# What formula is used for std in vertex_average and edge_average?

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()
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

0.3535 is sqrt(2)/4 which happens to be σ(n-1)/2, so it seems there is some
relation to that.

A little bigger graph.

t3=gt.Graph()
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 ?

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.

Hi there,

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()
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.

t3=gt.Graph()
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

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

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