10
Your question seems to miss some points in What topics can I ask about here?. This answer is for helping you to write the code by yourself and if you encounter any problem, feel free to post another question.
In Mathematica use:
(* comment *) syntax for commenting your code
LowerTriangularize function for accessing the lower triangular part of a matrix
...
7
Your integral has a closed form solution, so take that and give it a name:
g[x_] := 1/3 E^(-x/2) Sqrt[2/\[Pi]] Sqrt[x] (3 + x) + Erfc[Sqrt[x]/Sqrt[2]]
One way to find the value of x where g[x]==0.05 is to solve:
Minimize[Abs[(g[x] - 0.05)^2] && x \[Element] Reals, x]
{2.92244*10^-18, {x -> 11.0705}}
6
Clear["Global`*"]
Your integral is the complement of the CDF for a GammaDistribution.
dist = GammaDistribution[5/2, 2];
pdf[x_] = PDF[dist, x]
(* Piecewise[{{x^(3/2)/(E^(x/2)*(3*Sqrt[2*Pi])), x > 0}}, 0] *)
The integral is
int[x_] = Assuming[x > 0, 1 - CDF[dist, x] // Simplify]
(* 1 - GammaRegularized[5/2, 0, x/2] *)
While ...
5
dist = TransformedDistribution[x^2,
x \[Distributed] NormalDistribution[0, 1]]
(* ChiSquareDistribution[1] *)
CDF[dist, x]
PDF[dist, x]
Plot[{CDF[dist, x], PDF[dist, x]}, {x, 0, 5},
PlotRange -> {0, 1.1},
PlotLegends ->
Placed["Expressions", {0.7, 0.5}]]
EDIT: Alternatively,
cdf[x_] = Probability[y^2 <= x,
y \[Distributed] ...
3
(* English words, lowercase, length > 2 *)
words = ToLowerCase@Select[WordList[], StringLength[#] > 2 &];
(* Get first two chars, throw out any pairs containing non-letters/punctuation *)
pairs = Select[Characters[words][[All, ;; 2]], AllTrue[#, LetterQ] &];
(* Probability function *)
countProbs[list_] := Map[#/Length[list] &, Counts[list]...
3
Here is a pedestrian way to achieve the same that also works with older MMA:
We first define the normalized ProbabilityDistribution of the radius:p
r0 = 1;
pdf[x_] = Exp[-x^2/2]/Integrate[ Exp[-y^2/2], {y, 0, r0}]
p = ProbabilityDistribution[pdf[x], {x, 0, r0}];
With this distribution we can create r-values: r. The angle is uniformly distributed, what we ...
3
Assuming you're using the new spatial statistics from v12.2:
{x0, y0} = {1.5, 2};
σ = .7;
R = 3.0;
reg = Disk[{x0, y0}, R];
pts = RandomPointConfiguration[
InhomogeneousPoissonPointProcess[
Function[10000*Exp[-((#1 - x0)^2 + (#2 - y0)^2)/(2*σ)]],
2], reg];
Show[RegionPlot[reg], ListPlot[pts]]
2
If the "integer data" consists of counts that might be expected to follow a Poisson distribution given the predictor value, then a Poisson regression should be considered. That can be accomplished using GeneralizedLinearModelFit.
glm = GeneralizedLinearModelFit[dataHist5, t, t, ExponentialFamily -> "Poisson"]
Show[ListPlot[dataHist5], ...
2
(* Get the weekly closing prices *)
dataweekly = FinancialData["^SPX", "Close", {{2010, 7, 1}, Today, "Week"}];
(* 6 week rolling weekly return *)
rolling = MovingMap[100*(Last[#]/First[#] - 1) &, dataweekly, {6, "Week"}];
DateListPlot[rolling]
2
Not certain what your really want to do, but maybe the following will spark some ideas.
Manipulate[
Module[{dw, dpw},
dw = dataweekly[[All, 4]];
dpw = datapercweekly;
Grid[{
{ListLinePlot[dw, ImageSize -> 350],
ListLinePlot[dpw, ImageSize -> 350]},
{ListLinePlot[MovingAverage[dw, n], ImageSize -> 350],
ListLinePlot[...
2
dist = EmpiricalDistribution[
tableOfRandomVariable[[All, 2]] -> tableOfRandomVariable[[All, 1]]];
DiscretePlot[CDF[dist, x], {x, 0, 14}, ExtentSize -> Right,
ExtentMarkers -> {"Filled", "Empty"}]
2
I think this may be what you were looking for originally:
dist =
TransformedDistribution[
x^2 + y^2 + (1 - x - y)^2,
{x, y} \[Distributed] DirichletDistribution[{1, 1, 1}]
]
where $x+y+\underbrace{(1-x-y)}_{z}= 1$.
You can then get the mean and variance (for example)
Mean[dist] --> 1/2
Variance[dist] --> 1/60
You can confirm these ...
2
σ = 9;
L[s_, d_] = (1/(σ*Sqrt[2*π]))*
E^(-(1/2)*((s - (-50 - 11*Log[d]))/σ)^2);
The probability distribution is
dist[s_] =
ProbabilityDistribution[L[s, d], {d, 0, Infinity},
Method -> "Normalize"];
r[s, d] is the PDF of the distribution
r[s_, d_] = PDF[dist[s], d]
Verifying the normalization,
Integrate[r[s, d], {d, 0, Infinity}]
(* ...
1
If the $d$ takes on values from 0 to $\infty$, then the normalizing constant is given by
Integrate[L[s, d], {d, 0, ∞}]
(* 1/11 E^(-(1019/242) - s/11) *)
So you could define the following function which is non-negative and integrates to 1 for a pdf associated with $d$:
r[s_, d_] := L[s, d]/(1/11 E^(-(1019/242) - s/11))
You should avoid using capital letters ...
1
data = (SeedRandom[1234]; RandomVariate[
NormalDistribution[2, 1/2], 20])
(* {1.74583, 1.9647, 1.20305, 2.7683, 3.33901, 1.34158, 1.45272, 2.07147, \
2.05208, 1.16345, 2.01788, 2.08509, 2.46734, 1.58657, 1.95452, 2.5167, \
1.81824, 2.0943, 1.93193, 1.95803} *)
dist = NormalDistribution[m, s];
param = FindDistributionParameters[data, dist]
(* {m -> ...
1
Clear["Global`*"]
k[h_, x_] = PDF[NormalDistribution[0, h], x]
(* E^(-(x^2/(2 h^2)))/(h Sqrt[2 π]) *)
Assuming that the first part takes the argument x - y rather than being distributed across x - y
t[x_, y_] = ((k[h, #]*k[g, #] - k[g, #])^2) &[x - y] // Simplify
(* (E^(-(((g^2 + h^2) (x - y)^2)/(
g^2 h^2))) (-1 +
E^((x - y)^2/(2 h^2))...
Only top voted, non community-wiki answers of a minimum length are eligible
