My thanks to Warren, Gareth, Gene --- I will digest their replies. In the meantime, I realised that I conflated discrete (Pascal triangle) and continuous (binomial distribution) problems: the answers are not quite the same. For the continuous case instead I find experimentally n_0 = k^2 - 4/3 - (32/45)/k^2 - (4/9)/k^4 + O(1/k^6) . WFL On 6/28/14, Eugene Salamin via math-fun <math-fun@mailman.xmission.com> wrote:
In the Gaussian approximation, normalized to mean 0 and variance 1, we have
exp(-(x-s/2)^2/2)) + exp(-(x+s/2)^2/2)) = 2 exp(-(x^2+s^2)/2) cosh sx.
This has extrema where the derivative of the log is 0.
s tanh sx = x.
When s < 1, there is a single maximum at 0. When s > 1, there is a minimum at 0 and 2 maxima. The critical point is s = 1, when there is a triple root. That is, the critical shift s is 1 standard deviation, which for the binomial(n,k) is sqrt(n)/2.
-- Gene
________________________________ From: Fred Lunnon <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Friday, June 27, 2014 11:22 AM Subject: [math-fun] Pascal triangle row: shift-and-add
What happens when row n of the Pascal triangle is shifted by k then added to itself? Or in continuous terms, two binomial distributions, identical apart from the shift along the x-axis, are summed?
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 ;
indeed this bound estimate is remarkably good, being only <0.1 too large at k = 2 , and <0.001 too large at k = 20 .
No doubt probability wonks know all about this stuff already: so can anyone point me at a reference to a proof of the bound; or alternatively explain why it's perfectly obvious?
Fred Lunnon
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