- #1
fab13
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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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