Calculating the variance of integrated Poisson noise on a defined quantity

[fab13]

@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.

 

[kimbyd]

@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.