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What is the physical or statistical meaning of this integral

  1. Mar 8, 2009 #1
    What is the physical or statistical meaning of the following integral

    [tex]\int^{a}_{o} g(\vartheta) d(\vartheta)[/tex] = [tex]\int^{\infty}_{a} g(\vartheta) d(\vartheta)[/tex]

    where [tex]g(\vartheta)[/tex] is a Gaussian in [tex]\vartheta[/tex] describing the transition frequency fluctuation in a gaseous system (assume two-level and inhomogeneous) .

    [tex]\vartheta = \omega_{0} -\omega[/tex], where [tex]\omega_{0}[/tex] is the peak frequency and [tex]\omega[/tex] the running frequency.

    I can see that the integral finds a point [tex]\vartheta = a[/tex] for which the area under the curve (the Gaussian) between 0 to a and a to [tex]\infty[/tex] are equal.

    But is there a statistical meaning to this integral? Does it find something like the most-probable value [tex]\vartheta = a[/tex]? But the most probable value should be [tex]\vartheta = 0[/tex] in my understanding! So what does the point [tex]\vartheta = a[/tex] tell us?

    I will be grateful if somebody can explain this and/or direct me to a reference.


  2. jcsd
  3. Mar 8, 2009 #2


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    Hi Jamy! :smile:

    It's the 50% likelihood range …

    50% likely that the result will be between ±a

    (and it works for any symmetric probability distribution, not just Gaussian :wink:)
  4. Mar 8, 2009 #3
    Thanks a ton for the reply. Could you please provide a reference/book. I want to see how its 50% likely.
  5. Mar 8, 2009 #4
    Well I suppose the domain is on positive values. For it to be a probability distribution its norm should be unity.

    [tex] \int^{\infty}_{o} g(\vartheta) d(\vartheta) = 1 [/tex]

    The 50% follows from 11th grade math, by the rules of how to add integrals over different intervals.
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