- #36

fab13

- 312

- 6

@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

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)

##

##\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.

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: