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Deriving the MGF for the Weibull Distribution

by donald17
Tags: deriving, distribution, weibull
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donald17
#1
Apr12-11, 02:55 PM
P: 6
I'm attempting to derive the MGF for the Weibull distribution. I know that E([tex]e^{tx}[/tex]), which equals the integral shown here:

http://www.wolframalpha.com/input/?i=Integrate%5Be^%28t*x%29*%28k%2F%CE%BB%29*%28x%2F%CE%BB%29^%28k-1%29*e^-%28x%2F%CE%BB%29^k%2Cx%5D

where the parameters are k and λ.

The answer is found here:

http://www.wolframalpha.com/input/?i=Sum%5B%28t^n+%CE%BB^n%29%2Fn%21%2C+{n%2C+0%2C+Infinity}%5D*gamma%281 %2B%281%2Fk%29%29

So I see that I need to get the gamma function and the series representation for e^(t*λ) to show up in order to get the right answer. I've been trying to use a change of variable such as u = (x/λ)^k, and I feel like I've been getting close, but can't exactly get it right. Can someone guide me along with this? Thank you.

*For some reason it keeps putting a space in the URL, so just take them out
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Stephen Tashi
#2
Apr12-11, 07:12 PM
Sci Advisor
P: 3,300
Those links you gave might be temporary URLs. I didn't get the first one to work.
donald17
#3
Apr12-11, 08:16 PM
P: 6
In the first url try copy and pasting the whole thing, but taking out the space between the 2 and the F

Similarly, for the second url take out the space between the I and the nfinity. If this doesn't work let me know and I'll attempt to repost what I'm trying to show in another format.


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