Probability distribution of a multiple variable sum

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Rémy Bretin
Rémy Bretin 2019-5-10
回答： Rémy Bretin 2019-5-14
Hi everyone,
I’m coming here for really advance statistic/probability advice, which I'm a beginner in this field.
I would like to know the probability of a variable TAU_total such as TAU_total=TAU1+TAU2+….+TAU129.
The variables TAUi are independent of each other.
For each one of them, I have a sample of 20,000 values which you can see some examples of their distribution on the histograms in the attachment.
My question is the following: I would like to be able to determine the probability of TAU_total to be superior to a certain value Xmax.
Thank you for your help,
Regards,
Rémy
stat.png
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Walter Roberson
Walter Roberson 2019-5-11
The Wikipedia article about CLT talks about extensions beyond iid.
I did a quick simulation of size equal to the original question, using randn with a range of standard deviations. std() of the totals was roughly 10% larger than sum() of the individual std, divided by sqrt(129) . The calculations for iid where thus not exactly applicable, but they were pretty close. hist() of the total looks like model illustrations of a drawing a pure normal distribution until I got up to 56 bins in the histogram, at which point you could finally start to see statistical differences compared to a perfect curve.
John D'Errico
John D'Errico 2019-5-11
Admittedly, when I first saw this question, I read it as the sum of 12 terms, not 129. 129 terms will cause pretty much anything to look as if it is normally distributed. :)
N = 129;
alph = rand(1,N)*2 + .5;
bet = rand(1,N)*2 + .5;
betamean = alph./(alph + bet);
betavar = alph.*bet./((alph + bet).^2.*(alph+ bet+1));
CLTmean = sum(betamean);
CLTvar = sum(betavar);
CLTstd = sqrt(CLTvar);
nsim = 100000;
X = zeros(nsim,N);
for i = 1:N
X(:,i) = betarnd(alph(i),bet(i),[1,nsim]);
end
MCsum = sum(X,2);
MCmean = mean(MCsum);
MCvar = var(MCsum);
MCstd = std(MCsum);
[CLTmean, MCmean;CLTvar,MCvar;CLTstd,MCstd]
ans =
61.6267682855143 61.6366984653151
7.56366115010423 7.53757246937701
2.75021111009759 2.74546398071018
Comparing the histograms, we see:
histogram(MCsum,100,'norm','pdf')
hold on
fplot(@(x) normpdf(x,CLTmean,CLTstd),[min(MCsum),max(MCsum)],'r')
untitled.jpg
So I don't see any problem using either approach. With only 12 terms in the sum, I'd probably go with the Monte Carlo.

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采纳的回答

Torsten
Torsten 2019-5-10
编辑：Walter Roberson 2019-5-10
Take samples from your empirical data and use Monte-Carlo-simulation to determine the above probability.
This code should help to take the samples:
https://de.mathworks.com/matlabcentral/fileexchange/7976-random-number-from-empirical-distribution

更多回答（1 个）

Rémy Bretin
Rémy Bretin 2019-5-14
Thank you everyone for your support.
I decided to go for the MonteCarlo method which gave me a gaussien at the endas you expected.

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