From: Donald Darling <d.darling@home.com>
Subject: Re: Q : comp proba ??
Date: 9 May 1999 23:30:02 -0500
Newsgroups: sci.math.research
Keywords: probability that all initial sums are less than mean

The probability is 1/2^(2n)*BinomialCoefficient(2n,n).  This result is
true for any random variables having a symmetric density. Cf. D. A.
Darling "Sums of symmetrical random variables" Proc. Amer. Math. Soc. 2
(1951) PP 511-517.

Rune Kleveland wrote:
> 
> tayeb.amegroud@creditlyonnais.co.uk (Tayeb Amegroud) writes:
> 
> > let X1,X2, ... Xn be n Gaussian Random Variables iid N(0,1)
> > I'm wondering if there is a closed formula to compute the probability
> > of the following event :
> > { X1 < 0, X1+X2 < 0, X1+X2+X3 < 0 , .... , X1+X2+...+Xn < 0 }
> > I performed a Monte Carlo Simulation, and get an asymptotic formula
> > when n goes to infinity : 1/pow(n,0.6) !!!!!
> 
> The normal distribution has the well-known property that the
> corresponding n-dimensional distribution is rotation invariant.  It
> follows that the problem is equivalent to computing the amount of the
> n-dimensional sphere that satisfies
> 
> x_1<0, x_1+x_2<0, ... , x_1+...x_n<0
> 
[...]
> This led me to the conjecture that the probabilities are rational, and
> that
> 
> p_n = (1-1/2)*(1-1/4)*...*(1-1/2n) = gamma(n+1/2)/sqrt(pi)*gamma(n+1)
==============================================================================
[Remark: 1/2^(2n)*BinomialCoefficient(2n,n)  is, by Stirling's approximation,
 about  (1/sqrt(pi))*(1/pow(n,0.5))  in the first poster's notation. --djr]














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