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4

pspF = With[{dt=#,ca=ConstantArray[1,{#,#}&@Dimensions[#][[2]]], opts=Sequence[Axes->False,AspectRatio->1]}, Grid[ReplacePart[ca, {{i_,i_}:>Histogram[dt[[All,i]],opts], {i_,j_}:>ListPlot[dt[[All,{i,j}]],opts]}], Dividers->All]]&; data1 = RandomReal[MultinormalDistribution[{0, 0}, ...


1

Normalize the coefficients: pjj = Normalize[pj, Total]; data = {{3, 56, 19, 28, 18, 24, 14, 9}}; LogLikelihood[MultinomialDistribution[n, pjj], data] (* -252.89 *)


-1

I recently asked WRI about a similar behaviour on probability distributions. The answer was that the output IS generated, but erroneously the result IS NOT displayed. Try dist = ProbabilityDistribution[1 - Abs[x], {x, -1, 1}] with Mean[dist] (* out 0 *) or cdist = CopulaDistribution[{"FGM", .2}, {NormalDistribution[-1, 2], NormalDistribution[1, ...


5

The question is of some interest because it captures rather nicely the difference between: A. mathematical statistics ... where we work with characterisations of distributions, such as starting with a pdf, or cdf, or cf ... e.g. Let $X$ be a random variable with pdf $f(x)$: $$f(x) = 1 -|x| \quad \text{for}\quad x\in(-1,1)$$ and B. Mathematica's ...


2

LogLikelihood[ MultinomialDistribution[n, {p1, p2, p3, 1 - p1 - p2 - p3}], {{x1, x2, x3, x4}} ] This can be slightly cleaned up using FunctionExpand and FullSimplify to get rid of the binomials: FullSimplify[% // FunctionExpand, n == x1 + x2 + x3 + x4 && x1 >= 0 && x2 >= 0 && x3 >= 0 ...


1

No error nv 10.0.2. On windows 7, 64 bit SeedRandom[0]; x = RandomReal[{-5, 5}, 100]; y = 2 x + 1 + RandomReal[{-0.1, 0.1}, 100]; X = Transpose[{x, y}]; ListPlot[X] Correlation[X] // MatrixForm CorrelationTest[X, 99995/100000, "PearsonCorrelation"] No error messages.


1

There is an add-on for Mathematica, called MathStatica, which does exactly what you want, if I understood you right. Find an example on www.mathstatica.com/examples and click on the left side on the item "Person Fitting." Ths is an excellent add-on which fits seamless into mma and extends its statistical capabilities. Best regards Volker


0

Question If $X \sim N(0,1)$ and $Y \sim N(0,1)$ are independent, how come: $$ P(X<Y) = \frac12 \quad \text{while} \quad P(\frac{X}{Y}<1) = \frac34$$ Answer Nothing to do with Mma. The answer is that the rules of standard algebra do NOT apply to the algebra of random variables, so you cannot simply divide both parts of $P(X<Y)$ by $Y$ ... ...


2

As this is simple enough, you could use definitions : dist = MultinormalDistribution[{0, 0, 0}, IdentityMatrix[3]] ; prob[a_] = Integrate[PDF[dist, {z1, z2, z3}], {z2, -Infinity, Infinity}, {z3, -Infinity, Infinity}, {z1, a Sqrt[z2^2 + z3^2], Infinity}] (* ConditionalExpression[1/4 (2 - ...


3

First, let's find a in functional form: Probability[Abs[x] < a, x \[Distributed] NormalDistribution[]] which shows that f[a_] := Erf[a/Sqrt[2]] Now you want to solve for the a that has some probability, say 0.1. Then Solve[f[a] == 0.1, a] {{a -> 0.125661}} A little more generally: Solve[f[a] == p, a] {{a -> Sqrt[2] InverseErf[p]}} ...



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