Skip to main content

probability or statistics - Finding distribution parameters of a gaussian mixture distribution


Short version: how to estimate the parameters of a mixture of multivariate normal distributions (i.e.: Gaussian mixture model)?




Long version.


I am trying to estimate the parameters of a mixture of multivariate Gaussian distribution.


I know how to do it for a single multivariate normal distribution:


dist = MultinormalDistribution[{0, 0}, {{1, 0}, {0, 1}}];
dataSet = RandomVariate[dist, 300];
estDist = EstimatedDistribution[dataSet, MultinormalDistribution[{m1, m2}, {s11, s12}, {s12, s22]}}]]


Plot[{PDF[dist, {x, 0}], PDF[estDist, {x, 0}]}, {x, -4, 4}, Filling -> Axis]
SmoothDensityHistogram[dataSet]
Plot3D[PDF[estDist, {x, y}], {x, -2, 2}, {y, -2, 2}]

Similarity of PDF at y=0, density of dataset and estimated density.


All fine and dandy. However, the same approach does not work for me with mixture distributions. In particular, I am interested in mixtures of Gaussian distribution (Gaussian Mixture Model).


I generate a sample dataset:


targetDist = MixtureDistribution[{1/3, 2/3}, {MultinormalDistribution[{0, 0}, {{1, 0}, {0, 1}}], MultinormalDistribution[{3, 3}, {{1, 0}, {0, 1}}]}];
dataSet = RandomVariate[targetDist, 500];

ListPlot[dataSet]
SmoothHistogram3D[dataSet]

Mixture of Gaussians


I try to find the estimated distribution with:


estimatedDist = EstimatedDistribution[dataSet,
MixtureDistribution[{w1, w2}, {
MultinormalDistribution[{m11, m12}, {{s111, s112}, {s112, s122}}],
MultinormalDistribution[{m21, m22}, {{s211, s212}, {s212, s222}}]
}]]


But the evaluation always fails with:


NMaximize::cvdiv: Failed to converge to a solution. The function may be unbounded. >>

For some reason, it works if, instead of using MultinormalDistribution I use BinormalDistribution with $\rho$=0.


I know how to estimate these parameters using the Expectation Maximization algorithm, but I was wondering if there is a Mathematica-friendly way to do it.




Edit.


Giving initial estimates of the parameters does not really improve much. Even when giving the exact parameters like this:


estimatedDist = EstimatedDistribution[dataSet, 

MixtureDistribution[
{w1, w2},
{MultinormalDistribution[{m11, m12}, {{s111, s112}, {s121, s122}}],
MultinormalDistribution[{m21, m22}, {{s211, s212}, {s221, s222}}]}],
{{w1, 1/3}, {w2, 2/3}, {m11, 0}, {m12, 0}, {s111, 1}, {s112, 0}, {s121, 0}, {s122, 1}, {m21, 3}, {m22, 3}, {s211, 1}, {s212, 0}, {s221, 0}, {s222, 1}}]

EstimatedDistribution seems to take much more time than what it would be reasonable (and, since the estimates are exact, "reasonable" means 0.1 sec).


After about 15 minutes of processing on a 3.3GHz Xeon, I got this error:


FindMaximum::eit: The algorithm does not converge to the tolerance of
4.806217383937354`*^-6 in 500 iterations. The best estimated solution,

with feasibility residual, KKT residual, or complementary residual of
{2.1536*10^-12,0.00200273,5.6281*10^-13}, is returned. >>

Then a popup message:


INTERNAL SELF-TEST ERROR: MLParseStream|c|297
Click here to find out if this problem is known, and to help improve
Mathematica by reporting it to Wolfram Research.

Answer



You need to make sure that the constraints on the parameters are satisfied. In this case, these are


1) The mixture weights sum to one. 2) The covariance matrices of the two bivariate normals need to be positive definite.



The first constraint can be guaranteed by specifying the weights as follows:


w1 = Exp[w]/(1 + Exp[w]);

where w is unconstrained.


The second constraint can be enforced by using the Cholesky Decomposition of the covariance matrices as below.


c1 = {{c111, 0}, {c112, c122}};

where c1 is a lower triangular matrix of unrestricted elements. The resulting covariance matrix is given by s1 below.


s1 = c1.Transpose[c1]


Out[9]= {{c111^2, c111 c112}, {c111 c112, c112^2 + c122^2}}

Similarly, we have for the other covariance,


s2 = c2.Transpose[c2];

Let the two mean vectors be


m1 = {m11, m12};

m2 = {m21, m22};


The mixture pdf can be written as


In[15]:= mixPDF = MixtureDistribution[{w1, 1 - w1}, 
{MultinormalDistribution[m1, s1],
MultinormalDistribution[m2, s2]
}
]

Out[15]= MixtureDistribution[{E^w/(1 + E^w),
1 - E^w/(1 + E^w)}, {MultinormalDistribution[{m11,
m12}, {{c111^2, c111 c112}, {c111 c112, c112^2 + c122^2}}],

MultinormalDistribution[{m21,
m22}, {{c211^2, c211 c212}, {c211 c212, c212^2 + c222^2}}]}]

Then one may get what you are looking for, depending upon starting values.. The first try here does not give the correct answer.


In[16]:= est1 = EstimatedDistribution[dataSet, mixPDF]

Out[16]= MixtureDistribution[{0.0172133,
0.982787}, {MultinormalDistribution[{1.37871,
0.821044}, {{4.88591, 1.35155}, {1.35155, 0.373881}}],
MultinormalDistribution[{1.91595,

1.88312}, {{3.0206, 2.01692}, {2.01692, 3.08025}}]}]

Specifying starting values helps.


In[17]:= est2 = 
EstimatedDistribution[dataSet,
mixPDF, {{m11, - 0.5}, {m12, 0.8}, {m21, 1.5}, {m22, 2.0}, {c111,
1.5}, {c112, 0.0}, {c122, 1.0}, {c211, 1.5}, {c212, 0.0}, {c222,
1.0}, {w, 0.2}}]

Out[17]= MixtureDistribution[{0.39772,

0.60228}, {MultinormalDistribution[{0.110727,
0.110757}, {{1.05393, 0.060291}, {0.060291, 1.07923}}],
MultinormalDistribution[{3.0927,
3.02315}, {{0.84418, -0.148054}, {-0.148054, 0.982491}}]}]

An alternative approach can use FindDistributionParameters as follows


In[18]:= 
est3 = FindDistributionParameters[dataSet,
mixPDF, {{m11, 0.0}, {m12, 0.0}, {m21, 2.5}, {m22, 3.0}, {c111,
1.5}, {c112, 0.0}, {c122, 1.0}, {c211, 1.5}, {c212, 0.0}, {c222,

1.0}, {w, 0.5}}]

Out[18]= {m11 -> 0.110727, m12 -> 0.110757, m21 -> 3.0927,
m22 -> 3.02315, c111 -> 1.02661, c112 -> 0.0587281, c122 -> 1.0372,
c211 -> 0.918792, c212 -> -0.16114, c222 -> 0.978021, w -> -0.414975}

The original parameters can be obtained as


In[19]:= {w1, 1 - w1, m1, s1, m2, s2} /. est3

Out[19]= {0.39772, 0.60228, {0.110727,

0.110757}, {{1.05393, 0.060291}, {0.060291, 1.07923}}, {3.0927,
3.02315}, {{0.84418, -0.148054}, {-0.148054, 0.982491}}}

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 - 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.