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- TL;DR Summary
- I want to estimate the variance expression of Poisson Noise of the quantity ##\sum_{\ell=1}^{N} \sum_{m=-\ell}^{\ell} a_{\ell m}^{2}##

It is in cosmology context but actually, but it is also a mathematics/statistical issue.

From spherical harmonics with Legendre deccomposition, I have the following definition of

the standard deviation of a ##C_\ell## noised with a Poisson Noise ##N_p## :

##

\begin{equation}

\sigma({C_\ell})(\ell)=\sqrt{\frac{2}{(2 \ell+1) f_{sky}}}\left[C_\ell(\ell)+N_{p}(\ell)\right]\quad(1)

\end{equation}

##

Now I consider the quantity : ##\sum_{\ell=1}^{N} \sum_{m=-\ell}^{\ell} a_{\ell m}^{2}##

I want to estimate the variance expression of Poisson Noise of this qantity.

For that, I take the definition of ##a_{lm}## following a normal distribution with mean equal to zero and take also the definition of a ##C_\ell=\langle a_{lm}^2 \rangle=\dfrac{1}{2\ell+1}\sum_{m=-\ell}^{\ell}\,a_{\ell m}^2 = \text{Var}(a_{lm})##.

I use ##\stackrel{d}{=}## to denote equality in distribution :

##

\begin{align}

Z&\equiv \sum_{\ell=\ell_{min}}^{\ell_{max}} \sum_{m=-\ell}^\ell a_{\ell,m}^2 \\[6pt]

&= \sum_{\ell=\ell_{min}}^{\ell_{max}} \sum_{m=-\ell}^\ell C_\ell \cdot \bigg( \frac{a_{\ell,m}}{\sqrt{C_\ell}} \bigg)^2 \\[6pt]

&\stackrel{d}{=}\sum_{\ell=\ell_{min}}^{\ell_{max}} \sum_{m=-\ell}^\ell C_\ell \cdot \chi^{2}(1) \\[6pt]

&= \sum_{\ell=\ell_{min}}^{\ell_{max}} C_\ell \sum_{m=-\ell}^\ell \chi^{2}(1) \\[6pt]

&= \sum_{\ell=\ell_{min}}^{\ell_{max}} C_\ell \cdot \chi^{2}(2 \ell + 1). \\[6pt]

\end{align}

##

So Finally we have :

##

\begin{aligned}

\sum_{\ell=1}^{N} \sum_{m=-\ell}^{\ell} a_{\ell m}^{2} & \stackrel{d}{=} \sum_{\ell=1}^{N}\,C_\ell \chi^{2}\left(2\ell+1)\right)

\end{aligned}

##

To compute the Poisson Noise of each ##a_{\ell m}^2##, a colleague suggests me to take only the quantity :

##\dfrac{2}{f_{sky}\,N_p^2}\quad(2)##

and to do the summation to get the variance of Poisson Noise (to make the link with formula (1) and be consistent with it) :

##\text{Var}(N_{p,int})=\sum_{\ell=1}^{N} \dfrac{2}{f_{sky}N_{p}^2}\quad(3)##

I wrote above ##\text{Var}(N_{p,int})##, make caution, this is just to express the "integrated Poisson variance" over ##\ell## in the quantity ##\sum_{\ell=1}^{N} \sum_{m=-\ell}^{\ell} a_{\ell m}^{2}##.

Unfortunately, when I do numerical computation with this variance formula ##(3)##, I don't get the same variance than with another valid method.

Correct results are got by introducing a factor ##\sqrt{2\ell+1}## into formula (3).

But, under the condition this factor is right, I don't know how to justify it, it is just fined-tuned from my part for the moment and it is not rigorous I admit.

QUESTION :

Is foruma ##(2)## that expresses the variance of Shot Noise on the quantity ##\sum_{\ell=\ell_{min}}^{\ell_{max}} \sum_{m=-\ell}^\ell a_{\ell,m}^2## correct ?

