A348913 Number of N(0,1) variables needed for a 1/2 chance that their maximum exceeds the maximum of n N(1,1) variables.

4, 11, 19, 29, 40, 52, 64, 78, 92, 106, 122, 137, 154, 171, 188, 205, 224, 242, 261, 280, 300, 320, 340, 360, 381, 402, 424, 446, 468, 490, 512, 535, 558, 581, 605, 629, 652, 677, 701, 726, 750, 775, 801, 826, 852, 877, 903, 929, 956, 982, 1009

Offset: 1

Author

Matt Frank, Nov 03 2021

Keywords

• nonn
• more

Comments

The above data has only 51 entries because the Mathematica code below would give numerically unstable results for the integrals with n>51.

Examples

			For n = 2, a(2) = 11 because if X_i are standard normal variables, and Y_j are normal variables with mean and variance 1, then P[max(X_1, ... X_10) > max(Y_1, Y_2)] = 49% but P[max(X_1, ... X_11) > max(Y_1, Y_2)] = 51%.
		

Links

• Matt Frank, Answer to "Numerical superiority necessary to beat in L^oo a population one standard deviation ahead", on CrossValidated (aka stats.stackexchange.com)

Programs

Mathematica

cdf[x_] := CDF[NormalDistribution[], x];
prob[m_] := NIntegrate[D[cdf[x]^m, x] cdf[x - 1]^n , {x, -Infinity, Infinity}]
Table[FindArgMin[(prob[m] - 1/2)^2, {m, n^1.75}] // Ceiling, {n, 51}]