#StackBounty: #probability #distributions #normal-distribution #chi-squared-distribution Properties of $Gamma$ distribution function t…

Bounty: 50

In the context of harmonic spherical decomposition where $C_ell$ is the variance of $a_{lm}$ for a given multipole $ell$, i.e $C_ell = langle |a_{lm}^2|rangle$, I need help about the $chi^2$ properties.

Actually, I am posting initially this question since I want to compute the mean of obervable $O$, assuming the $a_{lm}$ follows a Normal distribution $mathcal{N}(0,C_ell$) since $C_ell$ is the variance of $a_{lm}$ for a given $ell$ :

$$O=frac{sum_{ell=1}^{N} sum_{m=-ell}^{ell} a_{ell m}^{2}}{sum_{ell=1}^{N} sum_{m=-ell}^{ell}a_{ell m}^{‘2}}$$

UPDATE : Maybe there is a solution for the expectation calculation of ratio. Considering the upper and lower terms independent, we could write : E[A/B] = E[A] E[1/B], such that :

$$begin{aligned}
langle Orangle &=leftlanglesum_{ell=1}^{N} sum_{m=-ell}^{ell}left(a_{ell m}right)^{2}rightrangleleftlangleleft(sum_{ell=1}^{N} sum_{m=-ell}^{ell}left(a_{ell m}^{prime}right)^{2}right)^{-1}rightrangle \
&=leftlanglesum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C_{ell}right)rightrangleleftlangleleft(sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C_{ell}^{prime}right)right)^{-1}rightrangle
end{aligned}$$

1) Given this, by taking the relation between $C_ell$ and $C’_ell$ which is : $C_ell=bigg(dfrac{b}{b’}bigg)^{2},C’_ell$

where $b$ and $b’$ are constants. Warning, rigorously, I should compute for expectation : E[A/B] = E[A] E[1/B]

$$langle Orangle = left(sum_{ell=1}^{N} (2ell+1),C_{ell}right),langleleft(sum_{ell=1}^{N} sum_{m=-ell}^{ell}left(a’_{ell m}right)^{2}right)^{-1}rangle$$

So the expectation E[1/B] is pretty difficult to compute :

If I take : $C_{ell}=bigg(dfrac{b}{b’}bigg)^{2},C’_{ell}$, then :

$$langle Orangle simeq dfrac{sum_{ell=1}^{N} (2ell+1),bigg(dfrac{b}{b’}bigg)^{2},C’_{ell}}{sum_{ell=1}^{N} (2ell+1),C’_{ell}}=bigg(dfrac{b}{b’}bigg)^{2}$$

Do you think approximation $simeq$ is not too large or the expression below with Gamma function is more exact ?

2) If I express directly the observable $O$ using $Gamma$ function, I can get maybe an exact calcultion of $langle Orangle$ (I am not totally sure) :

$$langle Orangle =dfrac{sum_{ell=1}^{N} bigg(dfrac{b}{b’}bigg)^{2} C_{ell}^{prime} ,Gammaleft((2 ell+1)/2,2right)}{sum_{ell=1}^{N} C_{ell}^{prime} ,Gammaleft((2 ell+1)/2,2right)}=bigg(dfrac{b}{b’}bigg)^{2}$$

So, the mean of observable $O$ would be equal to a constant=$bigg(dfrac{b}{b’}bigg)^{2}$ : is this calculation correct ?

3) To come back on the expression $leftlangleleft(sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C_{ell}^{prime}right)right)^{-1}rightrangle$, the issue is that I can’t write :

$$sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C’_{ell}right)=Gammaleft(sum_{ell=1}^{N}(2 ell+1) / 2,2 C’_{ell}right)$$

which could have made things simplier.

Is there a way to simplify this expression $sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C’_{ell}right)$ without removing the dependence in $ell$ between the shape $(dfrac{(2ell+1)}{2}$ ans the scale ($2,C’_ell$) parameters ?


Get this bounty!!!

#StackBounty: #probability #distributions #normal-distribution #chi-squared-distribution Properties of $Gamma$ distribution function t…

Bounty: 50

In the context of harmonic spherical decomposition where $C_ell$ is the variance of $a_{lm}$ for a given multipole $ell$, i.e $C_ell = langle |a_{lm}^2|rangle$, I need help about the $chi^2$ properties.

