ |_0^\infty +\int_0^\infty \frac{e^{-Ax^2}}{2xA}(n+1)x^n dx \ \}<br />

<br /> Mn=0 +(n+1)\int_0^\infty x^{n-1} \ e^{-Ax^2} dx<br />

So, How you get rid of the (n+1) term.
Thanks

 

[Tide]

This should get you where you want to go:

\int_0^{\infty}x^{n+1} e^{-Ax^2} dx = - \frac {1}{2A} <br /> \int_0^{\infty}x^n \frac {d}{dx}e^{-Ax^2}dx

 

[dbb04]

Thanks Tide,

appreciate your patience