probability or statistics - Efficient maximisation of log-likelihood

I would like to efficiently find the maximum $(\sigma,\lambda)$ for the log-likelihood of the derived distribution below. I only need $\sigma$ and $\lambda$ to one decimal place - so not very precise. I have tried lowering both the AccuracyGoal and PrecisionGoal to as low as 1, or 2, but this doesn't appear to affect the rate at which the solution is obtained.

The below code creates the function, the some test data, then attempts its maximisation:

aDist[σ_, λ_] := TruncatedDistribution[{0, ∞}, 

  MixtureDistribution[{1, 1}, {NormalDistribution[0, σ], ExponentialDistribution[λ]}]];

data = If[# > 0, #, 0] & /@ RandomVariate[aDist[4, 1/3], {20}];

NMaximize[{LogLikelihood[aDist[σ, λ], data], 10 > σ > 0, 1 > λ > 0}, {σ, λ}]

Whilst for 20 data points this method returns a solution in a short time, for my actual dataset I need to find a solution for a dataset of size 1000+. As it currently stands, this would be untenable for the above method.

I have tried some of the different methods available to NMaximize, including "RandomSearch", and "DifferentialEvolution". However, the method I choose does not seem to make the maximisation run faster.

I have also tried FindMaximum where I start the solver close to the actual parameter values, however this calculation appears to hang forever. I have also tried differentiating, then using NSolve, and FindRoot, but again I am not having any success. Finally I tried a variant on the $EM$ algorithm here, where I alternate between maximising over $\sigma$ then $\lambda$, but again this doesn't help.

Does anyone have any ideas here? I know I could use MCMC if I turn the problem into a Bayesian one, but for reasons I won't mention here I don't want to do this.

As an aside, I can see why it is difficult to find a reasonable maximum to this function, since there is a high degree of correlation between the parameters of the distribution. However, I can't help but think there's a solution of which I'm neglecting to think.

Comments

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.

Blog

Search This Blog