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.

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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(((0-0.5)^2+(1-0.5)^2)/2)=0.5 ? also q(n-1)=sqrt((0.5^2+0.5^2)/(2-1))~=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.

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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(n-1)~=0.44721, so where does 0.1788854 come from ?

Again, 0.4 / sqrt(5) = 0.17888543819998318...

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