On 27/06/2014 19:22, Fred Lunnon wrote:
Consider k fixed and n increasing: initially
n_C_i + n_C_j where i+j = n+k
is bimodal; but at some point n_0 it becomes unimodal and remains so for n > n_0.
Numerical experiments suggest that
n_0 ~ k^2 - 5/3 ;
First thought: For large n we have the familiar approximation that says binomial(n,p) is close to normal(np,npq) and, in particular, binomial(n,1/2) is close to normal(n/2,n/4). (The second parameter being variance, not stddev.) Now, the sum of two standard normals separated by 2t is (up to an irrelevant multiplicative constant) f(x) = exp(-(x+t)^2/2) + exp(-(x-t)^2/2) so f'(x) = - (x+t) exp(-(x+t)^2/2) - (x-t) exp(-(x-t)^2/2) f''(x) = (x+t)^2 exp(-(x+t)^2/2) + (x-t)^2 exp(-(x-t)^2/2) - exp(-(x+t)^2/2) - exp(-(x-t)^2/2) so f''(0) = [2t^2 - 2] exp(-t^2/2) and therefore switches from unimodal to bimodal at t=1. So the sum of two normals with stddev sqrt(n/4) switches from unimodal to bimodal when the separation is 2sqrt(n/4) = sqrt(n). Or, holding the separation fixed at k, when the stddev is sqrt(k^2/4), i.e., when n=k^2. Of course this requires more rigour to verify that the approximation is close enough; perhaps supplying that rigour might naturally lead to some of your higher-order terms, though I wouldn't count on it. Second thought: Wait, we can do this by hand. The question is whether the term at i=j=(n+k)/2 is bigger or smaller than its (equal) neighbours. In other words, we need to compare 2nCi with nC(i-1)+nC(i+1) -- to see when nC(i-1)/nCi + nC(i+1)/nCi crosses 2. Well, that quantity is simply i/(n-i+1) + (n-i)/(i+1) and if I've done my algebraic scribblings correctly the condition simplifies to k^2 = n+2. That doesn't match your more complicated formula. Are you considering i,j to be continuous variables and asking when there's any dip in the middle -- i.e., when the second derivative at (n+k)/2 changes sign? In that case I suppose we get something a bit hairier. If some more dubious algebraic scribblings are correct, the condition is (f1((n+k)/2) - f1((n-k)/2))^2 = f2((n+k)/2) + f2((n-k)/2) where f1 is the logarithmic derivative of the factorial function (i.e., digamma offset by 1) and f2 is f1'. Those more fluent than I with "special functions" may find it all obvious at this point :-). -- g