# Tag Info

0

I advise you to define ALD in following way. This code works in Mathematica 10.0.0.0 You can also you EstimatedDistribution of FindDistributionParameters

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data10 = RandomVariate[f[13, 0.5], {10, 25}]; (* 10 data sets from distribution f *) lls= LogLikelihood[EstimatedDistribution[#, f[a, b], {{a, 1}, {b, 1}}], #] & /@data10 (*{-32.4994, -25.2268, -21.9671, -26.8963, -25.9164, -22.8958, -26.5247, -24.9622, -33.9319, -28.6512}*) maxll=Block[{k=1}, MaximalBy[Last][ With[{dist = ...

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The issue is that your specification of the pdf, f has so many cycles in it for certain values of x that Mathematica is unable to verify that you have specified a valid pdf, because it loses too much precision in the integration. I checked this using a simple Manipulate. Manipulate[ Plot[f[tt, m], {tt, 0, Pi}, PlotRange -> All, Epilog -> ...

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It might be that you require a parametric process, the transformed process might not fit the bill. As a workaround how about doing this?Take the data back to a parametric process. S = TransformedProcess[Exp[P[t]], P \[Distributed] ARProcess[0, {0.5}, 1], t]; test = RandomFunction[S, {0, 1000}]; EstimatedProcess[TimeSeriesMap[Log, test], ARProcess[c, ...

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In order to get a feeling about what is going on here I suggest to perform a simulation. Let rx := RandomVariate[NormalDistribution[10, 5]]; ry := RandomVariate[ExponentialDistribution[0.2]]; rphi := \[Pi] Random[]; be the simulated variables, and s := Module[{x = rx, y = ry, phi = rphi}, Sqrt[x^2 + y^2 + 2*x*y*Cos[phi]]]; the function whose ...

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Thanks to @Verbeia, I have managed to find an answer, although it is a bit inelegant, and involves construction of my own 'custom' Discrete Markov chain. Please see this below: First select relevant probability from matrix: SelectProb[aVPast_, TrM_] := Module[{tempProb}, tempProb = Switch[aVPast, {1, 0}, TrM[[1, 1]], {0, 1}, TrM[[2, 2]]]]; Then, ...

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As per the answer from kguler - the following works: Inner[LogLikelihood[#,{#2}]&,Custom[vData],vData]

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Although we have now the elegant solution by wolfies I shall post my solution which uses only standard functions of MMA version 8 and which might be of interest because of a strategy to overcome difficulties with integration. We calculate the probability distribution functions PDF ($fp$) and CDF ($fc$) of a product of $n$ independent random variables $x_1, ... 4 The Problem Let$(X_1, \dots, X_n)$denote independent and identically distributed variables, each with common Exponential pdf$f(x)$: f = a Exp[-a x]; domain[f] = {x, 0, Infinity} && {a > 0}; We seek the pdf of$\prod_{i=1}^n X_i$, for$n = 2, 3, \dots\$ Solution The pdf of the product of two Exponentials is simply: where I am ...

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Use E rather than e {xlow, xhigh} = {0, Infinity}; Clear[pdf, x, t]; pdf[x_] := 2 E^(-2 x) You can use xhigh (i.e., Infinity) in a function like Integrate Integrate[pdf[x], {x, xlow, xhigh}] 1 However, a Plot must be on a finite domain Plot[pdf[x], {x, xlow, 6}, PlotStyle -> {{Thickness[0.01], Purple}}, AxesLabel -> {"x", "pdf[x]"}, ...

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ProbabilityDistribution defines a distribution. The constraints on the parameters are entered as Assumptions. dist = ProbabilityDistribution[a*b*x^(b - 1) E^(-a*x^b), {x, 0, Infinity}, Assumptions -> {a > 0, b > 0}]; PDF[dist, x] Piecewise[{{(abx^(-1 + b))/ E^(a*x^b), x > 0}}, 0] dpa = DistributionParameterAssumptions[dist] ...

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