Skip to main content

calculus and analysis - How to deal with complicated Gaussian integrals in Mathematica?


As we know, for most Gaussian integrals, we can get the analytical result. Now I have many Gaussian integrals to treat, which have the following general form,


Integrate[
x1^n1 x2^n2 x3^n3 ... xd^
nd Exp[-{x1, x2, x3 ... xd}.matrix.{x1, x2, x3 ... xd} + c1 x1 +
c2 x2 ... cd xd], {x1, -Infinity, Infinity},
{x2, -Infinity, Infinity} ... {xd, -Infinity, Infinity}].

The matrix is a symmetrical positive definite matrix.


We can get the analytical integral result for this general form. However, it is a difficult task for Mathematica to deal with the similiar expressions, which takes so much time. I have written a package for treating this problem.



But, it is still not sufficient because I need to deal with lots of these kinds of integral expressions. So the computation time is important.


So have you ever seen the package for doing this kind of Gaussian integral? Or can you give me some suggestions about this problem? Thank you very much!


Thanks all of you. For the concept MultinormalDistribution, it is very helpful for me.


By the way, How can I transform the following integral expression to the standard form of MultinormalDistribution in a simple way?


vec1 := {r1x, r1y, r1z};
vec2 := {r2x, r2y, r2z};
vect := {r1x, r1y, r1z, r2x, r2y, r2z};
vcof := {c1, c2, c3, c4, c5, c6};
Integrate[Exp[-vect.mat.vect + vcof.vect] (vec1.vec2)^n (vec1.vec1)^
l (vec2.vec2)^m, {r1x, -Infinity, Infinity}, {r1y, -Infinity,

Infinity}, {r1z, -Infinity, Infinity}, {r2x, -Infinity,
Infinity}, {r2y, -Infinity, Infinity}, {r2z, -Infinity, Infinity}];

The mat is a symmetrical positive definite sparse matrix. And n, l, m are integers.
Thanks!




Comments

Popular posts from this blog

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1....