Actually, I am posting initially this question since I want to compute the mean of obervable $O$, assuming the $a_{lm}$ follows a Normal distribution $mathcal{N}(0,C_ell$) since $C_ell$ is the variance of $a_{lm}$ for a given $ell$ :

$$O=frac{sum_{ell=1}^{N} sum_{m=-ell}^{ell} a_{ell m}^{2}}{sum_{ell=1}^{N} sum_{m=-ell}^{ell}a_{ell m}^{‘2}}$$

UPDATE : Maybe there is a solution for the expectation calculation of ratio. Considering the upper and lower terms independent, we could write : E[A/B] = E[A] E[1/B], such that :

$$begin{aligned}
langle Orangle &=leftlanglesum_{ell=1}^{N} sum_{m=-ell}^{ell}left(a_{ell m}right)^{2}rightrangleleftlangleleft(sum_{ell=1}^{N} sum_{m=-ell}^{ell}left(a_{ell m}^{prime}right)^{2}right)^{-1}rightrangle \
&=leftlanglesum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C_{ell}right)rightrangleleftlangleleft(sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C_{ell}^{prime}right)right)^{-1}rightrangle
end{aligned}$$

1) Given this, by taking the relation between $C_ell$ and $C’_ell$ which is : $C_ell=bigg(dfrac{b}{b’}bigg)^{2},C’_ell$

where $b$ and $b’$ are constants. Warning, rigorously, I should compute for expectation : E[A/B] = E[A] E[1/B]

$$langle Orangle = left(sum_{ell=1}^{N} (2ell+1),C_{ell}right),langleleft(sum_{ell=1}^{N} sum_{m=-ell}^{ell}left(a’_{ell m}right)^{2}right)^{-1}rangle$$

So the expectation E[1/B] is pretty difficult to compute :

If I take : $C_{ell}=bigg(dfrac{b}{b’}bigg)^{2},C’_{ell}$, then :

$$langle Orangle simeq dfrac{sum_{ell=1}^{N} (2ell+1),bigg(dfrac{b}{b’}bigg)^{2},C’_{ell}}{sum_{ell=1}^{N} (2ell+1),C’_{ell}}=bigg(dfrac{b}{b’}bigg)^{2}$$

Do you think approximation $simeq$ is not too large or the expression below with Gamma function is more exact ?

2) If I express directly the observable $O$ using $Gamma$ function, I can get maybe an exact calcultion of $langle Orangle$ (I am not totally sure) :

$$langle Orangle =dfrac{sum_{ell=1}^{N} bigg(dfrac{b}{b’}bigg)^{2} C_{ell}^{prime} ,Gammaleft((2 ell+1)/2,2right)}{sum_{ell=1}^{N} C_{ell}^{prime} ,Gammaleft((2 ell+1)/2,2right)}=bigg(dfrac{b}{b’}bigg)^{2}$$

So, the mean of observable $O$ would be equal to a constant=$bigg(dfrac{b}{b’}bigg)^{2}$ : is this calculation correct ?

3) To come back on the expression $leftlangleleft(sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C_{ell}^{prime}right)right)^{-1}rightrangle$, the issue is that I can’t write :

$$sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C’_{ell}right)=Gammaleft(sum_{ell=1}^{N}(2 ell+1) / 2,2 C’_{ell}right)$$

which could have made things simplier.

Is there a way to simplify this expression $sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C’_{ell}right)$ without removing the dependence in $ell$ between the shape $(dfrac{(2ell+1)}{2}$ ans the scale ($2,C’_ell$) parameters ?


Get this bounty!!!

#StackBounty: #probability #distributions #normal-distribution #chi-squared-distribution Properties of $Gamma$ distribution function t…

Bounty: 50

In the context of harmonic spherical decomposition where $C_ell$ is the variance of $a_{lm}$ for a given multipole $ell$, i.e $C_ell = langle |a_{lm}^2|rangle$, I need help about the $chi^2$ properties.

Actually, I am posting initially this question since I want to compute the mean of obervable $O$, assuming the $a_{lm}$ follows a Normal distribution $mathcal{N}(0,C_ell$) since $C_ell$ is the variance of $a_{lm}$ for a given $ell$ :

$$O=frac{sum_{ell=1}^{N} sum_{m=-ell}^{ell} a_{ell m}^{2}}{sum_{ell=1}^{N} sum_{m=-ell}^{ell}a_{ell m}^{‘2}}$$

UPDATE : Maybe there is a solution for the expectation calculation of ratio. Considering the upper and lower terms independent, we could write : E[A/B] = E[A] E[1/B], such that :

$$begin{aligned}
langle Orangle &=leftlanglesum_{ell=1}^{N} sum_{m=-ell}^{ell}left(a_{ell m}right)^{2}rightrangleleftlangleleft(sum_{ell=1}^{N} sum_{m=-ell}^{ell}left(a_{ell m}^{prime}right)^{2}right)^{-1}rightrangle \
&=leftlanglesum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C_{ell}right)rightrangleleftlangleleft(sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C_{ell}^{prime}right)right)^{-1}rightrangle
end{aligned}$$

1) Given this, by taking the relation between $C_ell$ and $C’_ell$ which is : $C_ell=bigg(dfrac{b}{b’}bigg)^{2},C’_ell$

where $b$ and $b’$ are constants. Warning, rigorously, I should compute for expectation : E[A/B] = E[A] E[1/B]

$$langle Orangle = left(sum_{ell=1}^{N} (2ell+1),C_{ell}right),langleleft(sum_{ell=1}^{N} sum_{m=-ell}^{ell}left(a’_{ell m}right)^{2}right)^{-1}rangle$$

So the expectation E[1/B] is pretty difficult to compute :

If I take : $C_{ell}=bigg(dfrac{b}{b’}bigg)^{2},C’_{ell}$, then :

$$langle Orangle simeq dfrac{sum_{ell=1}^{N} (2ell+1),bigg(dfrac{b}{b’}bigg)^{2},C’_{ell}}{sum_{ell=1}^{N} (2ell+1),C’_{ell}}=bigg(dfrac{b}{b’}bigg)^{2}$$

Do you think approximation $simeq$ is not too large or the expression below with Gamma function is more exact ?

