I have the following formula which is the error on a $C_\ell$ :

$$\sigma_(C_{\ell})=\sqrt{\frac{2}{(2 \ell+1)\Delta\ell}}\,C_{\ell}\quad(1)$$

where $\Delta\ell$ is the width of the multipoles bins used when computing the angular power spectra $C_\ell$.

I would like to understand why this factor appears. Normally, we can infer the expression of $\sigma_(C_{\ell})$ and get :

$$\sigma_(C_{\ell})=\sqrt{\frac{2}{(2 \ell+1)}}\,C_{\ell}$$

That is to say, without having the $\Delta\ell$ factor under the root.

In my case, I have only a finite number of multipoles $\ell$ (60 exactly) in a range between 10 and 5000 and 60 corresponding $C_\ell$. From my tutor, to compute this standard-deviation, I have to take :


But I would like to understand why I have to use this factor when we have a finite number of equidistant multipoles with associated $C_\ell$ values.

For example, my tutor told me that, ideally, we would have a $C_\ell$ for each $\ell=1..N$, i.e with $\ell=1,2,3,4,....N$ with the formula :

$$\sigma(C_{\ell})^2=\frac{2}{(2 \ell+1)}\,C_{\ell}^{2}$$

But he mentionned that, like we have only a finite number of equidistant multipoles $\ell$, we are obliged to use this factor $\Delta\ell$ in our case (equation$(1)$).

My tutor told me it is like an average on the expression with multipole $\ell$, i.e on the factor $\sqrt{\dfrac{2}{(2\ell+1)}}$, but I didn't fully understand this meaning.

We could naively write :

for multipole interval $[\ell, \ell+\Delta\ell]$, we have :

$C_b=\dfrac{1}{\Delta\ell)}\sum\limits_{\ell'=\ell}^{\ell'=\ell+\Delta\ell} C_{\ell}$

and Variance follows (from equation $(1)$) :



$\simeq \dfrac{1}{\Delta\ell^2}\,\dfrac{2}{2\ell_{mean}+1}\,(C_{\ell,mean})^2\,\Delta\ell$

$\simeq \dfrac{2}{(2\ell_{mean}+1)\,\Delta\ell}\,(C_{\ell,mean})^2$

But I don't really undertand how $\ell_{mean}$ and $C_{\ell,mean}$ are obtained.

Classically, for a function $f(x)$, the mean is : $<f> = \dfrac{1}{N}\sum\limits_{i=1}^{N}\,f(x_i)$, isn't it ? How to make the link with above equations in order to introduce this factor $\Delta\ell$.

Any help would be really appreciated.


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