I am curious what is being used to calculate the standard deviation of the average in gt.vertex_average and gt.edge_average
>>> t2=gt.Graph() >>> t2.add_vertex(2) >>> t2.add_edge(t2.vertex(0), t2.vertex(1)) >>> gt.vertex_average(t2, "in") (0.5, 0.35355339059327373) Now, shouldn't std be σ(n)=sqrt(((00.5)^2+(10.5)^2)/2)=0.5 ? also q(n1)=sqrt((0.5^2+0.5^2)/(21))~=0.70710 0.3535 is sqrt(2)/4 which happens to be σ(n1)/2, so it seems there is some relation to that. A little bigger graph. >>> t3=gt.Graph() >>> t3.add_vertex(5) >>> t3.add_edge(t3.vertex(0), t3.vertex(1)) >>> gt.vertex_average(t3, "in") (0.2, 0.17888543819998318) Now, we should have 0,1,0,0,0 series for vertex incoming degree. So Windows calc gives σ(n)=0.4 and σ(n1)~=0.44721, so where does 0.1788854 come from ? Reason, I am asking because, I have a large graph, where the average looks quite alright but the std makes no sense, as going by the histogram, degree values are quite a bit more distributed than the std would indicate. 
Administrator

Hi there,
On 05/21/2013 01:37 PM, VaSa wrote: > I am curious what is being used to calculate the standard deviation of the > average in gt.vertex_average and gt.edge_average These functions return the standard deviation of *the mean* not the standard deviation of the distribution, which is given by, \sigma_a = \sigma / sqrt(N) where \sigma is the standard deviation of the distribution, and N is the number of samples. >>>> t2=gt.Graph() >>>> t2.add_vertex(2) >>>> t2.add_edge(t2.vertex(0), t2.vertex(1)) >>>> gt.vertex_average(t2, "in") > (0.5, 0.35355339059327373) > > Now, shouldn't std be σ(n)=sqrt(((00.5)^2+(10.5)^2)/2)=0.5 ? > also q(n1)=sqrt((0.5^2+0.5^2)/(21))~=0.70710 The standard deviation of the mean is therefore: 0.5 / sqrt(2) = 0.35355339059327373... which is what you see. > A little bigger graph. >>>> t3=gt.Graph() >>>> t3.add_vertex(5) >>>> t3.add_edge(t3.vertex(0), t3.vertex(1)) >>>> gt.vertex_average(t3, "in") > (0.2, 0.17888543819998318) > > Now, we should have 0,1,0,0,0 series for vertex incoming degree. > So Windows calc gives σ(n)=0.4 and σ(n1)~=0.44721, so where does 0.1788854 > come from ? > Reason, I am asking because, I have a large graph, where the average looks > quite alright but the std makes no sense, as going by the histogram, degree > values are quite a bit more distributed than the std would indicate. If you want the deviation of the distribution to compare with the histogram, just multiply by sqrt(N). Cheers, Tiago  Tiago de Paula Peixoto <[hidden email]> _______________________________________________ graphtool mailing list [hidden email] http://lists.skewed.de/mailman/listinfo/graphtool signature.asc (567 bytes) Download Attachment

Tiago de Paula Peixoto <tiago@skewed.de> 
Hi, there
I had the same problem. This topic answered me what I wanted, but I have a doubt: Why this calculation is more importante/often then just standard deviation of the distribution? It is just a curiosity because I never saw that measurement :) Thanks, Éverton  Sent from: https://nabble.skewed.de/ _______________________________________________ graphtool mailing list [hidden email] https://lists.skewed.de/mailman/listinfo/graphtool 
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Am 16.07.20 um 21:45 schrieb Éverton Fernandes da Cunha:
> Hi, there > > I had the same problem. This topic answered me what I wanted, but I have a > doubt: Why this calculation is more importante/often then just standard > deviation of the distribution? Because we want to express the uncertainty of the mean, not of the distribution. > It is just a curiosity because I never saw that measurement :) https://en.wikipedia.org/wiki/Standard_deviation#Standard_deviation_of_the_mean  Tiago de Paula Peixoto <[hidden email]> _______________________________________________ graphtool mailing list [hidden email] https://lists.skewed.de/mailman/listinfo/graphtool

Tiago de Paula Peixoto <tiago@skewed.de> 
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