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 :

  1. $\mathcal{N}(0, C)^{2} \sim C\chi^{2}(1)=\Gamma\left(\frac{1}{2}, 2 C\right)$,
  2. $\langle\Gamma(k, \theta)\rangle=k \theta$ and $\operatorname{Var}(\Gamma(k, \theta))=k \theta^{2}$, and
  3. $\sum_{i=1}^{N} \Gamma\left(k_{i}, \theta\right) = \Gamma\left(\sum_{i=1}^{N} k_{i}, \theta\right)$ 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.


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 $\dfrac{b}{b'}$ :

$$\Bigg\langle\left(\sum_{\ell=1}^{N} \Gamma\left((2 \ell+1) / 2,2 C_{\ell}^{\prime}\right)\right)^{-1}\Bigg\rangle=\Bigg\langle\left(\sum_{\ell=1}^{N} \dfrac{b'}{b}\Gamma\left((2 \ell+1) / 2,2 C_{\ell}\right)\right)^{-1}\Bigg\rangle$$


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 ?

  • 1
    $\begingroup$ Chi-squared distributions are Gamma distributions. Although your notation is unclear, one can infer that the "$C_l$" are related to scale rather than shape. Regardless, linear combinations of Gamma distributions of different shapes are not Gamma distributions. See stats.stackexchange.com/questions/72479 for the calculation. $\endgroup$
    – whuber
    Commented Aug 6, 2021 at 12:49
  • 1
    $\begingroup$ Re the edit: your attempt looks invalid at the outset. (I have to make assumptions about your notation and what "$a^\prime$" might refer to.) Think about the simpler case of a ratio of random variables where the denominator can take on the values $\pm 1$ with equal probability, so that (a) the fraction is just the numerator with a random sign, making it clear that if the numerator has an expectation, then so does the fraction; but (b) the expectation of the denominator is zero. What formula would you try to write in this case that is analogous to the one in your attempt? $\endgroup$
    – whuber
    Commented Aug 6, 2021 at 15:38
  • 1
    $\begingroup$ You seem to be asking about the distribution of the ratio of two independent quantities, each of which is the sum of squares of independent Normal distributions of different variances. Would that be correct? If the variances were all the same, you would have an $F$ ratio distribution; but when they are not the same, you will get the messy result I linked to earlier (concerning sums of Gamma distributions) for both the numerator and the denominator. The distribution of the ratio will not then be calculable in a closed form. $\endgroup$
    – whuber
    Commented Aug 6, 2021 at 18:09
  • 4
    $\begingroup$ This question is unpleasantly long. It could be shorter — and more appealing to read and to answer — by removing the inner sums and normal variables entirely, and asking directly about a ratio of sums of $\Gamma(\ell+\frac12,2C_\ell)$ variables. $\endgroup$
    – Matt F.
    Commented Aug 10, 2021 at 10:24
  • 1
    $\begingroup$ I was really hoping to see if there was some way to sum gamma distributions having scale parameters that were not all equal and were not all integers. Apparently, the 1985 Moschopoulos paper and the Welch-Satterthwaite approximation are still the two best options. It would be nice to be proven wrong! $\endgroup$
    – Ed V
    Commented Aug 20, 2021 at 13:13

1 Answer 1


$$O=\frac{\sum_{\ell=1}^{N} \sum_{m=-\ell}^{\ell} a_{\ell m}^{2}}{\sum_{\ell=1}^{N} \sum_{m=-\ell}^{\ell}\left(a_{\ell m}^{\prime}\right)^{2}}$$

$a_{\ell m}\sim N(0,C_{\ell})$ so

Preface: I assume that $C_{\ell}={Var(a_{\ell m})}$.

Let $a_{\ell m}\sim N(0,C_{\ell})$ so $\frac{a_{\ell m}}{\sqrt{C_{\ell}}}\sim N(0,1)$. Using this, we get $a_{\ell m}^2=C_{\ell}\cdot \left(\frac{a_{\ell m}}{\sqrt{C_{\ell}}}\right)^2\sim\chi^2_{C_{\ell}}$, Summing up, we get $\sum_{m=-\ell}^{\ell} {a_{\ell m}^{2}}\sim\chi^2_{(2\ell+1)C_{\ell}}$, and for the overall sum we get

$$\sum_{\ell=1}^{N} {\sum_{m=-\ell}^{\ell} a_{\ell m}^{2}}\sim\chi^2\left(\sum_{\ell=1}^{N}{(2\ell+1)C_{\ell}}\right)$$ which can also be written as $\Gamma(\frac{1}{2}\sum_{\ell=1}^{N}{(2\ell+1)C_{\ell}},2)$. For simplicity, denote $K=\sum_{\ell=1}^{N}{(2\ell+1)C_{\ell}}$, so the numerator as $\Gamma(0.5K,2)$ distribution. We can also note that $$2\sum_{\ell=1}^{N} {\sum_{m=-\ell}^{\ell} a_{\ell m}^{2}}\sim\Gamma(K,1).$$

