Skip to main content

probability or statistics - Calculating the Mean of a Truncated Multinormal Distribution


First, I'm a little disappointed that Mathematica balks at:


Mean[TruncatedDistribution[{{0, Infinity}},MultinormalDistribution[{0}, {{1}}]]]

Second, is the numerical computation of means from truncated multinormal distributions so hard? Is anyone aware of a package that implements the algorithm of Leppard and Tallis (1989) (see here for FORTRAN code) or anything like it?


Edit: rm asked for an example that fails to compute:


Mean[TruncatedDistribution[{{0, Infinity}, {0, Infinity}},
MultinormalDistribution[{0.5, 1.5}, {{1., 0.3}, {0.3, 1.}}]]]


Answer



If you want to do serious statistical work, I would suggest to not use Mean and instead use specialized functions that work on distributions, such as Expectation and NExpectation. Although the documentation says that Mean[dist] gives the mean of the symbolic distribution, I suspect they meant it for basic distributions such as NormalDistribution, BinomialDistribution, etc., which were all there when Mean was written. Mean was last modified in version 6 and most probably is not aware of newer functions such as TruncatedDistribution, MultinormalDistribution, etc., which were all introduced in version 8.


So the equivalent code for your example is:


NExpectation[{x, y}, {x, y} \[Distributed] 
TruncatedDistribution[
{{0, Infinity}, {0, Infinity}},
MultinormalDistribution[{0.5, 1.5}, {{1., 0.3}, {0.3, 1.}}]
]
]
(* {1.02198, 1.74957} *)


Using Expectation offers more flexibility than Mean, because you can now calculate the expectations of arbitrary quantities:


NExpectation[{Sin[x], y^3}, {x, y} \[Distributed] 
TruncatedDistribution[
{{0, Infinity}, {0, Infinity}},
MultinormalDistribution[{0.5, 1.5}, {{1., 0.3}, {0.3, 1.}}]
]
]
(* {0.650673, 9.70065} *)


NExpectation still does not work with MultinormalDistribution with a single dimension... I don't know why exactly, but personally I would never use a Multi-something function to mean just 1 (which is the opposite of multi). I would suggest using a Switch and use NormalDistribution when you have a MultinormalDistribution of dimension 1.


Comments

Popular posts from this blog

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 - 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 - Adding a thick curve to a regionplot

Suppose we have the following simple RegionPlot: f[x_] := 1 - x^2 g[x_] := 1 - 0.5 x^2 RegionPlot[{y < f[x], f[x] < y < g[x], y > g[x]}, {x, 0, 2}, {y, 0, 2}] Now I'm trying to change the curve defined by $y=g[x]$ into a thick black curve, while leaving all other boundaries in the plot unchanged. I've tried adding the region $y=g[x]$ and playing with the plotstyle, which didn't work, and I've tried BoundaryStyle, which changed all the boundaries in the plot. Now I'm kinda out of ideas... Any help would be appreciated! Answer With f[x_] := 1 - x^2 g[x_] := 1 - 0.5 x^2 You can use Epilog to add the thick line: RegionPlot[{y < f[x], f[x] < y < g[x], y > g[x]}, {x, 0, 2}, {y, 0, 2}, PlotPoints -> 50, Epilog -> (Plot[g[x], {x, 0, 2}, PlotStyle -> {Black, Thick}][[1]]), PlotStyle -> {Directive[Yellow, Opacity[0.4]], Directive[Pink, Opacity[0.4]],