Introduction of a factor Δℓ when summing equal distants 𝐶ℓ

  • #1
fab13
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TL;DR Summary
I would like to demonstrate the formula that defines the standard deviation of a 𝐶ℓ and mostly justify the presence of the factor Δℓ when we have a finite number of 𝐶ℓ's angular power spectrum and associated finite number of multipole ℓ (generated by a code ). I tried at the end of my post to calculate the mean Cℓ for a single bin but it is missing a factor 2. Any help would be really appreciated. This is a problem about the theorical/numerical frontier given the fact I have not Δℓ=1.
Hello,

In the context of Legendre expansion with ##C_\ell## quantities, below 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.

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 :

##\Delta\ell=(4990/60)##.

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.

Could anyone help me to justify the using of this factor ##\Delta\ell## ?

UPDATE : Someone tried quickly to explain why we introduce this factor ##\Delta\ell## inside the square root, but I am still confused about this demonstration.

Here is his reasoning :

For the interval ##b##: ##b=[\ell, \ell+\Delta\ell]##, one has :

##C_{\ell,b}=\dfrac{1}{\Delta\ell}\,\sum_{\ell'=\ell}^{\ell'+\Delta\ell}\,C_{\ell'}##

If one takes the variance, then :

##\text{Var}(C_{\ell,b})=\dfrac{1}{\Delta\ell^2}\,\sum_{\ell'}\text{Var}(C_{\ell'})##

##\text{Var}(C_{\ell,b})=\dfrac{1}{\Delta\ell^2}\,\sum_{\ell'}\dfrac{2}{(2\ell'+1)}\,(C_{\ell'})\quad(2)##

Important step here, one considers the equality (which is actually an approximation) :

##\sum_{\ell'}\dfrac{2}{(2\ell'+1)}\,(C_{\ell'})=\dfrac{2}{(2\ell_{mean}+1)}\,(C_{\ell'_{mean}}^2)\,\Delta\ell\quad(3)##

This way, we have from equation ##(2)## :

##\text{Var}(C_{\ell,b})\simeq \dfrac{1}{\Delta\ell^2}\,\dfrac{2}{(2\ell_{mean}+1)}\,(C_{\ell'_{mean}}^2)\,\Delta\ell##

Finally the expression of the variance for the interval ##b##:

##\text{Var}(C_{\ell,b})\simeq \dfrac{1}{\Delta\ell}\,\dfrac{2}{(2\ell_{mean}+1)}\,(C_{\ell'_{mean}}^2)##

I still didn't understand the step between ##(2)## and ##(3)##.

For me, an average is under the form :

##C_{\ell'_{mean}}=\dfrac{1}{2}\,(C_{\ell'}+C_{\ell'+\Delta\ell})## but I can't see this factor ##\dfrac{1}{2}## in this reasoning.

If someone could explain this critical step between ##(2)## and ##(3)##.
 
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  • #2
The average ##\bar x## of ##N## different quantities ##x_i## satisfies ##\sum x_i = \bar x N##. This is what is being used.
 
  • #3
@Orodruin : thanks for your quick answer.

Could you be please more explicit with the notations I took ? your formula is trivial but I have difficulties to apply it with my notations.

Moreover, do you give a reasoning on a single bin of width ##\Delta\ell## or on the total interval ##[\ell_{min},\ell_{max}]## ?

In the formula ##\sum x_i = \bar x N##, does ##N## correspond to ##\Delta\ell## in my formula ?

Thanks in advance, Regards
 
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  • #4
Isn't really there nobody who could explain me my problem of understanding about my last message above ?
 
  • #5
fab13 said:
Summary:: I would like to demonstrate the formula that defines the standard deviation of a 𝐶ℓ and mostly justify the presence of the factor Δℓ when we have a finite number of 𝐶ℓ's angular power spectrum and associated finite number of multipole ℓ (generated by a code ). I tried at the end of my post to calculate the mean Cℓ for a single bin but it is missing a factor 2. Any help would be really appreciated. This is a problem about the theorical/numerical frontier given the fact I have not Δℓ=1.

For me, an average is under the form :

Cℓmean′=12(Cℓ′+Cℓ′+Δℓ) but I can't see this factor 12 in this reasoning.

If someone could explain this critical step between (2) and (3).
Can you not work it "backwards" to see what N must be in this situation? You have the answer provided to you, It is worrisome that this step eludes you.
 
  • #6
Hi everyone,

It seems that I may have a "demonstration" about my issue of understanding :

If i take the expression for the bin "##b##" ##[\ell_b,\ell_b+\Delta\ell_b]## :

##C_{\ell,b}=\dfrac{1}{\Delta\ell_b}\,\sum_{\ell=\ell_b}^{\ell_b+\Delta\ell_{b}}\,(C_{\ell})##

Then :

##\text{Var}(C_{\ell,b})=\dfrac{1}{\Delta\ell_b^2}\,\sum_{\ell=\ell_b}^{\ell_b+\Delta\ell_{b}}\,\text{Var}(C_{\ell})\quad(2)##

By taking ##\Delta\ell=1##, we can have :

##\text{Var}(C_{\ell_b})=\dfrac{1}{\Delta\ell_{b}^2}\,\sum_{\ell=\ell_b}^{\ell+\Delta\ell}\dfrac{2}{(2\ell+1)}\,(C_{\ell}^2)\Delta\ell \simeq \int_{\ell=\ell_{b}}^{\ell_{b}+\Delta\ell_b} \dfrac{1}{\Delta\ell_{b}^2} \dfrac{2}{2 \ell+1}\,C_\ell^2\text{d}\ell \quad(3)##

and this integral can be approximated by rectangular integration method :

##\int_{\ell=\ell_{b}}^{\ell_{b}+\Delta\ell_b} \dfrac{1}{\Delta\ell_{b}^2} \dfrac{2}{2 \ell+1}\,C_\ell^2\text{d}\ell \simeq \dfrac{1}{\Delta\ell_{b}^{2}} \dfrac{2}{(2\ell_{b}+1)} \,C_{\ell_{b}}^{2} \Delta{\ell_{b}} =\quad(4)##

where ##\Delta{\ell_{b}}## is the delta width of bin "##b##" :

then, I could write :

##\text{Var}(C_{\ell,b})=\dfrac{1}{\Delta\ell_{b}} \dfrac{2}{(2 \ell_{b}+1)}\,C_{\ell_b}^{2} \quad(5)##

Do you think my reasoning is right ?
 
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