New answers tagged probability-or-statistics
5
We can replace NIntegrate by Integrate and use FindRoot.
F[x_] := 2*Pi^(-1/2)*Integrate[Exp[-y^2], {y, 0, x}]
FindRoot[F[x] == 0.5, {x, 0}]
{x -> 0.476936}
Another way is again use NDSolve.
NDSolve[{F'[x] == 2*Pi^(-1/2)*Exp[-x^2], F[0] == 0,
WhenEvent[F[x] == 0.5, {Print[x], "StopIntegration"}]},
F[x], {x, 0, ∞}]
0.476936
0
You need to improve your patterns:
pair1[{___, x_, ___, x_, ___, y_, ___, y_, ___} /; x != y] := False;
pair1[{___, x_, ___, x_, ___, y_, ___, y_, ___} /; x != y] := False;
pair1[{___, x_, ___, x_, ___, x_, ___, x_, ___}] := False;
pair1[{___, x_, ___, x_, ___, x_, ___}] := False;
pair1[{___, x_, ___, x_, ___}] := True;(*a pair*)
pair1[{___}] := False;
pairQ[...
2
I agree, all the forms should result in the same behaviour. That the Switch/Which forms fail for RandomVariate I regard as a bug.
The whole point of If,Which and Switch possessing this dual nature of being both control-flow conditionals and potentially functions is that they can be treated like the latter in cases like this. In other words, the purpose of ...
0
Many thanks for all the helpful comments. On the variance, if you don't subtract Sigma^2/2 then the mean will not be mu. The variance I meant is the variance of log returns not the variance of returns, so that is OK.
I tried the numerical integration and it gives me useful results which I can plot and make sense of, which is very helpful. An analytical ...
1
This can be done numerically:
f[\[Mu]_?NumericQ, \[Sigma]_?NumericQ] := NExpectation[(2 Sqrt[x])/(x + 1) - 1,
x \[Distributed] LogNormalDistribution[ (\[Mu] - \[Sigma]^2/2), \[Sigma]]];
f[1, 2]
-0.295453
0
It's very slow, but this is my brute force way (ignores draw order otherwise remove the Sort):
cards = Flatten[Outer[List, Range@13, {♡, ♠, ♢, ♣}], 1];
numbers = cards[[All, 1]];
consecutive[hand_, n_] := Count[Differences@Sort@hand, 1] >= n - 1
numconscv = CountsBy[Subsets[numbers, {6}],consecutive[#,3]&];
numconscv[True]/Total[Values@numconscv]
(* ...
0
Try first the simpler problem: solve for p, for example
NSolve[CDF[BinomialDistribution[50, p], 35] == 0.025, p] // Quiet
The solution is obtained right away {{p -> 0.821382}}
1
By default, Mathematica appears to use the "MethodOfMoments" method to calculate ARProcess parameters. This is involves calculating the lagged auto-covariances of the input data. However, these are likely calculated by calling the Mathematica "CovarianceFunction" function. The definition given for the covariance calculation (under "...
1
A faster option than the one in the comment
flushQ[hand_] := MatchQ[hand/100 // Floor, {___, x_, x_, x_, x_, x_, ___}]
Count[sortedHands, _?flushQ]/Length[sortedHands]
(* 20889/1059380 *)
7
Very similar:
Probability[
control < var, {control \[Distributed] BetaDistribution[24, 141],
var \[Distributed] BetaDistribution[30, 151]}]
1191614106688032995829016253297371 / 1700223091652404809206230640390474
(roughly 0.7)
Verify numerically:
control = BetaDistribution[24, 141]
var = BetaDistribution[30, 151]
Count[Thread[
RandomVariate[control, ...
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