## #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!!!