New answers tagged probability-or-statistics
3
Couldn't you model this as a DiscreteMarkovProcess?
(* states are start(1), stage1(2), stage2(3), stage3(4), missing(5) *)
proc = DiscreteMarkovProcess[{1, 0, 0, 0, 0},
{{0, 0.253, 0, 0, 0.747},
{0, 0, 0.507, 0, 0.493},
{0, 0, 0, 0.705, 0.295},
{0, 0, 0, 1, 0},
{0, 0, 0, 0, 1}
}];
Graph[proc]
PDF[StationaryDistribution[proc], 4]
(* 0....
2
This method generalises easily to larger sums:
k = 2;
vec = Array[x, k];
dist = EmpiricalDistribution[{2/9, 5/9, 2/9} -> {-1, 0, 1}];
pdf = PDF @ TransformedDistribution[
Total[vec],
vec \[Distributed] ProductDistribution[
{dist, k}
]
]
DiscretePlot[pdf[n], {n, -10, 10}, PlotRange -> All]
1
Two random variables, x and y, each with the same distribution:
dist = EmpiricalDistribution[{2/9, 5/9, 2/9} -> {-1, 0, 1}];
ya[t_] := PDF[TransformedDistribution[x + y,
{x \[Distributed] dist, y [Distributed] dist}], t];
ya /@ Range[-10, 10]
(*Output:{0,0,0,0,0,0,0,0,4/81,20/81,11/27,20/81,4/81,0,0,0,0,0,0,0,0}*)
3
Amplifying on answer by @bRost03
distX = TruncatedDistribution[{0, Pi/2}, NormalDistribution[μ, σ]];
distY = TransformedDistribution[h/Tan[x], x \[Distributed] distX,
Assumptions -> h > 0];
g = Assuming[{y > 0, h > 0, σ > 0, μ ∈ Reals},
PDF[distY, y] // FullSimplify]
(* (E^(-((μ + 2 ArcTan[(y - Sqrt[h^2 + y^2])/h])^2/(
2 σ^2))) h ...
4
EDIT
It seems to work now. The key is to use the TruncatedDistribution so $h \cot(x)$ is monotonic.
transF=TransformedDistribution[h Cot[x],
x\[Distributed]TruncatedDistribution[{0,π/2},NormalDistribution[μ,σ]],
Assumptions->μ∈Reals&&σ>0&&h>0];
FullSimplify[PDF[transF,y],{y>0,h>0,μ∈Reals,σ>0}]
gives
(E^(-((μ + 2 ArcTan[(...
2
It works if you boost the Working precision.
h[x_] := (x*CDF[NormalDistribution[0, 1], x]*
Log[CDF[NormalDistribution[0, 1], x]])/
PDF[NormalDistribution[0, 1], x];
Then
Plot[h[x], {x, -5, 14}, WorkingPrecision -> 80]
Note that for a WorkingPrecision -> 40 one gets a smooth output with a sharp discontinuity near 13.5 which is somewhat ...
7
The slowest part of the code is clearly Zeta. You have to compute it only on a small parameter range and you also do not much very high accuracy (as Monte Carlo won't deliver much accuracy, too). So you can just sample Zeta over a fixed grid and use an interpolation function instead. Mathematica's interpolation functions are notoriously slow, so I implement ...
3
Just an extended comment stating that the one-step can be made to work but it's a very fragile approach. The list z below is a list of equivalent formulations of the desired random variable:
z = {10^(1/5 (25 + 5 x + y)), Exp[5 Log[10] + Log[10] x + Log[10] y/5],
Exp[Log[10] (5 + x + y/5)], Exp[a (5 + x + y/5)]};
TableForm[results = {#, Mean[...
0
randomNoodle[start_: {0, 0}, angle_: Pi/4, length_: 1, pieces_: 20] :=
Line[AnglePath[start, Thread[{length/pieces, RandomReal[{-angle, angle}, pieces]}]]]
Examples:
SeedRandom[4444]
noodles = Table[randomNoodle[RandomReal[1, 2]], 20];
MinMax[ArcLength /@ noodles]
{1., 1.}
g = Graphics[{RandomColor[], Thick, #} & /@ noodles, ImageSize -> Large]
...
Top 50 recent answers are included
