Probability density of an exponential probability function

  • #1
eXorikos
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I have a model where the probability is spherically symmetric and follows an exponential law. Now I need the probability density function of this model. The problem is the singularity at the origin. How can I handle this?

P(r) = ∫p(r) dr = exp(-μr)
p(r) = dP(r)/(4πr²dr)

One way I tried to handle this is numerically in Matlab by having the probability at 0 such that the total probability is 1. The problem there is that this depends highly on the mesh you chose, because of the steepness of the pdf close to the origin.

Is there a mathematical way to handle this analytically?

Afterwards I need to combine this pdf with different gaussians in a convolution to get a combined probability map. Obviously I would love to extend this to non-isotropic in carthesian coordinates as a final step. Can this be done with homothety?
 

Answers and Replies

  • #2
mathman
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Could you clarify p(r). The expressions don't seem to agree.
 
  • #3
eXorikos
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p(r) is the pdf I want to calculate starting from P(r). What is wrong with the expressions?
 
  • #4
mjc123
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P(r) is not ∫p(r)dr. P(r)dr = ∫p(r)dV between r and r+dr, i.e. P(r)dr = 4πr2p(r)dr
i.e. P(r) = 4πr2p(r), or p(r) = P(r)/4πr2
Note that your P(r) is not normalised.
 
  • #5
eXorikos
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Thanks for the correction. I haven't done any real mathematics in years, so I'm sure I'm missing a lot.

1-P(r) is the cummulative probability for the sphere of radius r.

p(r) = P(r)/4πr2 and the singularity at r=0 is a problem. I need to convolve this pdf with a guassian pdf.
 
  • #6
mjc123
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Oh right, I misunderstood you. I assumed p(r) was the pdf (i.e. p(r)dV is the probability of being in a volume element dV) and P(r) was the radial probability function (i.e. P(r)dr is the probability of being between r and r+dr).
Now you say 1-P(r) is the cumulative probability of being within a sphere of radius r. In that case e-μr is acceptable, it doesn't need to be normalised.
Now what do you mean by p(r)? Is it the pdf, as defined above? Or is it the differential radial probability function, which I though you meant by P(r) before?
If the latter, then p(r) = -dP(r)/dr = μe-μr. No 4πr2 factor.
If the former, let's call it q(r), then q(r) = p(r)/4πr2
 
  • #7
eXorikos
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But if the pdf is: μe-μr/4πr2, then the integral r=0 to infinity is undefined, where it should be unity if it is a pdf.
 
  • #8
mjc123
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The pdf is not integrated from r = 0 to infinity; it is integrated over the whole volume. With spherical symmetry, ∫pdV = ∫p4πr2dr, which is easily integrable.
That is, if you mean the volume probability density function - the probability of being in a volume element dV; equivalent to ΨΨ* in the case of an atomic orbital.
If you meant the radial probability density function - the probability of being between r and r+dr - then this function is just μe-μr, as stated above.
The problem is that you have never stated precisely what distribution function you are looking for.
 

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