1) Introduction :
I am interested in computing the variance of an observable $$ O=\frac{\sum_{\ell=1}^{N} \sum_{m=-\ell}^{\ell} a_{\ell m}^{2}}{\sum_{\ell=1}^{N} \sum_{m=-\ell}^{\ell}\left(a_{\ell m}^{\prime}\right)^{2}} $$ where $\left(a_{\ell m}, \ell \in\{1, \cdots, N\},|m| \leq \ell\right)$ and $\left(a_{\ell m}^{\prime}, \ell \in\{1, \cdots, N\},|m| \leq \ell\right)$ are independent random variables, with $a_{\ell m} \sim \mathcal{N}\left(0, C_{\ell}\right)$ for each $|m| \leq \ell$ and $a_{\ell m}^{\prime} \sim \mathcal{N}\left(0, C_{\ell}^{\prime}\right)$ for each $|m| \leq \ell .$ We recall the properties of a few basic distributions. We have :
- $\mathcal{N}(0, C)^{2} \sim C\chi^{2}(1)=\Gamma\left(\frac{1}{2}, 2 C\right)$,
- $\langle\Gamma(k, \theta)\rangle=k \theta$ and $\operatorname{Var}(\Gamma(k, \theta))=k \theta^{2}$, and
- $\sum_{i=1}^{N} \Gamma\left(k_{i}, \theta\right) = \Gamma\left(\sum_{i=1}^{N} k_{i}, \theta\right)$ for independent summands.
2) Important precision : for each $\ell$, I have the relation $C_\ell=\dfrac{b}{b'}C'_\ell$ with $b$ and $b'$ being constants, I wonder how it could help for the rest of post.
3) Partial solution not finished (only mean $\langle O\rangle$ ) :
We have by points 1 and 3 $$ \begin{aligned} \sum_{\ell=1}^{N} \sum_{m=-\ell}^{\ell} a_{\ell m}^{2} & = \sum_{\ell=1}^{N} \sum_{m=-\ell}^{\ell} \Gamma\left(1 / 2,2 C_{\ell}\right) \\ & = \sum_{\ell=1}^{N} \Gamma\left((2 \ell+1) / 2,2 C_{\ell}\right) \end{aligned}\quad(1) $$ where the summands are independent. Similarly, using points 1 and 3 again, we obtain $$ \begin{aligned} \sum_{\ell=1}^{N} \sum_{m=-\ell}^{\ell}\left(a_{\ell m}^{\prime}\right)^{2} & = \sum_{\ell=1}^{N} \sum_{m=-\ell}^{\ell} \Gamma\left(1 / 2,2 C_{\ell}^{\prime}\right) \\ & = \sum_{\ell=1}^{N} \Gamma\left((2 \ell+1) / 2,2 C_{\ell}^{\prime}\right) \end{aligned}\quad(2) $$ where the summands are independent. By independence of the sequences $\left(a_{\ell m}, \ell \in\{1, \cdots, N\},|m| \leq \ell\right)$ and $\left(a_{\ell m}^{\prime}, \ell \in\{1, \cdots, N\},|m| \leq \ell\right)$, equations (1) and (2), we obtain $$ \begin{aligned} \langle O\rangle &=\left\langle\sum_{\ell=1}^{N} \sum_{m=-\ell}^{\ell}\left(a_{\ell m}\right)^{2}\right\rangle\left\langle\left(\sum_{\ell=1}^{N} \sum_{m=-\ell}^{\ell}\left(a_{\ell m}^{\prime}\right)^{2}\right)^{-1}\right\rangle \\ &=\left\langle\sum_{\ell=1}^{N} \Gamma\left((2 \ell+1) / 2,2 C_{\ell}\right)\right\rangle\left\langle\left(\sum_{\ell=1}^{N} \Gamma\left((2 \ell+1) / 2,2 C_{\ell}^{\prime}\right)\right)^{-1}\right\rangle \end{aligned} $$
The first factor simplifies :
$$\left\langle\sum_{\ell=1}^{N} \Gamma\left((2 \ell+1) / 2,2 C_{\ell}\right)\right\rangle=\sum_{\ell=1}^{N}(2 \ell+1) C_{\ell}$$
As you can see, I can't conclude on the second factor (expectation of the inverse of sum of Gamma distributions), especially since I can't manage to simplify it.
I have looked for a solution on the web but none solution for the instant.
UPDATE 1:
From the following link Expectation of inverse of sum of random variables, if we have $X_i$'s ($i=1,..,n$) be i.i.d. random variables with mean $\mu$ and variance $\sigma^2$, there is a method that can be used to compute $\mathbb{E}[1/(X_1+...+X_n)]$ :
Assuming the expectation does exist, and further assuming $X$ to be positive random variables: $$ \mathbb{E}\left(\frac{1}{X_{1}+\cdots+X_{n}}\right)=\mathbb{E}\left(\int_{0}^{\infty} \exp \left(-t\left(X_{1}+\cdots+X_{n}\right)\right) \mathrm{d} t\right) $$ Interchanging the integral over $t$ with expectation: $$ \mathbb{E}\left(\int_{0}^{\infty} \exp \left(-t\left(X_{1}+\cdots+X_{n}\right)\right) \mathrm{d} t\right)=\int_{0}^{\infty} \mathbb{E}\left(\exp \left(-t\left(X_{1}+\cdots+X_{n}\right)\right)\right) \mathrm{d} t $$ Using iid property: $$ \int_{0}^{\infty} \mathbb{E}\left(\exp \left(-t\left(X_{1}+\cdots+X_{n}\right)\right)\right) \mathrm{d} t=\int_{0}^{\infty} \mathbb{E}(\exp (-t X))^{n} \mathrm{~d} t $$ So should you know the Laplace generating function $\mathcal{L}_{X}(t)=\mathbb{E}\left(\mathrm{e}^{-t X}\right)$ we have: $$ \mathbb{E}\left(\frac{1}{X_{1}+\cdots+X_{n}}\right)=\int_{0}^{\infty} \mathcal{L}_{X}(t)^{n} \mathrm{~d} t $$
How could I apply it in my case with $\Gamma$ distribution, i.e for the expectation $\Bigg\langle\left(\sum_{\ell=1}^{N} \Gamma\left((2 \ell+1) / 2,2 C_{\ell}^{\prime}\right)\right)^{-1}\Bigg\rangle$ ?
From 2) Important precision, the only thing I can reformulate is about the scale parameter $\dfrac{b}{b'}$ :
$$\Bigg\langle\left(\sum_{\ell=1}^{N} \Gamma\left((2 \ell+1) / 2,2 C_{\ell}^{\prime}\right)\right)^{-1}\Bigg\rangle=\Bigg\langle\left(\sum_{\ell=1}^{N} \dfrac{b'}{b}\Gamma\left((2 \ell+1) / 2,2 C_{\ell}\right)\right)^{-1}\Bigg\rangle$$
UPDATE 2:
I wonder if I should rather write only :
$$\begin{aligned} \sum_{\ell=1}^{N} \sum_{m=-\ell}^{\ell}\left(a_{\ell m}^{\prime}\right)^{2} & \sim \sum_{\ell=1}^{N} \sum_{m=-\ell}^{\ell} \Gamma\left(1 / 2,2 C_{\ell}\right) \\ & \sim \sum_{\ell=1}^{N} \Gamma\left((2 \ell+1) / 2,2 C_{\ell}\right) \end{aligned}$$
? what do you think about this slight modification but with important consequences on the following ?