[Rivet] Rivet "collaboration meeting"?

[Leif Lönnblad]

On 2013-05-21 16:50, Frank Siegert wrote:
> Why is this not using the normal variance a la
> s^2 = 1/(n-1) <(w-<w>)^2>
>     = 1/(n-1) (<w^2> - <w>^2)
>     = 1/(n-1) (sum(w^2)/n - sum(w)^2/n^2)


If we use this, the error will be zero for uniform weights...

For uniform weights, the number of entries in a bin is typically given 
by a Poissonian distribution , which means that the variance is the same 
as the mean, so the error on the number of entries is given by 
sqrt(number of entries).

For weighted events, one way of thinking is the following: Imagine there 
is a discrete set of weights w_i. For each of these there are n_i 
entries giving the error w_i*sqrt(n_i) for the sum of weight, n_i*w_i. 
The total height of the histogram is sum_i(n_i*w_i), and the error in 
that number is the square root of the sum of the squared errors for each 
w_i: sqrt(sum_i(n_i*w_i^2)). This generalizes to the error 
sqrt(sum(w^2)) for the bin height sum(w).

Of course, this "derivation" only holds for a large number of entries in 
a large number of bins, but it gives a reasonable error estimate also 
for a few entries per bin. But maybe there is a better estimate out there...

/Leif