2) If I express directly the observable $O$ using $Gamma$ function, I can get maybe an exact calcultion of $langle Orangle$ (I am not totally sure) :

$$langle Orangle =dfrac{sum_{ell=1}^{N} bigg(dfrac{b}{b’}bigg)^{2} C_{ell}^{prime} ,Gammaleft((2 ell+1)/2,2right)}{sum_{ell=1}^{N} C_{ell}^{prime} ,Gammaleft((2 ell+1)/2,2right)}=bigg(dfrac{b}{b’}bigg)^{2}$$

So, the mean of observable $O$ would be equal to a constant=$bigg(dfrac{b}{b’}bigg)^{2}$ : is this calculation correct ?

3) To come back on the expression $leftlangleleft(sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C_{ell}^{prime}right)right)^{-1}rightrangle$, the issue is that I can’t write :

$$sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C’_{ell}right)=Gammaleft(sum_{ell=1}^{N}(2 ell+1) / 2,2 C’_{ell}right)$$

which could have made things simplier.

Is there a way to simplify this expression $sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C’_{ell}right)$ without removing the dependence in $ell$ between the shape $(dfrac{(2ell+1)}{2}$ ans the scale ($2,C’_ell$) parameters ?


Get this bounty!!!

#StackBounty: #probability #distributions #normal-distribution #chi-squared-distribution Properties of $Gamma$ distribution function t…

Bounty: 50

In the context of harmonic spherical decomposition where $C_ell$ is the variance of $a_{lm}$ for a given multipole $ell$, i.e $C_ell = langle |a_{lm}^2|rangle$, I need help about the $chi^2$ properties.

Actually, I am posting initially this question since I want to compute the mean of obervable $O$, assuming the $a_{lm}$ follows a Normal distribution $mathcal{N}(0,C_ell$) since $C_ell$ is the variance of $a_{lm}$ for a given $ell$ :

$$O=frac{sum_{ell=1}^{N} sum_{m=-ell}^{ell} a_{ell m}^{2}}{sum_{ell=1}^{N} sum_{m=-ell}^{ell}a_{ell m}^{‘2}}$$

UPDATE : Maybe there is a solution for the expectation calculation of ratio. Considering the upper and lower terms independent, we could write : E[A/B] = E[A] E[1/B], such that :

$$begin{aligned}
langle Orangle &=leftlanglesum_{ell=1}^{N} sum_{m=-ell}^{ell}left(a_{ell m}right)^{2}rightrangleleftlangleleft(sum_{ell=1}^{N} sum_{m=-ell}^{ell}left(a_{ell m}^{prime}right)^{2}right)^{-1}rightrangle \
&=leftlanglesum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C_{ell}right)rightrangleleftlangleleft(sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C_{ell}^{prime}right)right)^{-1}rightrangle
end{aligned}$$

1) Given this, by taking the relation between $C_ell$ and $C’_ell$ which is : $C_ell=bigg(dfrac{b}{b’}bigg)^{2},C’_ell$

where $b$ and $b’$ are constants. Warning, rigorously, I should compute for expectation : E[A/B] = E[A] E[1/B]

$$langle Orangle = left(sum_{ell=1}^{N} (2ell+1),C_{ell}right),langleleft(sum_{ell=1}^{N} sum_{m=-ell}^{ell}left(a’_{ell m}right)^{2}right)^{-1}rangle$$

So the expectation E[1/B] is pretty difficult to compute :

If I take : $C_{ell}=bigg(dfrac{b}{b’}bigg)^{2},C’_{ell}$, then :

$$langle Orangle simeq dfrac{sum_{ell=1}^{N} (2ell+1),bigg(dfrac{b}{b’}bigg)^{2},C’_{ell}}{sum_{ell=1}^{N} (2ell+1),C’_{ell}}=bigg(dfrac{b}{b’}bigg)^{2}$$

Do you think approximation $simeq$ is not too large or the expression below with Gamma function is more exact ?

