I was reading this page: Sample and Cosmic Variance. The section states that

The multipoles $C_\ell$ can be related to the expected value of the spherical harmonic coefficients by
$$ \Bigg\langle \sum_{m=-\ell}^{\ell} a^2_{\ell m}\Bigg\rangle = (2\ell+1)C_\ell $$ since there are $(2\ell + 1)$ $a_{\ell m}$ for each $\ell$ and each has an expected autocorrelation of $C_\ell$. In a theory such as inflation, the temperature fluctuations follow a Gaussian distribution about these expected ensemble averages. This makes the $a_{\ell m}$ Gaussian random variables, resulting in a $\chi^2_{2\ell+1}$ distribution for $\sum_m a_{\ell m}^2$. The width of this distribution leads to a cosmic variance in the estimated $C_{\ell}$ of
$$ \bigg(\frac{\Delta C_\ell}{C_\ell}\bigg)_\mathrm{cosmic\,variance}=\;\sqrt{\frac{2}{2\ell+1}} $$

I don't get how the cosmic variance is derived. Can someone explain it?


1 Answer 1


The key statement is that the $a_{\ell,m}$ are independent Gaussian random variables. For each $\ell$, there are $2\ell+1$ of them. So their sum is, essentially by definition, a chi-squared distribution with $2\ell+1$ degrees of freedom. Now, it is a known fact that the variance of a chi-squared distribution with $k$ degrees of freedom is just $2k$, so the variance of $\sum_m a_{\ell,m}^2$ is therefore $2(2\ell+1)$. This is a basic result from statistics, and you should be able to find a derivation in any textbook.

  • $\begingroup$ Thank you, but where does the square root come from? $\endgroup$
    – snowball
    Commented Jun 21, 2016 at 20:21
  • $\begingroup$ Well, your notation is a little unclear, so I can't say precisely, but basically it comes from the normalization that you have to do to get this result. If you look at the standard normal distribution, there's a $1/\sqrt{2\pi \sigma^2}$ out front to normalize it. This would seem to result in something like your last expression, but I don't know exactly what you mean by $\Delta C_\ell$. $\endgroup$
    – Mike
    Commented Jun 21, 2016 at 21:31
  • $\begingroup$ @Mike : Does the expression $\bigg(\frac{\Delta C_\ell}{C_\ell}\bigg)_\mathrm{cosmic\,variance}$ refer simply to the standard deviation of Cosmic variance ? (but under the form of a relative error ?) $\endgroup$
    – user87745
    Commented May 15, 2021 at 6:14

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