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  1. I manage to compute the Poisson noise of a density field like this :

If I take $N$ the density of galaxies and compute the Shot noise with a Poisson distribution, I get :

$\langle N^2\rangle - \langle N\rangle^2 = \langle N\rangle$ so :

$\langle N^2\rangle = \langle N\rangle + \langle N\rangle^2$

Let's take the variable : $X=\dfrac{N}{\langle N\rangle}-1$

So, I get $\langle X\rangle=0$.

Then : $\langle X^2\rangle = \left\langle\left (\dfrac{N}{\langle N\rangle}-1\right )^2\right\rangle=\dfrac{\langle N^2\rangle}{\langle N\rangle^2}-2+1 = \dfrac{\langle N^2\rangle}{\langle N\rangle^{2}}-1$

Finally, I get : $\sigma_x^2 = \langle X^2\rangle=\dfrac{1}{\langle N\rangle}+1-1$

$$\Rightarrow\quad \sigma_x^2=\dfrac{1}{\langle N\rangle}$$

  1. Now, I would like to do the same kind of calculation with variable $Y=\left(\left(\dfrac{N}{\langle N\rangle}\right)-1\right)^2$, and conclude normally that :

$$\sigma_y^2 = \dfrac{1}{\langle N\rangle^2}\quad(1)$$

But following the same reasoning with $N$ following the Poisson distribution, I can't manage to get this expression $(1)$.

UPDATE : i think that I have the proof :

$$\langle X^4\rangle = \Bigg\langle\bigg(\dfrac{N}{\langle N\rangle}-1\bigg)^2\,\bigg(\dfrac{N}{\langle N\rangle}-1\bigg)^2\Bigg\rangle$$

$$=\dfrac{\langle N^2\rangle^2}{\langle N\rangle^4}-2\dfrac{\langle N^2\rangle}{\langle N\rangle^2}+1$$

I) with $\langle N^2\rangle = \langle N\rangle + \langle N\rangle^2$

II) and squared :

$$\langle N^2\rangle^2 = \langle N\rangle^2 + 2\langle N\rangle^3 + \langle N\rangle^4$$

We conclude :

$$\langle X^4\rangle=\dfrac{1}{\langle N\rangle^2}+\dfrac{2}{\langle N\rangle} +1 - \dfrac{2}{\langle N\rangle} -2 + 1 = \dfrac{1}{\langle N\rangle^2}$$

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  • $\begingroup$ It's not clear to me what you mean by "Shot Noise", particularly as this is usually a synonym for noise that follows Poisson statistics. (e.g., en.wikipedia.org/wiki/Shot_noise) $\endgroup$ Jun 1 at 12:51
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    $\begingroup$ @uhoh A Poisson distribution has (using the symbols from the Wiki page) an expectation value of $\lambda$ (not $k$) and a variance of $\lambda$. For the "density of galaxies" problem, $\lambda = \bar{n}_{\rm gal}$, so I don't understand why the OP says the "Shot Noise" should be $1/\bar{n}_{\rm gal}$. $\endgroup$ Jun 2 at 9:26
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    $\begingroup$ @uhoh . Thanks for your answer. Unfortunately, I don't know yet how to justify that the expectation of Shot Noise $\bar{N}_{sp}$ should be equal to 0 like a white noise. I am also interested in the variance of Shot Noise : if I take a fixed value for the spectroscopic Shot Noise, then, the variance is equal to : $<(N_{sp}-\bar{N}_{sp})^{2}>=1/\bar{n}_{\rm gal}^{2}$. Could anyone confirm this result ? Regards $\endgroup$
    – youpilat13
    Jun 2 at 10:34
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    $\begingroup$ @youpilat13 I'm afraid I have no idea what you're trying to get at. Your last comment mentions "spectroscopic Shot Noise" -- whatever that's supposed to mean -- and then tries to relate that to the mean density of galaxies. That makes no sense. $\endgroup$ Jun 3 at 11:39
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    $\begingroup$ @youpilat13 Have you tried asking about this on statistics.stackexchange? That seems like a much more appropriate place, since your question appears to be about estimating variances by calculations and manipulation of expectation values, and not really anything astronomical. $\endgroup$ Aug 11 at 13:03

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