[Date Index]
[Thread Index]
[Author Index]
Convolution of a Gaussian Distribution and a discontinuous function
*To*: mathgroup at smc.vnet.net
*Subject*: [mg97011] Convolution of a Gaussian Distribution and a discontinuous function
*From*: Dan629 <dangerrity at gmail.com>
*Date*: Mon, 2 Mar 2009 05:20:53 -0500 (EST)
This is a repost of a reply on an earlier thread but I think it got
buried so I'm putting it up as a new topic with a more relevant
subject line.
-------
I don't know if this is a forum where math questions are posted (as
opposed to strictly Mathematica questions), but I'll ask anyway.
The PDF of the circular uniform distribution in one dimension has two
discontinuities at x = radius. This makes the required integration to
do a convolution fail, and it fails in Fourier space as well. If I do
a numeric convolution with ListConvolve, of course, I can easily get a
result. Alternatively I can use a summation and get a symbolic result
consisting of hundreds of summands. These work, but neither solution
can be readily integrated or differentiated when the other independent
variables are complicated as they are in my case. I'd really like to
get a closed form symbolic formula for the convolution. Am I dreaming
to think that Mathematica, with all it's rich complexity, can do
something like this? Or it is simply a fact of math that a
discontinuous function cannot be convolved with a Gaussian
Distribution and no trick (DiracDelta, HeavisideTheta) or
computational genius will make it work?
Given a failed convolution, I can use the summation solution and then
fit the resulting curve. I've used the Taylor and polynomial
expansions along with FFTs, but I'm not able to get a close enough
approximation. The convolution is a beautiful, continuous curve based
on a sum of Gaussians and so intuitively it seems as though it would
be easy to fit, but I can't seem to get it. It seems that an
approximation of either the Gaussian Distribution or the circular
uniform distribution would make the integration possible, too. But I
struggle with getting suitable non-exponential approximations to
either of those.
So, for the illustration, look at this:
vars = { ll -> .27, mm -> 0, ss -> .06 };
pdf1[ xx_ ] := PDF[ NormalDistribution[ mm, ss ], xx ];
pdf2[ xx_ ] :=
Piecewise[
{ { 1/(Pi * Sqrt[ ll^2 - xx^2 ]), Abs[ xx ] < ll } }
];
Plot[ { pdf1[ xx ], pdf2[ xx ] } /. vars, { xx, -.3, .3 } ]
conv = Sum[
(pdf1[ xx - zz ] pdf2[ zz ]), { zz, -.6, .6, 0.02 } ] / 60;
Plot[ Evaluate[ conv /. vars ], { xx, -.5, .5 } ]
I'm seeking a symbolic representation of the second curve.
(All variables are real, and both ll and ss are always positive.)
Please be kind Jens -- I'm admittedly an amateur!
Prev by Date:
**Re: Delayed function does not behave as original input**
Next by Date:
**Re: "Do What I Mean" - a suggestion for improving**
Previous by thread:
**Re: Show problem: combining ListPlot and Plot**
Next by thread:
**Re: Convolution of a Gaussian Distribution and a discontinuous function**
| |