New answers tagged probability-or-statistics
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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