If yes, how could integrate it correctly by summing it on multipole ##\ell## and to remain consistent with the standard deviation formula ##(1)## ? (I talk about the pre-factor ##\sqrt{\frac{2}{(2 \ell+1) f_{sky}}}## that disturbs me in the expression of integrated Poisson Noise)

From spherical harmonics with Legendre deccomposition, I have the following definition of

the standard deviation of a ##C_\ell## noised with a Poisson Noise ##N_p## :

##

\begin{equation}

\sigma({C_\ell})(\ell)=\sqrt{\frac{2}{(2 \ell+1) f_{sky}}}\left[C_\ell(\ell)+N_{p}(\ell)\right]\quad(1)

\end{equation}

##

Now I consider the quantity : ##\sum_{\ell=1}^{N} \sum_{m=-\ell}^{\ell} a_{\ell m}^{2}##

I want to estimate the variance expression of Poisson Noise of this qantity.

For that, I take the definition of ##a_{lm}## following a normal distribution with mean equal to zero and take also the definition of a ##C_\ell=\langle a_{lm}^2 \rangle=\dfrac{1}{2\ell+1}\sum_{m=-\ell}^{\ell}\,a_{\ell m}^2 = \text{Var}(a_{lm})##.

I use ##\stackrel{d}{=}## to denote equality in distribution :

##

\begin{align}

Z&\equiv \sum_{\ell=\ell_{min}}^{\ell_{max}} \sum_{m=-\ell}^\ell a_{\ell,m}^2 \\[6pt]

&= \sum_{\ell=\ell_{min}}^{\ell_{max}} \sum_{m=-\ell}^\ell C_\ell \cdot \bigg( \frac{a_{\ell,m}}{\sqrt{C_\ell}} \bigg)^2 \\[6pt]

&\stackrel{d}{=}\sum_{\ell=\ell_{min}}^{\ell_{max}} \sum_{m=-\ell}^\ell C_\ell \cdot \chi^{2}(1) \\[6pt]

&= \sum_{\ell=\ell_{min}}^{\ell_{max}} C_\ell \sum_{m=-\ell}^\ell \chi^{2}(1) \\[6pt]

&= \sum_{\ell=\ell_{min}}^{\ell_{max}} C_\ell \cdot \chi^{2}(2 \ell + 1). \\[6pt]

\end{align}

##

So Finally we have :

##

\begin{aligned}

\sum_{\ell=1}^{N} \sum_{m=-\ell}^{\ell} a_{\ell m}^{2} & \stackrel{d}{=} \sum_{\ell=1}^{N}\,C_\ell \chi^{2}\left(2\ell+1)\right)

\end{aligned}

##

To compute the Poisson Noise of each ##a_{\ell m}^2##, a colleague suggests me to take only the quantity :

##\dfrac{2}{f_{sky}\,N_p^2}\quad(2)##

and to do the summation to get the variance of Poisson Noise (to make the link with formula (1) and be consistent with it) :

##\text{Var}(N_{p,int})=\sum_{\ell=1}^{N} \dfrac{2}{f_{sky}N_{p}^2}\quad(3)##

I wrote above ##\text{Var}(N_{p,int})##, make caution, this is just to express the "integrated Poisson variance" over ##\ell## in the quantity ##\sum_{\ell=1}^{N} \sum_{m=-\ell}^{\ell} a_{\ell m}^{2}##.

Unfortunately, when I do numerical computation with this variance formula ##(3)##, I don't get the same variance than with another valid method.

Correct results are got by introducing a factor ##\sqrt{2\ell+1}## into formula (3).

But, under the condition this factor is right, I don't know how to justify it, it is just fined-tuned from my part for the moment and it is not rigorous I admit.

QUESTION :

Is foruma ##(2)## that expresses the variance of Shot Noise on the quantity ##\sum_{\ell=\ell_{min}}^{\ell_{max}} \sum_{m=-\ell}^\ell a_{\ell,m}^2## correct ?

If yes, how could integrate it correctly by summing it on multipole ##\ell## and to remain consistent with the standard deviation formula ##(1)## ? (I talk about the pre-factor ##\sqrt{\frac{2}{(2 \ell+1) f_{sky}}}## that disturbs me in the expression of integrated Poisson Noise)

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