A288347 Median of X^2 + Y^2 where X and Y are independent random variables with B(n, 1/2) distributions.

1, 2, 5, 9, 13, 20, 25, 34, 41, 52, 61, 74, 85, 100, 116, 130, 149, 164, 185, 202, 225, 244, 269, 290, 317, 340, 369, 394, 425, 452, 485, 520, 549, 585, 617, 653, 689, 730, 765, 808, 845, 890, 929, 976, 1017, 1066, 1109, 1160, 1205, 1258, 1305, 1360, 1409

Offset: 1

Author

Matt Frank, Jun 08 2017

Keywords

• nonn

Comments

Interpretation: Start at the origin, and flip a pair of coins. Move right one unit if the first coin is heads, and otherwise stay in place. Then move up one unit if the second coin is heads, and otherwise stay in place. This sequence gives your median squared-distance from the origin after n pairs of coin flips.

Although a median of integers can be a half-integer, as an empirical observation only integers appear in this sequence.

The mean of X^2 + Y^2 is (n^2+n)/2, or A000217.

From Robert Israel, Oct 04 2017: (Start)

To avoid the possibility of half-integer values, the median can be taken as the least integer v such that Probability(X^2 + Y^2 <= v) >= 1/2.

All terms are in A001481.

Using the Central Limit Theorem, 4*(X^2+Y^2)/n has approximately a noncentral chi-square distribution with 2 degrees of freedom and noncentrality parameter 2*n. Thus integral_{t=0..4*a(n)/n} exp(-n-t/2) BesselI(0,sqrt(2*n*t)) dt is approximately 1.

Since a random variable not too far from normal has median approximately mu - gamma*sigma/6 where mu, sigma and gamma are the mean, standard deviation and skewness, we should expect a(n) to be approximately n^2/2 + n/4.

(End)

Links

• Robert Israel, Table of n, a(n) for n = 1..800

Crossrefs

Cf. A288416, which is similar, with shifted coordinates; and also A288346, which is multiplicative rather than additive.

Programs

Maple

f:= proc(n) local S,P,i,j,q;
  S:= sort( [seq(seq([i,j],i=0..n),j=0..n)],(a,b) -> a[1]^2 + a[2]^2 < b[1]^2 + b[2]^2);
  P:= ListTools:-PartialSums(map(t -> binomial(n,t[1])*binomial(n,t[2])/2^(2*n), S));
  q:= ListTools:-BinaryPlace(P,1/2);
  if P[q] = 1/2 then S[q][1]^2 + S[q][2]^2
  else S[q+1][1]^2 + S[q+1][2]^2
  fi
end proc:
map(f, [$1..80]); # Robert Israel, Oct 04 2017

Mathematica

Squared[x_] := x^2;
WeightsMatrix[n_] := Table[Binomial[n, i] Binomial[n, j], {i, 0, n}, {j, 0, n}]/2^(2 n);
ValuesMatrix[n_, f_] := Table[f[i] + f[j], {i, 0, n}, {j, 0, n}];
Distribution[n_, f_] := EmpiricalDistribution[Flatten[WeightsMatrix[n]] -> Flatten[ValuesMatrix[n, f]]];
NewMedian[n_, f_] := Mean[Quantile[Distribution[n, f], {1/2, 1/2 + 1/2^(2 n)}]];
Table[NewMedian[n, Squared], {n, 53}]