2) If I express directly the observable $O$ using $Gamma$ function, I can get maybe an exact calcultion of $langle Orangle$ (I am not totally sure) :

$$langle Orangle =dfrac{sum_{ell=1}^{N} bigg(dfrac{b}{b’}bigg)^{2} C_{ell}^{prime} ,Gammaleft((2 ell+1)/2,2right)}{sum_{ell=1}^{N} C_{ell}^{prime} ,Gammaleft((2 ell+1)/2,2right)}=bigg(dfrac{b}{b’}bigg)^{2}$$

So, the mean of observable $O$ would be equal to a constant=$bigg(dfrac{b}{b’}bigg)^{2}$ : is this calculation correct ?

3) To come back on the expression $leftlangleleft(sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C_{ell}^{prime}right)right)^{-1}rightrangle$, the issue is that I can’t write :

$$sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C’_{ell}right)=Gammaleft(sum_{ell=1}^{N}(2 ell+1) / 2,2 C’_{ell}right)$$

which could have made things simplier.

Is there a way to simplify this expression $sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C’_{ell}right)$ without removing the dependence in $ell$ between the shape $(dfrac{(2ell+1)}{2}$ ans the scale ($2,C’_ell$) parameters ?


Get this bounty!!!

#StackBounty: #probability #distributions #normal-distribution #chi-squared-distribution Properties of $Gamma$ distribution function t…

Bounty: 50

In the context of harmonic spherical decomposition where $C_ell$ is the variance of $a_{lm}$ for a given multipole $ell$, i.e $C_ell = langle |a_{lm}^2|rangle$, I need help about the $chi^2$ properties.

Actually, I am posting initially this question since I want to compute the mean of obervable $O$, assuming the $a_{lm}$ follows a Normal distribution $mathcal{N}(0,C_ell$) since $C_ell$ is the variance of $a_{lm}$ for a given $ell$ :

$$O=frac{sum_{ell=1}^{N} sum_{m=-ell}^{ell} a_{ell m}^{2}}{sum_{ell=1}^{N} sum_{m=-ell}^{ell}a_{ell m}^{‘2}}$$

UPDATE : Maybe there is a solution for the expectation calculation of ratio. Considering the upper and lower terms independent, we could write : E[A/B] = E[A] E[1/B], such that :

$$begin{aligned}
langle Orangle &=leftlanglesum_{ell=1}^{N} sum_{m=-ell}^{ell}left(a_{ell m}right)^{2}rightrangleleftlangleleft(sum_{ell=1}^{N} sum_{m=-ell}^{ell}left(a_{ell m}^{prime}right)^{2}right)^{-1}rightrangle \
&=leftlanglesum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C_{ell}right)rightrangleleftlangleleft(sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C_{ell}^{prime}right)right)^{-1}rightrangle
end{aligned}$$

1) Given this, by taking the relation between $C_ell$ and $C’_ell$ which is : $C_ell=bigg(dfrac{b}{b’}bigg)^{2},C’_ell$

where $b$ and $b’$ are constants. Warning, rigorously, I should compute for expectation : E[A/B] = E[A] E[1/B]

$$langle Orangle = left(sum_{ell=1}^{N} (2ell+1),C_{ell}right),langleleft(sum_{ell=1}^{N} sum_{m=-ell}^{ell}left(a’_{ell m}right)^{2}right)^{-1}rangle$$

So the expectation E[1/B] is pretty difficult to compute :

If I take : $C_{ell}=bigg(dfrac{b}{b’}bigg)^{2},C’_{ell}$, then :

$$langle Orangle simeq dfrac{sum_{ell=1}^{N} (2ell+1),bigg(dfrac{b}{b’}bigg)^{2},C’_{ell}}{sum_{ell=1}^{N} (2ell+1),C’_{ell}}=bigg(dfrac{b}{b’}bigg)^{2}$$

Do you think approximation $simeq$ is not too large or the expression below with Gamma function is more exact ?

2) If I express directly the observable $O$ using $Gamma$ function, I can get maybe an exact calcultion of $langle Orangle$ (I am not totally sure) :

$$langle Orangle =dfrac{sum_{ell=1}^{N} bigg(dfrac{b}{b’}bigg)^{2} C_{ell}^{prime} ,Gammaleft((2 ell+1)/2,2right)}{sum_{ell=1}^{N} C_{ell}^{prime} ,Gammaleft((2 ell+1)/2,2right)}=bigg(dfrac{b}{b’}bigg)^{2}$$

So, the mean of observable $O$ would be equal to a constant=$bigg(dfrac{b}{b’}bigg)^{2}$ : is this calculation correct ?

3) To come back on the expression $leftlangleleft(sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C_{ell}^{prime}right)right)^{-1}rightrangle$, the issue is that I can’t write :

$$sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C’_{ell}right)=Gammaleft(sum_{ell=1}^{N}(2 ell+1) / 2,2 C’_{ell}right)$$

which could have made things simplier.

Is there a way to simplify this expression $sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C’_{ell}right)$ without removing the dependence in $ell$ between the shape $(dfrac{(2ell+1)}{2}$ ans the scale ($2,C’_ell$) parameters ?


Get this bounty!!!