Let's observe the denominator:

$$\sum_{\ell=1}^{N} \sum_{m=-\ell}^{\ell}\left(a_{\ell m}^{\prime}\right)^{2}\sim\Gamma(\frac{1}{2}\sum_{\ell=1}^{N}{(2\ell+1)C'_{\ell}},2)$$ as $C'_{\ell}=\frac{b'}{b}C_{\ell}$, we get $$2\sum_{\ell=1}^{N} {\sum_{m=-\ell}^{\ell} \left(a'_{\ell m}\right)^{2}}\sim\Gamma(\frac{b'}{b}K,1).$$

As these are two Gamma variables, the ratio $O=\frac{2\sum_{\ell=1}^{N} {\sum_{m=-\ell}^{\ell} a_{\ell m}^{2}}}{2\sum_{\ell=1}^{N} {\sum_{m=-\ell}^{\ell} \left(a'_{\ell m}\right)^{2}}}$ as a beta prime distribution:

$$\frac{2\sum_{\ell=1}^{N} {\sum_{m=-\ell}^{\ell} a_{\ell m}^{2}}}{2\sum_{\ell=1}^{N} {\sum_{m=-\ell}^{\ell} \left(a'_{\ell m}\right)^{2}}}\sim\beta'\left(K, \frac{b'}{b}K\right)$$

So, according to the properties of the beta prime distribution, $$E\left[ \frac{\sum_{\ell=1}^{N} \sum_{m=-\ell}^{\ell} a_{\ell m}^{2}}{\sum_{\ell=1}^{N} \sum_{m=-\ell}^{\ell}\left(a_{\ell m}^{\prime}\right)^{2}} \right]=\frac{K}{\frac{b'}{b}K-1}=\frac{bK}{b'K-b}=\frac{b\sum_{\ell=1}^{N}{(2\ell+1)C_{\ell}}}{b'\sum_{\ell=1}^{N}{(2\ell+1)C_{\ell}} - b}$$

and if $\frac{b'}{b}K > 2$,

$$Var\left( \frac{\sum_{\ell=1}^{N} \sum_{m=-\ell}^{\ell} a_{\ell m}^{2}}{\sum_{\ell=1}^{N} \sum_{m=-\ell}^{\ell}\left(a_{\ell m}^{\prime}\right)^{2}} \right)=\frac{K(K+\frac{b'}{b}K-1)}{(\frac{b'}{b}K-1)^2(\frac{b'}{b}K-2)}=\frac{b^2K(bK+b'K-b)}{(b'K-b)^2(b'K-2b)}=\frac{b^3K^2+b^2K(b'k-b)}{(b'K-b)^3-b(b'k-b)^2}.$$

Dirty, but that's it.

  • $\begingroup$ Hi Spätzie ! Thanks for your detailled answer. I have a difficulty on this step : Can we have : $\begin{aligned} \sum_{\ell=1}^{N} \sum_{m=-\ell}^{\ell} a_{\ell m}^{2} & \sim \sum_{\ell=1}^{N}(2 \ell+1) \chi^{2}\left(C_{\ell}\right) \\ & \sim \chi^{2}\left(\sum_{\ell=1}^{N}(2 \ell+1) C_{\ell}\right) \end{aligned}$ if I have a different value of $C_\ell$ for each different $\ell$. I wonder we can write this above when we have a linear combination of $\chi^2$. $\endgroup$
    – user226073
    Commented Aug 17, 2021 at 3:22
  • $\begingroup$ .When i say "linear combination", I talk about the sum : $\sum_{\ell=1}^{N} \sum_{m=-\ell}^{\ell} a_{\ell m}^{2} \stackrel{d}{=} \sum_{\ell=1}^{N}(2 \ell+1) \chi^{2}\left(C_{\ell}\right)$ $\endgroup$
    – user226073
    Commented Aug 17, 2021 at 3:22
  • $\begingroup$ Recall that $\chi^2(N)$ is the sum $\sum_{i=1}^{N}{Z_i^2}$ where $Z_i\sim N(0,1)$. As long as you're summing over standard normal variables (which is the case, as I've shown above) you're ok to add up the $C_{\ell}$'s. $\endgroup$
    – Spätzle
    Commented Aug 17, 2021 at 5:00
  • $\begingroup$ thanks for your support. Actually, the normal distribution is $\dfrac{a_{\ell m}}{\sqrt{C_{\ell}}} \sim N(0,1)$ and you write after : $a_{\ell m}^{2}=C_{\ell} \cdot\left(\dfrac{a_{\ell m}}{\sqrt{C_{t}}}\right)^{2} \sim \chi_{C_{\ell}}^{2}$ : how do you get rif of the first factor $C_\ell$ in the right hand side of this expression, i.e how to conclude that : $C_{\ell} \cdot\left(\dfrac{a_{\ell m}}{\sqrt{C_{t}}}\right)^{2} \sim \chi_{C_{\ell}}^{2}$ knowing we have a $C_\ell$ different for each multipole $\ell$ : $\endgroup$
    – user226073
    Commented Aug 17, 2021 at 21:54
  • 2
    $\begingroup$ This attempted solution errs at the outset: the sum of chi-squared variates with (potentially) different scale factors does not have a chi-squared distribution. $\endgroup$
    – whuber
    Commented Aug 20, 2021 at 13:55

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