# Tag Info

12

iqrc = "IQRCoefficient" /. ChartElementData["BoxWhisker", "Options"]; 1.5 fences = Quartiles[data][[{1, 3}]] + {-1, 1} iqrc InterquartileRange[data] outliers = Select[Not @* Between[fences]] @ data {-3.0844, -3.4005} which matches the vertical coordinates of outlier glyphs: Cases[BoxWhiskerChart[data, "Outliers", PerformanceGoal -> "Speed"], ...

10

So given distribution: dis=MixtureDistribution[ {0.43999526215889906,0.5600047378411006}, {NormalDistribution[58745.18201580311,15134.735419294597], NormalDistribution[78262.1217564796,2797.225012305396]}]; with the PDF: Plot[PDF[dis, x], {x, 0, 10^5}, PlotRange -> All,PlotTheme -> "Detailed"] You can compute probability via built in function:...

9

When you have just a single additive error term with a known distributional form, you should avoid thinking about "weights" (except for some distributions where using weights and least squares gives you good starting values for troublesome datasets and models). For the example you give using maximum likelihood is pretty straightforward. If you have some ...

9

It is actually in the help documentation as the last item. Outliers are 1.5 InterQuartileRanges (IQR) away from the nearest quartile and Far Outliers are 3 IQR distant.

6

The Silverman rule is $0.9 \min({\hat{\sigma},IQR/1.34})n^{-1/5}$ and the rule $1.06\hat{\sigma} n^{-1/5}$ is the rule that minimizes the mean integrated square error when the underlying distribution is normal. The default for SmoothKernelDistribution is to use the Silverman rule. That bandwidth (in Mathematica code) is 0.9*Min[StandardDeviation[x], ...

5

This is a quite standard way. a[t_?NumericQ] := CDF[MixtureDistribution[{0.43999526215889906, 0.5600047378411006}, {NormalDistribution[58745.18201580311, 15134.735419294597], NormalDistribution[78262.1217564796, 2797.225012305396]}], t] a - a 0.991409

5

From @user64494 's comments and the mentioned Wiki page the expectation of $$\sqrt{(x_0-x_1)^2+(x_2-x_3)^2+(x_4-x_5)^2}$$ will be or in Wolfram Language Sqrt[π] σ* LaguerreL[1/2, 1/2, 1/2 (-((p0 - p1)^2/(2 σ^2)) - (p2 - p3)^2/(2 σ^2) - (p4 - p5)^2/(2 σ^2))] (where I've substituted the more commonly used $\sigma$ for $s$). As a simple check: p = {σ -...

4

There is some hidden MCMC (Markov Chain Monte Carlo) functionality in Mathematica that can come in handy, though there are obviously no guarantees about how well it works since it's unexposed. Here's a simple example: dist = ProbabilityDistribution[ Beta[3/4, 1/2]/(Sqrt Pi^2) 1/(1 + x^4 + y^4), {x, -Infinity, Infinity}, {y, -Infinity, Infinity}];...

4

The following applies JimB's suggestion; it would work if you have enough data to be representative of the "old" distribution: ClearAll[standardizeTo] standardizeTo[oldData_, newData_] := Module[ {oldMean, oldStdev}, {oldMean, oldStdev} = Through[{Mean, StandardDeviation}[oldData]]; oldStdev Standardize[newData] + oldMean ] Let's create some ...

4

The issue is to do with RandomFunction For continuous-time processes with jumps, such as WhiteNoiseProcess[] the step dt is random and given by the process itself. e.g this works RandomFunction[WhiteNoiseProcess[1/3], {0, 50}] This doesn't RandomFunction[WhiteNoiseProcess[1/3], {0, 50,0.1}] For continuous-time processes without jumps, such as ...

3

This is an extended comment. What you have described for each $z$ is a mixture distribution and with a different but known probability for each observation. If you know the form for each of the two distributions (i.e., everything known about the distribution except the specific parameters), then you could use maximum likelihood to estimate the parameters. ...

3

I don't think it is a bug. I executed your Code in M 11.1.1. Using ClassifierInformation[cls] you get some Information About the Classifier. Now I use M 12.0.0 and pass the regularization coefficients explicitly to the Classifier. data = {{1, a}, {2, a}, {3, b}, {4, a}, {5, b}, {6, c}, {7, c}, {8, b}, {9, c}}; cls = Classify[Rule @@@ data, Method -> {"...

3

Better use exact values, because they have infinite precision. Try: NSolve[Pr[52, 0, p] == 5/100, p, Reals] (* {{p -> 0.0559822}, {p -> 1.94402}} *) Or: (* exact numbers *) Solve[Pr[52, 0, p] == 5/100, p, Reals] (* specifying 30-digit precision *) NSolve[Pr[52, 0, p] == 0.05`30, p, Reals] (* machine precision returns error message *) FindRoot[Pr[...

2

Edit Your numeric problems are induced because you are using machine precision arithmetic. The fix is to define the parameters as exact quantities and do the computations with Mathematica's arbitrary precision arithmetic. You also have some bad syntax in your code, I will correct those errors in presentation of the making of the computation with ...

2

Use exact constants so that the numerical integration can be done with a specified level of precision. p = 1/100; m = -1/2*Log[29/10]; s = Sqrt[Log[29/10]]; r = 1/2; tailF[x_] := 1 - CDF[NormalDistribution[0, 1], (Log[x] - m)/s] t = Quantile[LogNormalDistribution[m, s], p]; v1 = -r*NIntegrate[((tailF[x]/(1 - p))^r)*Log[tailF[x]/(1 - p)], {x, t, ...

2

Not a super great answer, but you can extract a rough distribution $P$ by removing the data points related to $Y$ as follows. First, bin your data. In this case, put it into 20 unique bins from $-1$ to $1$. f[{x_, p_}] := Floor@(10 x); bins = GatherBy[Sort@dat, f]; counts = Length /@ bins; (* {22, 22, 19, 17, 25, 27, 32, 56, 62, 222, 213, 62, 38, 31, 34, ...

1

This is an extended comment. I think the issue is that using {σ, 1, ∞} for the limits of integration doesn't allow NIntegrate to work well. (Yes, those are what the limits should be.) Here's an example. integrand[z_, μ_, σ_] := Sqrt[Sqrt*Pi]/2* Exp[-Sqrt[Sqrt*Pi] (z + σ)^2/μ*Sqrt[Pi*σ]*Erf[μ (z - σ)^2/(Sqrt[Pi*σ])]] Plot[integrand[4.75, 4, σ], {...

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