Need help about a demo with inverse weighted variance average

In summary, the conversation discusses the use of spectroscopic and photometric probes in cosmology, and how to estimate the ratio of their respective coefficients in the absence of Poisson noise. The conversation then moves on to discussing the variance and optimal estimator for this ratio, and the goal of proving a specific relation related to this estimator.
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fab13
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TL;DR Summary
I need help about the understanding of all the steps in a demonstration of the optimal variance by inverse-weighted variance average.
I have a problem of understanding in the following demo :

In a cosmology context with 2 probes (spectroscopic and photometric), let notice ##a_{\ell m, s p}## the spectroscopic and ##a_{\ell m, p h}## the photometric coefficients of the decomposition in spherical harmonics of the distributions of each population. In the absence of any Poisson noise we have:
##\dfrac{a_{\ell m, s p}^{2}}{a_{\ell m, p h}^{2}}=\left(\dfrac{b_{s p}}{b_{p h}}\right)^{2}\quad(1)##
Now let assume the spectroscopic sample is a Poisson realization of density ##N_{s p}## (the galaxy density of the spectroscopic sample) and that we have an unbiased estimator ##\hat{a}_{\ell m, s p}## of ##a_{\ell m, s p}##. We then have the average :
##
\left\langle\dfrac{\hat{a}_{\ell m, s p}^{2}}{a_{\ell m, p h}^{2}}\right\rangle=\dfrac{\left\langle\hat{a}_{\ell m, s p}^{2}\right\rangle}{a_{\ell m, p h}^{2}}=\left(\dfrac{b_{s p}}{b_{p h}}\right)^{2}\quad(2)
##
with its variance :
##
\dfrac{2}{f_{s k y} a_{\ell m, p h}^{4} N_{s p}^{2}}
##
We can therefore build an estimator ##\hat{O}## of ##\left(\dfrac{b_{s p}}{b_{p h}}\right)^{2}## by taking the optimal (inverse-variance weighted) average over all ##\ell## and ##m## :

##\hat{O}=\dfrac{\sum_{\ell=\ell_{\min }}^{\ell_{\max }} \sum_{m=-\ell}^{\ell} a_{\ell m, p h}^{2} \hat{a}_{\ell m, s p}^{2}}{\sum_{\ell=\ell_{\min }}^{\ell_{\max }} \sum_{m=-\ell}^{\ell} a_{\ell m, p h}^{4}}\quad(3)##
the variance of which being:

##\sigma_{\hat{o}}^{2}=\left(\sum_\limits{\ell=\ell_{\min }}^{\ell_{\max }}(2 \ell+1) C_{\ell, p h}^{2}\right)^{-1} \dfrac{2}{f_{s k y} N_{sp}^{2}}\quad(4)##

I don't understanding the passing between eq(3) and eq(4). Indeed, I can't make appear from eq(3) the existing factor ##2\ell+1## in eq(4). The goal is to prove the relation eq(4).

If someone could help me to detail the different necessary steps to obtain eq(4), this would be fine.

I recall that in general, ##C_{\ell}=\dfrac{1}{2\ell+1}\sum_{m=-\ell}^{+\ell} a_{\ell m}^{2}##

Best regards
 
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a little up to demonstrate the passing from eq(3) to eq(4) ?
 

FAQ: Need help about a demo with inverse weighted variance average

What is inverse weighted variance average?

Inverse weighted variance average is a statistical method used to calculate the average of a set of data points. It takes into account both the value and the variability of each data point, giving more weight to data points with lower variability.

How is inverse weighted variance average calculated?

The formula for inverse weighted variance average is:
IWVA = (∑(1/σ2 * x) / ∑(1/σ2))
Where σ is the standard deviation of each data point and x is the value of each data point.

When is inverse weighted variance average used?

Inverse weighted variance average is commonly used in scientific research, particularly in fields such as physics, biology, and economics. It is used when the data points have different variances and the researcher wants to give more weight to data points with lower variability.

What are the advantages of using inverse weighted variance average?

One advantage of inverse weighted variance average is that it takes into account the variability of data points, giving more weight to data points with lower variability. This can help to reduce the impact of outliers on the calculated average. Additionally, it can provide a more accurate representation of the data.

Are there any limitations to using inverse weighted variance average?

One limitation of inverse weighted variance average is that it assumes a normal distribution of the data points. If the data is not normally distributed, the calculated average may not accurately represent the data. Additionally, it can be more complex to calculate compared to other methods of calculating averages.

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