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You can use DistributionDomain to find the domain of a distribution, which will also tell you the dimension. I do not know where this is documented, but it does appear in some examples in the documentation. Usage examples: DistributionDomain[NormalDistribution[]] (* Interval[{-∞, ∞}] *) DistributionDomain[ParetoDistribution[xmin, alpha]] (* ...


1

Per my comment (this is not a fleshed out answer, just an example ): entX[p_] := With[{vars = Unique[] & /@ Range@Length@p}, Expectation[-Log[PDF[p, vars]], vars \[Distributed] p]] entX[nd2] (* 1+Log[2 π] *) Note, you'll want to use more sophistacated means to detrmine needed number, Length works here for your example, and is probably OK for some ...


5

How about f[x_, y_] = h - 3.07; Then, drawing candidates dat = {RandomVariate[UniformDistribution[{-2, 2}], np], RandomVariate[UniformDistribution[{-2, 2}], np]} // Transpose; and selecting Show[Select[dat, f @@ # > 0 &] // ListPlot[#, AspectRatio -> 1] &, S] the corresponding data can be exported as Export["test.dat",dat] ...


6

The integral is conditionally convergent. You can progress using substitution: $u=2^{\frac{r}{b}}\iff r= b\log_2 u $ Hence,$\frac{dr}{du}=\frac{b}{u\ln 2}$ You can do these substitutions in Mathematica: f[r_, b_, la_, k_] := 2^(r/b) Exp[k (2^(r/b) - 1)/la]/(b la) exp = f[x, a1, a2, a3] /. {2^(x/a1) -> u}; ex = D[a1 Log[2, u], u]; ans = Integrate[a1 ...



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