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LaplaceTransform attempts a symbolic evaluation of the transform, which Mathematica fails to do in this case. To get an answer, write the transform explicitly as an NIntegrate multiple integral. It still takes a while, but reducing the accuracy/precision demands will speed things up: chat = NIntegrate[ GumExpExp Exp[-Ab[s] x - Ar[s] y], {x, 0, ...


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With the data beeing data = RandomVariate[NormalDistribution[0, 1], 200]; the range of the box specified to be one sigma (approx. 68.3 %tile range) by sigma=Erf[1/Sqrt[2]] and a limit for the fences defined to be 10 % fencesLimit = 0.1 we can plot a BoxWhiskerChart using: BoxWhiskerChart[data, "Median", Method -> "BoxRange" -> (Quantile[#, ...


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p1 = Histogram[data, {0, 9, 1}, "Count", ChartLegends -> {"Experimental Result"}, ChartStyle -> "Pastel"]; p2 = DiscretePlot[ Length@data*PDF[PoissonDistribution[2.2766917293233084`], x], {x, 0, 10}, PlotStyle -> {Red, Medium}, PlotLegends -> {"Theoretical Poisson"}]; Show[{p1, p2}, AxesLabel -> {"Time (s)", "Counts"}] ...


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You can also depict the expected value nicely as a function of Mu and Sigma using Plot3D. px = Simplify[PDF[LogNormalDistribution[Mu, Sigma], x], x > 0] E^(-((Mu - Log[x])^2/(2 Sigma^2)))/(Sqrt[2 \[Pi]] Sigma x) Plot3D[NIntegrate[Log2[1 + x] px, {x, 0, \[Infinity]}], {Mu, -5, 10}, {Sigma, .2, 2}, AxesLabel -> {Mu, Sigma, EY}] PS: Reminds me ...


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I don't think there is anything unusual here that can't be found in the documentation. Normalization parameters in the x and y dimensions (b, and a, respectively) can be added into the model, as can the x-axis shift (parameter c). Parameter estimates are taken from a look at the original data. nlm = NonlinearModelFit[data, a Erfc[b (x + c)], {{a, 8 ...


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I am a little unsure what the aim is here. Looking at the data only the first third of the plot could be fitted to Erfc like function. I post this, perhaps, as motivation. d = Transpose@data; ListPlot[ds = SortBy[d, #[[1]] &], Joined -> True] Extracting the part of plot I referred to: fd = ds[[1 ;; 35]] fdlp = ListPlot[fd] Now as a rather ...


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Using the function positionDuplicates from this answer by @Szabolcs positionDuplicates[list_] := GatherBy[Range@Length[list], list[[#]] &]; dispatchF = Thread[DeleteDuplicates[#] -> positionDuplicates[#]] &; positionBins = Union @@@ (BinLists[#, #2] /. dispatchF[#1]) &; data = RandomInteger[5, 17] (* {5, 4, 1, 3, 0, 3, 0, 0, 5, 5, 0, 0, ...


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I suspect there may not be a closed form expression (I have not looked at it hard enough). If the aim is to not analytical but numerical, I post the following for illustration (apologies if not the intent of question): f[x_, y_] := NIntegrate[ Log2[t + 1] Exp[-(x - Log[t])^2/(2 y^2)]/(Sqrt[2 Pi] t y), {t, 0, Infinity}] rv[a_, b_, n_] := ...


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Updated with working code (tnx @rasher @mfvonh) Let’s start by importing Fisher’s classic dataset on Iris flower measurements… Fisher’s classic paper can be found here…. Needs["MultivariateStatistics`"] (*Import Data*) irisData = Import["http://aima.cs.berkeley.edu/data/iris.csv", "CSV"]; plotLabels = {"Sepal.Length", "Sepal.Width", "Petal.Length", ...


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SeedRandom[123]; data = 5 + RandomVariate[BinormalDistribution[.5], 20]; TableForm[data,TableHeadings->{Range@Length@data,{"value1", "value2"}}]//Style[#, 16] & Concordant and discordant pairs: From the documentation on KendallTau Generalizing @eldo's post we can get the concordant and discordant pairs of observations as follows: cdpairsF = ...


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m = {{{1, 1}, {2, 4}}, {{1, 5}, {2, 4}}, {{1, 1}, {1, 1}}}; Times @@@ (Sign /@ Subtract @@@ m) /. (0 | 1) -> "Concordant" /. -1 -> "Discordant" {"Concordant", "Discordant", "Concordant"}



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