# Mean Deviation.

• Nov 2nd 2009, 06:41 AM
ardam
Mean Deviation.
I am doing some work with moment generating functions and have been given the question ''calculate the mean deviation'', this is defined to be E(lX-E(X)l).

My m.g.f is (1-t/lambda)^-1 and was generated from f(x)=(lambda)e^-(lambda)x.

I have no idea how to find MD from a m.g.f.

cheers
• Nov 2nd 2009, 01:52 PM
mr fantastic
Quote:

Originally Posted by ardam
I am doing some work with moment generating functions and have been given the question ''calculate the mean deviation'', this is defined to be E(lX-E(X)l).

My m.g.f is (1-t/lambda)^-1 and was generated from f(x)=(lambda)e^-(lambda)x.

I have no idea how to find MD from a m.g.f.

cheers

Here is a blunt approach:

Get $\displaystyle E(X) = \frac{1}{\lambda}$ from the mgf. Now use the pdf to calculate $\displaystyle E\left(\left |X - \frac{1}{\lambda} \right|\right)$.
• Nov 3rd 2009, 03:45 AM
ardam
Quote:

Originally Posted by mr fantastic
Here is a blunt approach:

Get $\displaystyle E(X) = \frac{1}{\lambda}$ from the mgf. Now use the pdf to calculate $\displaystyle E\left(\left |X - \frac{1}{\lambda} \right|\right)$.

Got $\displaystyle E(X) = \frac{1}{\lambda}$ by differentiating m.g.f once then setting t=0.

So do i manipulate the expectation of a random variable formula to look like:
$\displaystyle E\left(\left |X - \frac{1}{\lambda} \right|\right)$ = intergral |X - (1\lambda)| (lambda)e^-(lambda)x. dx?
• Nov 3rd 2009, 09:43 PM
mr fantastic
Quote:

Originally Posted by ardam
Got $\displaystyle E(X) = \frac{1}{\lambda}$ by differentiating m.g.f once then setting t=0.

So do i manipulate the expectation of a random variable formula to look like:
$\displaystyle E\left(\left |X - \frac{1}{\lambda} \right|\right)$ = intergral |X - (1\lambda)| (lambda)e^-(lambda)x. dx?

Yes. And note that $\displaystyle \left |X - \frac{1}{\lambda} \right| = X - \frac{1}{\lambda}$ when $\displaystyle X > \frac{1}{\lambda}$ and $\displaystyle \left |X - \frac{1}{\lambda} \right| = \frac{1}{\lambda} - X$ when $\displaystyle X < \frac{1}{\lambda}$.