#StackBounty: #probability #distributions #normal-distribution #chi-squared-distribution Properties of $Gamma$ distribution function t…

Bounty: 50

In the context of harmonic spherical decomposition where $C_ell$ is the variance of $a_{lm}$ for a given multipole $ell$, i.e $C_ell = langle |a_{lm}^2|rangle$, I need help about the $chi^2$ properties.

Actually, I am posting initially this question since I want to compute the mean of obervable $O$, assuming the $a_{lm}$ follows a Normal distribution $mathcal{N}(0,C_ell$) since $C_ell$ is the variance of $a_{lm}$ for a given $ell$ :

$$O=frac{sum_{ell=1}^{N} sum_{m=-ell}^{ell} a_{ell m}^{2}}{sum_{ell=1}^{N} sum_{m=-ell}^{ell}a_{ell m}^{‘2}}$$

UPDATE : Maybe there is a solution for the expectation calculation of ratio. Considering the upper and lower terms independent, we could write : E[A/B] = E[A] E[1/B], such that :

$$begin{aligned}
langle Orangle &=leftlanglesum_{ell=1}^{N} sum_{m=-ell}^{ell}left(a_{ell m}right)^{2}rightrangleleftlangleleft(sum_{ell=1}^{N} sum_{m=-ell}^{ell}left(a_{ell m}^{prime}right)^{2}right)^{-1}rightrangle \
&=leftlanglesum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C_{ell}right)rightrangleleftlangleleft(sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C_{ell}^{prime}right)right)^{-1}rightrangle
end{aligned}$$

1) Given this, by taking the relation between $C_ell$ and $C’_ell$ which is : $C_ell=bigg(dfrac{b}{b’}bigg)^{2},C’_ell$

where $b$ and $b’$ are constants. Warning, rigorously, I should compute for expectation : E[A/B] = E[A] E[1/B]

$$langle Orangle = left(sum_{ell=1}^{N} (2ell+1),C_{ell}right),langleleft(sum_{ell=1}^{N} sum_{m=-ell}^{ell}left(a’_{ell m}right)^{2}right)^{-1}rangle$$

So the expectation E[1/B] is pretty difficult to compute :

If I take : $C_{ell}=bigg(dfrac{b}{b’}bigg)^{2},C’_{ell}$, then :

$$langle Orangle simeq dfrac{sum_{ell=1}^{N} (2ell+1),bigg(dfrac{b}{b’}bigg)^{2},C’_{ell}}{sum_{ell=1}^{N} (2ell+1),C’_{ell}}=bigg(dfrac{b}{b’}bigg)^{2}$$

Do you think approximation $simeq$ is not too large or the expression below with Gamma function is more exact ?

2) If I express directly the observable $O$ using $Gamma$ function, I can get maybe an exact calcultion of $langle Orangle$ (I am not totally sure) :

$$langle Orangle =dfrac{sum_{ell=1}^{N} bigg(dfrac{b}{b’}bigg)^{2} C_{ell}^{prime} ,Gammaleft((2 ell+1)/2,2right)}{sum_{ell=1}^{N} C_{ell}^{prime} ,Gammaleft((2 ell+1)/2,2right)}=bigg(dfrac{b}{b’}bigg)^{2}$$

So, the mean of observable $O$ would be equal to a constant=$bigg(dfrac{b}{b’}bigg)^{2}$ : is this calculation correct ?

3) To come back on the expression $leftlangleleft(sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C_{ell}^{prime}right)right)^{-1}rightrangle$, the issue is that I can’t write :

$$sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C’_{ell}right)=Gammaleft(sum_{ell=1}^{N}(2 ell+1) / 2,2 C’_{ell}right)$$

which could have made things simplier.

Is there a way to simplify this expression $sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C’_{ell}right)$ without removing the dependence in $ell$ between the shape $(dfrac{(2ell+1)}{2}$ ans the scale ($2,C’_ell$) parameters ?


Get this bounty!!!

#StackBounty: #probability #distributions #normal-distribution #chi-squared-distribution Properties of $Gamma$ distribution function t…

Bounty: 50

In the context of harmonic spherical decomposition where $C_ell$ is the variance of $a_{lm}$ for a given multipole $ell$, i.e $C_ell = langle |a_{lm}^2|rangle$, I need help about the $chi^2$ properties.

