A Expression of Shot noise when expanding ##a_{\ell m}## coefficients

AI Thread Summary
The discussion focuses on deriving the expression for the quantity o_ℓ in the context of dark matter and Poisson noise. The proposed formula is o_ℓ = b_sp² C_ℓ^{DM} + B_sp, where B_sp represents the Poisson noise defined as B_sp = 1/𝑛̄. The expansion of o_ℓ leads to the conclusion that the term <(a_ℓm^P)²> must be justified as equal to 1/𝑛̄, raising questions about its relationship to variance. Clarification is needed on how B_sp relates to the variance of a non-centered Poisson distribution, given that <a_ℓm^P> does not equal zero. Ultimately, the total signal C_ℓ is expressed as C_ℓ = b_sp² C_ℓ^{DM} + B_sp.
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TL;DR Summary
I would like to prove that Shot noise follows a Poisson distribution.
I would like to arrive at the following expression for the quantity ##o_{\ell}## ( with "DM" for Dark Matter ):

##o_{\ell}=b_{s p}^2 C_{\ell}^{D M}+B_{s p}##

with Poisson noise ##B_{s p}=\frac{1}{\bar{n}}(\bar{n}## being the average number of galaxies observed). the index "sp" is for spectro. I think for now that ##B_{s p}## is the variance of a Poisson noise but see the following below to really confirm: To arrive at this same expression, I would like to start from ##{ }_{\ell m}^{a D M}## (DM for Dark matter) and ##a_{\ell m}^P## (" ##\mathrm{P}## " for fish).
So I start from the fact that ##C_{\ell}=\operatorname{Var}\left(a_{\ell m}\right)## :

##o_{\ell}=<\left(b_{s p} a_{\ell m}^{D M}+a_{\ell m}^P\right)^2>##

If we expand, we have: ##o_{\ell}=<b_{s p}^2\left(a_{\ell m}^{D M}\right)^2+2 b_{s p} a_{\ell m}^{D M}+\left(a_{\ell m}^P\right)^2>##

##o_{\ell}=b_{s p}^2 C_{\ell}^{D M}+2 b_{s p}<a_{\ell m}^{D M}><a_{\ell m}^P>+<\left(a_{\ell m}^P\right)^2>##

##=b_{s p}^2 C_{\ell}^{D M}+<\left(a_{\ell m}^P\right)^2>##

because we have ##<a_{\ell_m}^{D M}>=0##

The problem comes from the term ##<\left(a_{\ell m}^P\right)^2>## : I don't know how to justify that this term is equal to ##\frac{1}{\bar{n}}##

Indeed, if ##B_{s p}## is a fish noise, we should have, to make the correspondence, ##B_{s p}=<\left(a_{\ell m}^P\right)^2>-<## ##a_{\ell m}^P>2## which is different from: ##B_{s p}=<\left(a_{\ell m}^P\right)^2>=\operatorname{Var}\left(a_{\ell m}^P\right)##.

How to obtain the quantity ##B_{s p}## which seems a priori equal to ##\frac{1}{\bar{n}}## ?

If ##B_{s p}## is equal to ##<\left(a_{\ell m}^P\right)^2>##, how to make the link with a variance since a Poisson law is not centered ( I mean ##<a_{\ell m}^P>\neq 0## ?
 
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There is a typo showed in attachment : the factor "2" is acutally an exponent in ##<a_{\ell m}^P>^2##.
 

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Also (sorry), the initial quantity at the beginning ( ##o_{\ell}##) is simply the total signal ##C_\ell## :

##o_{\ell}=b_{s p}^2 C_{\ell}^{D M}+B_{s p}##

is equal to :

##C_{\ell}=b_{s p}^2 C_{\ell}^{D M}+B_{s p}##
 
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