- #1
RJLiberator
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Homework Statement
The Lorentz Distribution is given by
[tex]p(x) = \frac{1}{pi}\frac{ϒ}{(x-a)^2+ϒ^2}[/tex]
for -∞≤x≤∞.
a) Sketch the distribution for a=0, ϒ=1, and compare the form to the Gaussian Distribution.
b) Calculate the first moment of the distribution assuming again that a =0 and ϒ =1.
c) Does the second moment exist?
Homework Equations
Definition of moment: https://en.wikipedia.org/wiki/Moment_(mathematics)
Cauchy (Lorentz) distribution: https://en.wikipedia.org/wiki/Cauchy_distribution#Mean
First moment is the mean
Second moment is the variance
The Attempt at a Solution
The first moment, being mean, is thus equated by
[tex]
\lim_{\substack{a\rightarrow ∞}}
\int_{-2a}^{a} xp(x) dx = \lim_{\substack{a\rightarrow ∞}}\int_{-2a}^{a} \frac{x}{pi}\frac{1}{x^2+1} dx = \frac{-log(4)} {2*pi}[/tex]By looking at part a (I have this easily graphed), the mean should be something like 0.3 where this value gave me -0.22. Am I off by a negative? Am I even doing this correctly?
Thank you.