Actually, I am posting initially this question since I want to compute the mean of obervable $O$, assuming the $a_{lm}$ follows a Normal distribution $mathcal{N}(0,C_ell$) since $C_ell$ is the variance of $a_{lm}$ for a given $ell$ :

$$O=frac{sum_{ell=1}^{N} sum_{m=-ell}^{ell} a_{ell m}^{2}}{sum_{ell=1}^{N} sum_{m=-ell}^{ell}a_{ell m}^{‘2}}$$

UPDATE : Maybe there is a solution for the expectation calculation of ratio. Considering the upper and lower terms independent, we could write : E[A/B] = E[A] E[1/B], such that :

$$begin{aligned}
langle Orangle &=leftlanglesum_{ell=1}^{N} sum_{m=-ell}^{ell}left(a_{ell m}right)^{2}rightrangleleftlangleleft(sum_{ell=1}^{N} sum_{m=-ell}^{ell}left(a_{ell m}^{prime}right)^{2}right)^{-1}rightrangle \
&=leftlanglesum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C_{ell}right)rightrangleleftlangleleft(sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C_{ell}^{prime}right)right)^{-1}rightrangle
end{aligned}$$

1) Given this, by taking the relation between $C_ell$ and $C’_ell$ which is : $C_ell=bigg(dfrac{b}{b’}bigg)^{2},C’_ell$

where $b$ and $b’$ are constants. Warning, rigorously, I should compute for expectation : E[A/B] = E[A] E[1/B]

$$langle Orangle = left(sum_{ell=1}^{N} (2ell+1),C_{ell}right),langleleft(sum_{ell=1}^{N} sum_{m=-ell}^{ell}left(a’_{ell m}right)^{2}right)^{-1}rangle$$

So the expectation E[1/B] is pretty difficult to compute :

If I take : $C_{ell}=bigg(dfrac{b}{b’}bigg)^{2},C’_{ell}$, then :

$$langle Orangle simeq dfrac{sum_{ell=1}^{N} (2ell+1),bigg(dfrac{b}{b’}bigg)^{2},C’_{ell}}{sum_{ell=1}^{N} (2ell+1),C’_{ell}}=bigg(dfrac{b}{b’}bigg)^{2}$$

Do you think approximation $simeq$ is not too large or the expression below with Gamma function is more exact ?

2) If I express directly the observable $O$ using $Gamma$ function, I can get maybe an exact calcultion of $langle Orangle$ (I am not totally sure) :

$$langle Orangle =dfrac{sum_{ell=1}^{N} bigg(dfrac{b}{b’}bigg)^{2} C_{ell}^{prime} ,Gammaleft((2 ell+1)/2,2right)}{sum_{ell=1}^{N} C_{ell}^{prime} ,Gammaleft((2 ell+1)/2,2right)}=bigg(dfrac{b}{b’}bigg)^{2}$$

So, the mean of observable $O$ would be equal to a constant=$bigg(dfrac{b}{b’}bigg)^{2}$ : is this calculation correct ?

3) To come back on the expression $leftlangleleft(sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C_{ell}^{prime}right)right)^{-1}rightrangle$, the issue is that I can’t write :

$$sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C’_{ell}right)=Gammaleft(sum_{ell=1}^{N}(2 ell+1) / 2,2 C’_{ell}right)$$

which could have made things simplier.

Is there a way to simplify this expression $sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C’_{ell}right)$ without removing the dependence in $ell$ between the shape $(dfrac{(2ell+1)}{2}$ ans the scale ($2,C’_ell$) parameters ?


Get this bounty!!!

#StackBounty: #probability #distributions #normal-distribution #chi-squared-distribution Properties of $Gamma$ distribution function t…

Bounty: 50

In the context of harmonic spherical decomposition where $C_ell$ is the variance of $a_{lm}$ for a given multipole $ell$, i.e $C_ell = langle |a_{lm}^2|rangle$, I need help about the $chi^2$ properties.

Actually, I am posting initially this question since I want to compute the mean of obervable $O$, assuming the $a_{lm}$ follows a Normal distribution $mathcal{N}(0,C_ell$) since $C_ell$ is the variance of $a_{lm}$ for a given $ell$ :

$$O=frac{sum_{ell=1}^{N} sum_{m=-ell}^{ell} a_{ell m}^{2}}{sum_{ell=1}^{N} sum_{m=-ell}^{ell}a_{ell m}^{‘2}}$$

UPDATE : Maybe there is a solution for the expectation calculation of ratio. Considering the upper and lower terms independent, we could write : E[A/B] = E[A] E[1/B], such that :

$$begin{aligned}
langle Orangle &=leftlanglesum_{ell=1}^{N} sum_{m=-ell}^{ell}left(a_{ell m}right)^{2}rightrangleleftlangleleft(sum_{ell=1}^{N} sum_{m=-ell}^{ell}left(a_{ell m}^{prime}right)^{2}right)^{-1}rightrangle \
&=leftlanglesum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C_{ell}right)rightrangleleftlangleleft(sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C_{ell}^{prime}right)right)^{-1}rightrangle
end{aligned}$$

1) Given this, by taking the relation between $C_ell$ and $C’_ell$ which is : $C_ell=bigg(dfrac{b}{b’}bigg)^{2},C’_ell$

where $b$ and $b’$ are constants. Warning, rigorously, I should compute for expectation : E[A/B] = E[A] E[1/B]

$$langle Orangle = left(sum_{ell=1}^{N} (2ell+1),C_{ell}right),langleleft(sum_{ell=1}^{N} sum_{m=-ell}^{ell}left(a’_{ell m}right)^{2}right)^{-1}rangle$$

So the expectation E[1/B] is pretty difficult to compute :

If I take : $C_{ell}=bigg(dfrac{b}{b’}bigg)^{2},C’_{ell}$, then :

$$langle Orangle simeq dfrac{sum_{ell=1}^{N} (2ell+1),bigg(dfrac{b}{b’}bigg)^{2},C’_{ell}}{sum_{ell=1}^{N} (2ell+1),C’_{ell}}=bigg(dfrac{b}{b’}bigg)^{2}$$

Do you think approximation $simeq$ is not too large or the expression below with Gamma function is more exact ?

