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I want to clarify a point that disturbs me among different cases.

I am interested in formulate correctly in a general case when we know the distribution of different random variables and we want to calculate the distribution of the sum of these random variables.

For example, the distribution of 2 random variables following a different uniform distribution is not the sum of the 2 distributions, we are ok ? (by the way, how to compute the distribution of this sum ?).

On another side, when we take 2 random variables following a different normal distribution, the PDF of the sum of both is a PDF which is the sum of 2 Gaussians ? I think that it would logical but I am not sure (regarding my first example above), especially for the normalization of the PDF of sum which has to be equal to 1 when integrate over all the domain.

Now, in my case, I have to compute the distribution of the following quantity :

$$\sum_{\ell=\ell_{\min }}^{\ell_{\max }} \sum_{m=-\ell}^{\ell} a_{\ell m}^{2}$$

with the random variable $a_{\ell m}$ following a normal centered on 0 and with a variance equal to $C_{\ell}$.

So, I decided to begin firstly by the quantity $\sum_{m=-\ell}^{\ell} a_{\ell m}^{2}$ from a distribution point of view :

We recall the properties of a few basic distributions.

  1. $\mathcal{N}(0, C_{\ell})^2$ distribution is equivalent to $C_{\ell}\,\chi^2(1)=\Gamma(\frac{1}{2}, 2C_{\ell})$ distribution.
  2. The distribution $\sum_{i=1}^N\Gamma(k_i, \theta)$ is equivalent to a $\Gamma (\sum_{i=1}^N k_i, \theta)$ distribution for independent summands.

Let us formulate the distribution followed by this random variable. Using previous points 1 and 2, we obtain : $$ \begin{align} \sum_{m=-\ell}^\ell (a_{\ell m})^2&= \sum_{m=-\ell}^{\ell} C_{\ell} \cdot\left(\frac{a_{\ell, m}}{\sqrt{C_{\ell}}}\right)^{2} \label{sum_alm} \end{align} $$

We can develop the distribution of this observable like this :

  1. $a_{\ell m}$ follows a $\mathcal{N}(0, C_{\ell})$ distribution.
  2. $\sum_{m=-\ell}^{\ell} C_{\ell} \cdot\left(\dfrac{a_{\ell, m}}{\sqrt{C_{\ell}}}\right)^{2}$ follows a $\sum_{m=-\ell}^{\ell} C_{\ell} \, \mathrm{\chi^2}(1)$ distribution.
  3. $\sum_{m=-\ell}^{\ell} C_{\ell}\,\mathrm{\chi^2}(1)$ distribution is equivalent to a $C_{\ell}\,\sum_{m=-\ell}^{\ell}\, \mathrm{\chi^2(1)}$ distribution.
  4. $C_{\ell} \sum_{m=-\ell}^{\ell} \mathrm{\chi^2}(1)$ is equivalent to a $C_{\ell} \,\mathrm{\chi^2}(2\ell+1)$ distribution.
  5. $C_{\ell}\,\mathrm{\chi^2}(2 \ell+1)$ distribution is equivalent to $C_{\ell}\,\mathrm{Gamma}((2\ell+1)/2, 2)$ distribution.
  6. $C_{\ell}\,\mathrm{Gamma}((2\ell+1)/2, 2)$ is equivalent to a $\mathrm{Gamma}((2\ell+1)/2, 2C_\ell)$ distribution.

We have taken the convention (shape,scale) parameters for $\mathrm{Gamma}$ distribution. Given the fact that we consider the random variable :

$$ \begin{equation} \sum\limits_{\ell=\ell_{min}}^{\ell_{max}}\sum\limits_{m=-\ell}^{\ell} a_{\ell m}^2 \end{equation} $$

This sum of random variables $\sum\limits_{m=-\ell}^{\ell} a_{\ell m}^2$ follows a Moschopoulos distribution : it represents the distribution of the sum of random variables each one following a $Gamma$ distribution with different shape and scale parameters.

Is this reasoning correct ? I mean, about the sum of random variables $\sum\limits_{m=-\ell}^{\ell} a_{\ell m}^2$ which follows a Moschopoulos distribution ?

Indeed, if it is correct, the random variables $\sum\limits_{m=-\ell}^{\ell} a_{\ell m}^2$ follow a $\text{Gamma}$ distribution with different shape and scale parameters.

As conclusion, my main questions :

1) Is my reasoning above correct ? I mean does the random variable $\sum_{\ell=\ell_{\min }}^{\ell_{\max }} \sum_{m=-\ell}^{\ell} a_{\ell m}^{2}$ follows a Moschopoulos distribution ?

2) How to express correctly the things when I want to say for example, the distribution of sum of random variable "$X_i$" is equivalent to a "given" distribution or the sum of random variables "$X_i$" follows a "given" distribution ?

3) How to demonstrate that the convolution operation preserve the area of convolution to be equal to 1 when we integrate from all the domain ?

