# Tag Info

## New answers tagged distributions

2

Alternative 2D Solution A simpler and more readily generalizable alternative to the solution presented in my previous Answer can be obtained by noting that a UniformDistribution on the curve shown in the first figure is equivalent to a ProbabilityDistribution in p weighted by dsdp, ProbabilityDistribution[dsdp/arcmax, {p, 0, 1}] as can be can be ...

3

Perhaps this is what you need? r = RandomReal[{0, 1}, {1000, 1000}]; ev = Eigenvalues@r; ListPlot[{Re@#, Im@#} & /@ ev, AspectRatio -> 1, ImageSize -> 300, PlotRange -> 10*{{-1, 1}, {-1, 1}}, AxesLabel -> {"Re", "Im"}]

5

Solving the one-dimensional problem is a first step toward solving the three-dimensional problem described in the question and comments. Without loss of generality, it can be written after renormalization as 1 == p^2 + Sqrt[x^2/4] Noting that phase space is mirror-symmetric about both axes, we solve for the phase curve in the quadrant p > 0, x > 0. ...

4

Let $Z \sim N(\mu,s^2)$. By definition, $X = e^Z$ has a Lognormal distribution with pdf $f(x)$: f[x_, mu_, s_] = (1/(x s Sqrt[2 Pi]))*Exp[-((Log[x] - mu)^2/(2 s^2))]; In your case, the variance-covariance matrix has zero correlation, so (given Normality) your two variables are in fact independent. Thus, the joint pdf of two Lognormals is just the product ...

1

Here is how I would do it. First, import the data using Import. This is straightforward from the documentation, using either Excel or txt files, so for the purposes of this exercise, I'm going to generate the data in Mathematica instead. myD = MixtureDistribution[{2, 1}, {NormalDistribution[2, 2], NormalDistribution[2, 1/2]}]; For reference, here is ...

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