2) If I express directly the observable $O$ using $Gamma$ function, I can get maybe an exact calcultion of $langle Orangle$ (I am not totally sure) :

$$langle Orangle =dfrac{sum_{ell=1}^{N} bigg(dfrac{b}{b’}bigg)^{2} C_{ell}^{prime} ,Gammaleft((2 ell+1)/2,2right)}{sum_{ell=1}^{N} C_{ell}^{prime} ,Gammaleft((2 ell+1)/2,2right)}=bigg(dfrac{b}{b’}bigg)^{2}$$

So, the mean of observable $O$ would be equal to a constant=$bigg(dfrac{b}{b’}bigg)^{2}$ : is this calculation correct ?

3) To come back on the expression $leftlangleleft(sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C_{ell}^{prime}right)right)^{-1}rightrangle$, the issue is that I can’t write :

$$sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C’_{ell}right)=Gammaleft(sum_{ell=1}^{N}(2 ell+1) / 2,2 C’_{ell}right)$$

which could have made things simplier.

Is there a way to simplify this expression $sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C’_{ell}right)$ without removing the dependence in $ell$ between the shape $(dfrac{(2ell+1)}{2}$ ans the scale ($2,C’_ell$) parameters ?


Get this bounty!!!

#StackBounty: #probability #distributions #normal-distribution #chi-squared-distribution Properties of $Gamma$ distribution function t…

Bounty: 50

In the context of harmonic spherical decomposition where $C_ell$ is the variance of $a_{lm}$ for a given multipole $ell$, i.e $C_ell = langle |a_{lm}^2|rangle$, I need help about the $chi^2$ properties.

Actually, I am posting initially this question since I want to compute the mean of obervable $O$, assuming the $a_{lm}$ follows a Normal distribution $mathcal{N}(0,C_ell$) since $C_ell$ is the variance of $a_{lm}$ for a given $ell$ :

$$O=frac{sum_{ell=1}^{N} sum_{m=-ell}^{ell} a_{ell m}^{2}}{sum_{ell=1}^{N} sum_{m=-ell}^{ell}a_{ell m}^{‘2}}$$

UPDATE : Maybe there is a solution for the expectation calculation of ratio. Considering the upper and lower terms independent, we could write : E[A/B] = E[A] E[1/B], such that :

$$begin{aligned}
langle Orangle &=leftlanglesum_{ell=1}^{N} sum_{m=-ell}^{ell}left(a_{ell m}right)^{2}rightrangleleftlangleleft(sum_{ell=1}^{N} sum_{m=-ell}^{ell}left(a_{ell m}^{prime}right)^{2}right)^{-1}rightrangle \
&=leftlanglesum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C_{ell}right)rightrangleleftlangleleft(sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C_{ell}^{prime}right)right)^{-1}rightrangle
end{aligned}$$

1) Given this, by taking the relation between $C_ell$ and $C’_ell$ which is : $C_ell=bigg(dfrac{b}{b’}bigg)^{2},C’_ell$

where $b$ and $b’$ are constants. Warning, rigorously, I should compute for expectation : E[A/B] = E[A] E[1/B]

$$langle Orangle = left(sum_{ell=1}^{N} (2ell+1),C_{ell}right),langleleft(sum_{ell=1}^{N} sum_{m=-ell}^{ell}left(a’_{ell m}right)^{2}right)^{-1}rangle$$

So the expectation E[1/B] is pretty difficult to compute :

If I take : $C_{ell}=bigg(dfrac{b}{b’}bigg)^{2},C’_{ell}$, then :

$$langle Orangle simeq dfrac{sum_{ell=1}^{N} (2ell+1),bigg(dfrac{b}{b’}bigg)^{2},C’_{ell}}{sum_{ell=1}^{N} (2ell+1),C’_{ell}}=bigg(dfrac{b}{b’}bigg)^{2}$$

Do you think approximation $simeq$ is not too large or the expression below with Gamma function is more exact ?

