# Formula for cosmic variance

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?

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.
• 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 $\Delta C_\ell$.
• @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 ?)