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5

Use TransformedDistribution distY = TransformedDistribution[1/x, x \[Distributed] UniformDistribution[]]; The PDF is then PDF[distY, y] EDIT: corrected error in syntax CDF[distY, y] Plot[{CDF[distY, y], PDF[distY, y]}, {y, 1, 5}, PlotLegends -> Placed["Expressions", {0.7, 0.4}]]

3

m = 1; ω = {10, 10, 10, 10, 10, 10, 10, 10, 10, 10}; Convert γ to exact values to avoid forcing machine precision calculations γ = {0.1202, 0.1413, 0.1862, 0.2399, 0.3090, 1.0000, 1.5849, 1.9055, 1.9953, 2.5704} // Rationalize; f0[k_, y_] := PDF[ChiSquareDistribution[ω[[k]]], y] f00[k_, z_] := Gamma[ω[[k]]/2, 0, z/2]/Gamma[ω[[k]]/2] f1[k_, y_] := ...

1

Edit: I've changed this from an extended comment to an answer. The title concerns estimating the difference between two $\chi^2$ distribution functions. The text has the appearance of not directly addressing that question because of the large amount of code that doesn't explicitly mention a distribution function (at least in my opinion). A brute force ...

1

To find PlotRange for each plot, run Options[ab, PlotRange] Options[bc, PlotRange] Then adjust your code as follows: spCnts1 = {12.5, 17.8, 19.9, 21.40, 25.6, 28.4, 29.5, 30.2, 35.1, 40.1, 46.4, 47.5, 48.3, 50.2, 53.2, 55.9, 56.3, 57.51, 60.3, 62.3, 65.4, 66.4, 68.60, 71.2}; ab = ProbabilityScalePlot[spCnts1, "LogNormal", ImageSize -> Medium, ...

0

Since $x$ is a count variable, just state it in the assumptions. Then everything is hunky dory (whatever hunky dory is): Assuming[{\[Lambda] > 0, \[Kappa] > 0, 0 < \[Theta] < 1, x >= 0, x \[Element] Integers}, PDF[aDist, x]] which evaluates to: \[Kappa]^\[Kappa] (\[Theta] \[Lambda])^x (\[Kappa] + \[Theta] \ \[Lambda])^(-x - \[...

7

For simple processes this can be done by using the function PDF: proc = ItoProcess[{\[Mu], \[Sigma]}, {x, 0}, t]; PDF[proc[t], x] E^(-((x - t [Mu])^2/(2 t [Sigma]^2)))/(Sqrt[2 [Pi]] Sqrt[ t [Sigma]^2]) If that doesn't work for your process, it's most likely that the analytic PDF simply cannot be computed. In that case you're pretty much down to ...

1

This may not be efficient but could be a good starting point. It works for data up to 26 columns. If you need more columns, you can modify the function. tableView[data_] := Module[{rowheadings, columnheading, output}, rowheadings = Flatten[{Range[1, Length[data]]}]; columnheading =Flatten[{"", CharacterRange["A", "Z"][[1 ;; Dimensions[data] ...

0

Update: My original answer which constructed the estimate of the sum of all of the correlations one-at-a-time was much, much slower than the method provided by the OP. (I should have followed the RTFM advice I keep giving others.) But below I give what I hope are relevant comments: Parameters and estimates should have separate notations such as using \$...

0

Looking at the options of TableView it seems like it's not possible. Options[TableView] {Alignment -> {Automatic, Automatic}, AllowedDimensions -> Automatic, Appearance -> Automatic, AppearanceElements -> All, Background -> Automatic, BaselinePosition -> Automatic, BaseStyle -> {}, ContentPadding -> True, DefaultBaseStyle -> "TableView", ...

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