any logic to do this programming on random number
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Hi,
I have generated 1000 random numbers from a normal distribution with std deviation=0.10 and mean=0.36
like
r=0.36+0.10*randn(1,1000)
now, I want to select 10 numbers from this random numbers, the summation of which will be in between 3.55 to 3.65..
I need 30 sets of such 10 numbers which sum will lie between this..3.55 to 3.65
any logic to do so
thanks
2 commentaires
Geoff Hayes
le 11 Juin 2014
Modifié(e) : Geoff Hayes
le 11 Juin 2014
Note that the above equation is not quite correct (or your standard deviation and mean are reversed). If std is 0.1 and the mean is 0.36, then
r = 0.36+0.1*randn(1,1000);
Try std(r) and mean(r) to verify this.
joy
le 11 Juin 2014
Réponse acceptée
Plus de réponses (1)
Sean de Wolski
le 11 Juin 2014
Modifié(e) : Sean de Wolski
le 11 Juin 2014
I would take a different approach. First generate the 30 numbers who you must sum to:
xsum = rand(1,30)+3.55;
Example:
%%Desired sums
xsum = rand(1,30)+3.55;
%%build random numbers
rn = zeros(10,30);
for ii = 1:30
rn(:,ii) = randfixedsum(10,1,xsum(ii),0,xsum(ii));
end
%%verify
assert(norm(sum(rn)-xsum)<10^-14)
3 commentaires
John D'Errico
le 11 Juin 2014
Modifié(e) : John D'Errico
le 11 Juin 2014
A flaw here is the numbers won't be at all normally distributed. Given the bound constraint on the mean, they won't be normally distributed anyway, but this gives up on any semblance of normality.
Sean de Wolski
le 11 Juin 2014
Ahh! Good catch.
Sean de Wolski
le 11 Juin 2014
I'll have to think about it a little more.
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