What is the physical or statistical meaning of this integral

[jam_27]

What is the physical or statistical meaning of the following integral

\int^{a}_{o} g(\vartheta) d(\vartheta) = \int^{\infty}_{a} g(\vartheta) d(\vartheta)

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

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

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

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

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

Cheers

Jamy

 

[tiny-tim]

\int^{a}_{o} g(\vartheta) d(\vartheta) = \int^{\infty}_{a} g(\vartheta) d(\vartheta)

But is there a statistical meaning to this integral?

Hi Jamy!

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 )

 

[jam_27]

Hi Jamy!

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 )

Thanks a ton for the reply. Could you please provide a reference/book. I want to see how its 50% likely.
Cheers
Jamy

 

[0xDEADBEEF]

Well I suppose the domain is on positive values. For it to be a probability distribution its norm should be unity.

\int^{\infty}_{o} g(\vartheta) d(\vartheta) = 1

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