2) If I express directly the observable $O$ using $Gamma$ function, I can get maybe an exact calcultion of $langle Orangle$ (I am not totally sure) :

$$langle Orangle =dfrac{sum_{ell=1}^{N} bigg(dfrac{b}{b’}bigg)^{2} C_{ell}^{prime} ,Gammaleft((2 ell+1)/2,2right)}{sum_{ell=1}^{N} C_{ell}^{prime} ,Gammaleft((2 ell+1)/2,2right)}=bigg(dfrac{b}{b’}bigg)^{2}$$

So, the mean of observable $O$ would be equal to a constant=$bigg(dfrac{b}{b’}bigg)^{2}$ : is this calculation correct ?

3) To come back on the expression $leftlangleleft(sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C_{ell}^{prime}right)right)^{-1}rightrangle$, the issue is that I can’t write :

$$sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C’_{ell}right)=Gammaleft(sum_{ell=1}^{N}(2 ell+1) / 2,2 C’_{ell}right)$$

which could have made things simplier.

Is there a way to simplify this expression $sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C’_{ell}right)$ without removing the dependence in $ell$ between the shape $(dfrac{(2ell+1)}{2}$ ans the scale ($2,C’_ell$) parameters ?


Get this bounty!!!

#StackBounty: #probability #distributions #normal-distribution #chi-squared-distribution Properties of $Gamma$ distribution function t…

Bounty: 50

In the context of harmonic spherical decomposition where $C_ell$ is the variance of $a_{lm}$ for a given multipole $ell$, i.e $C_ell = langle |a_{lm}^2|rangle$, I need help about the $chi^2$ properties.

Actually, I am posting initially this question since I want to compute the mean of obervable $O$, assuming the $a_{lm}$ follows a Normal distribution $mathcal{N}(0,C_ell$) since $C_ell$ is the variance of $a_{lm}$ for a given $ell$ :

$$O=frac{sum_{ell=1}^{N} sum_{m=-ell}^{ell} a_{ell m}^{2}}{sum_{ell=1}^{N} sum_{m=-ell}^{ell}a_{ell m}^{‘2}}$$

UPDATE : Maybe there is a solution for the expectation calculation of ratio. Considering the upper and lower terms independent, we could write : E[A/B] = E[A] E[1/B], such that :

$$begin{aligned}
langle Orangle &=leftlanglesum_{ell=1}^{N} sum_{m=-ell}^{ell}left(a_{ell m}right)^{2}rightrangleleftlangleleft(sum_{ell=1}^{N} sum_{m=-ell}^{ell}left(a_{ell m}^{prime}right)^{2}right)^{-1}rightrangle \
&=leftlanglesum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C_{ell}right)rightrangleleftlangleleft(sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C_{ell}^{prime}right)right)^{-1}rightrangle
end{aligned}$$

1) Given this, by taking the relation between $C_ell$ and $C’_ell$ which is : $C_ell=bigg(dfrac{b}{b’}bigg)^{2},C’_ell$

where $b$ and $b’$ are constants. Warning, rigorously, I should compute for expectation : E[A/B] = E[A] E[1/B]

$$langle Orangle = left(sum_{ell=1}^{N} (2ell+1),C_{ell}right),langleleft(sum_{ell=1}^{N} sum_{m=-ell}^{ell}left(a’_{ell m}right)^{2}right)^{-1}rangle$$

So the expectation E[1/B] is pretty difficult to compute :

If I take : $C_{ell}=bigg(dfrac{b}{b’}bigg)^{2},C’_{ell}$, then :

$$langle Orangle simeq dfrac{sum_{ell=1}^{N} (2ell+1),bigg(dfrac{b}{b’}bigg)^{2},C’_{ell}}{sum_{ell=1}^{N} (2ell+1),C’_{ell}}=bigg(dfrac{b}{b’}bigg)^{2}$$

Do you think approximation $simeq$ is not too large or the expression below with Gamma function is more exact ?

2) If I express directly the observable $O$ using $Gamma$ function, I can get maybe an exact calcultion of $langle Orangle$ (I am not totally sure) :

$$langle Orangle =dfrac{sum_{ell=1}^{N} bigg(dfrac{b}{b’}bigg)^{2} C_{ell}^{prime} ,Gammaleft((2 ell+1)/2,2right)}{sum_{ell=1}^{N} C_{ell}^{prime} ,Gammaleft((2 ell+1)/2,2right)}=bigg(dfrac{b}{b’}bigg)^{2}$$

So, the mean of observable $O$ would be equal to a constant=$bigg(dfrac{b}{b’}bigg)^{2}$ : is this calculation correct ?

3) To come back on the expression $leftlangleleft(sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C_{ell}^{prime}right)right)^{-1}rightrangle$, the issue is that I can’t write :

$$sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C’_{ell}right)=Gammaleft(sum_{ell=1}^{N}(2 ell+1) / 2,2 C’_{ell}right)$$

which could have made things simplier.

Is there a way to simplify this expression $sum_{ell=1}^{N} Gammaleft((2 ell+1) / 2,2 C’_{ell}right)$ without removing the dependence in $ell$ between the shape $(dfrac{(2ell+1)}{2}$ ans the scale ($2,C’_ell$) parameters ?


Get this bounty!!!