Calculating the variance of integrated Poisson noise on a defined quantity

In summary, the conversation discusses the estimation of the variance expression of Poisson noise on a quantity that is calculated through a sum of terms. The formula for the standard deviation of the noise includes a factor of (2l+1) which leads to discrepancies in the results. Various expressions are proposed and tested, but a definitive justification for the correct formula is not yet provided. Suggestions are made to use a symbolic algebra program to ensure reliable results.
  • #36
@Twigg . Thanks for your advices.

1) Unfortunately, I have already tested in my previous post your "third observable" like this :

##O=\dfrac{\sum_{\ell=1}^{N} \sum_{m=-\ell}^{\ell} a_{\ell m}^{2}}{\sum_{\ell=1}^{N} \sum_{m=-\ell}^{\ell}\left(a_{\ell m}^{\prime}\right)^{2}}=\dfrac{\sum_{\ell=1}^{N}(2\ell+1)\,C_{\ell,sp}}{\sum_{\ell=1}^{N}(2\ell+1)\,C_{\ell,ph}}##

But it gives a lower FoM (1443) than for the first observable :

##\begin{gathered}
O=\left(\frac{\mathcal{D}_{s p}}{\mathcal{D}_{p h}}\right) \\
\mathcal{D}_{s p}=\sum_{i=l}^{n} C_{\ell, s p}\left(\ell_{i}\right) \\
\mathcal{D}_{p h}=\sum_{i=l}^{n} C_{\ell, p h}\left(\ell_{i}\right)
\end{gathered}##

with which I get FoM = 1556.

2) Jus an important point to be more explicit :

From the beginning, I try to use Inverse-variance weighted :

Given a sequence of independent observations ##y_{i}## with variances ##\sigma_{i}^{2}##, the inversevariance weighted average is given by :
##
\hat{y}=\frac{\sum_{i} y_{i} / \sigma_{i}^{2}}{\sum_{i} 1 / \sigma_{i}^{2}}\quad(1)
##
The inverse-variance weighted average has the least variance among all weighted averages, which can be calculated as
##
\operatorname{Var}(\hat{y})=\dfrac{1}{\sum_{i} 1 / \sigma_{i}^{2}}\quad(2)
##
In our case, let us take the quantity ##\hat{y}=C_{\ell}## which is the average of ##a_{\ell m}^{2}## and so the variance of an ##a_{\ell m}## :
##
C_{\ell}=\left\langle a_{l m}^{2}\right\rangle=\dfrac{1}{2 \ell+1} \sum_{m=-\ell}^{\ell} a_{\ell m}^{2}=\operatorname{Var}\left(a_{l m}\right)\quad(3)
##
If we take all ##\sigma_{i}^{2}## equal to :
##
\sigma_{i}^{2}=\frac{2}{f_{sky} N_{p}^{2}}\quad(4)
##
Then we get an optimum variance regarding the Shot Noise. Let's demonstrate it with shot noise equal to the inverse of galaxies density : ##N_{p}=N_{gal}##. For a given multipole ##\ell## given, we have :

##
\operatorname{Var}(\text { Shot Noise optimal })(\ell)=\dfrac{1}{\sum_{m=-\ell}^{\ell} \dfrac{1}{\sigma_{i}^{2}}}
=\dfrac{1}{(2 \ell+1) \dfrac{f_{s k y} N_{p}^{2}}{2}}
=\dfrac{2}{(2 \ell+1)\left(f_{s k y} N_{p}^{2}\right)}\quad(5)
##

QUESTION : the choice made in eq##(4)## has been infered from the following relation expressing the standard deviation on a ##C_\ell## :

##\sigma_{C_\ell}(\ell)=\sqrt{\dfrac{2}{(2 \ell+1) f_{\mathrm{sky}}}}\left[C_\ell(\ell)+\dfrac{1}{N_{p}}\right]\quad(6)##

1) How can I justify the fact that I decide to assimilate ##\hat{y}## to a ##C_\ell## ?.

##C_\ell## is a quantity appearing in spherical harmonics but this is already an average of ##a_{\ell m}^2## : is it pertinent from your point of view ?

Moreover, I don't make appear nowhere the formula eq##(1)##, I take it as it was right but I am not sure in my case with the average ##C_\ell##.

2) Is the derivation of formula eq##(4)## correct from eq##(6)## ?

As you can see, I need clarifications.
 
Last edited:
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  • #37
fab13 said:
@Twigg##C_\ell## is a quantity appearing in spherical harmonics but this is already an average of ##a_{\ell m}^2## : is it pertinent from your point of view ?
That is extremely pertinent. Each ##C_\ell## represents ##2\ell+1## variables.
 

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