I make confusions between "equality from a distribution point of view", "the following distribution of a random variable" and the "sum of PDF".

Hoping have been enough clear.

youpilat13
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  • You *obviously* cannot get the pdf of a sum by summing pdfs, because the sum of two pdfs will integrate to 2, not 1, and hence cannot be a pdf. If the random variables are independent, the pdf of their sum is the *convolution* of their pdfs. The case resulting in the convolution of two different uniforms is discussed on site already, you might try some searches. – Glen_b Nov 05 '21 at 03:03
  • @Glen_b . I tried to look at some basic examples of "convolution of PDF". By the way, why one tells "convolution" : is it about the characteristic function (Fourier transform of PDF) ? If you could give me a simple example, I would be grateful. But surely I will launch a bounty to draw more attention on this not so much simple isssue. This may help other persons I think for a general approach. Regards – youpilat13 Nov 05 '21 at 17:02
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    This is standard stuff; it should be in any undergrad stats text that's designed for students with enough basic mathematics to do the calculations; I'd start there (eg Larsen&Marx). The density of a sum in the general case is computed *via integration* from the joint pdf. With independent variables that simplifies to a convolution; see https://en.wikipedia.org/wiki/Convolution_of_probability_distributions#Introduction starting from "If we start with random variables *X* and *Y*" to the end of that subsection, which does that exact simplification from the general case to the independent case. – Glen_b Nov 05 '21 at 23:37
  • @Glen_b . Ok thanks. Just a detail (though important) : Does the convolution operation preserve the area of convolution to be equal to 1 when we integrate from all the domain ? Regards – youpilat13 Nov 06 '21 at 08:29
  • Yes, the area of the result is as it should be. – Glen_b Nov 06 '21 at 12:21
  • @Glen_b . Thanks. Is it always gauranteed for sum of any random variables following any PDF. How to demonstrate it please ? Regards – youpilat13 Nov 06 '21 at 12:26
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    Your questions about convolution are addressed at https://stats.stackexchange.com/questions/331973, *inter alia.* – whuber Nov 06 '21 at 19:44

1 Answers1

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1) Is my reasoning above correct ? I mean does the random variable $\sum_{\ell=\ell_{\min }}^{\ell_{\max }} \sum_{m=-\ell}^{\ell} a_{\ell m}^{2}$ follows a Moschopoulos distribution ?

The reasoning is correct. You have a sum of variables $a_{\ell m}^2$ that are gamma distributed. That matches the description for the Moschopoulos distribution.

See also this question Generic sum of Gamma random variables

2) How to express correctly the things when I want to say for example, the distribution of sum of random variable "$X_i$" is equivalent to a "given" distribution or the sum of random variables "$X_i$" follows a "given" distribution ?

For example: "The distribution of the sum $ \sum\limits_{\ell=\ell_{min}}^{\ell_{max}}\sum\limits_{m=-\ell}^{\ell} a_{\ell m}^2$ follows a Moschopoulos distribution."

3) How to demonstrate that the convolution operation preserve the area of convolution to be equal to 1 when we integrate from all the domain ?

This is a property of convolution. If the convolution is

$$h(x) = \int_{-\infty}^\infty f(y)g(x-y)dy$$

Then the integral of $h(x)$ is

$$ \int_{-\infty}^\infty h(x) dx = \left( \int_{-\infty}^\infty f(x)dx \right) \cdot \left(\int_{-\infty}^\infty g(x)dx\right) = 1 \cdot 1 = 1$$

Which follows from Fubini's theorem which states that you can change the order of integration

$$ \begin{array}{} \int_{-\infty}^\infty h(x) dx & =& \int_{-\infty}^\infty \left(\int_{-\infty}^\infty f(y)g(x-y)dy\right)dx\\ &=& \int_{-\infty}^\infty \left(\int_{-\infty}^\infty f(y)g(x-y)dx\right)dy \\&=& \int_{-\infty}^\infty f(y)\underbrace{ \left(\int_{-\infty}^\infty g(x-y)dx\right)}_{=1}dy \\ &=& \int_{-\infty}^\infty f(y) dy \\ &=& 1 \end{array}$$

Sextus Empiricus
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  • Thanks for your answer. Just a detail, inner integral $\int_{-\infty}^\infty g(x-y)dx$ is always true ? Indeed, this is equivalent to a "shifted" PDF since the presence of $(x-y)$ but if we have a finite interval (for example $[a,b]$), doesn't it create an issue from a mathematical point of view ? Surely by taking $z=x-y$, we have $dz=dx$ and so : $\int_{-\infty}^\infty g(x-y)dx=\int_{-\infty}^\infty g(z)dz=1$ : do you agree with this little demo ? – youpilat13 Nov 07 '21 at 04:38
  • @youpilat13 you are right that it will be different for a finite interval. But in that case the convolution won't be nicely defined either. – Sextus Empiricus Nov 07 '21 at 07:08