# Nested sampling: estimate of bulk posterior support over prior

Going through the details of the Nested sampling Skilling paper, and I've encountered an estimate in Section 5 which I cannot reproduce. Rephrasing what's mentioned in the paper:

we assume to have a likelihood in $$C$$ dimensions, and we approximate it with a rank $$C$$ multivariate normal. Likelihood values $$L\propto\exp(-\frac{r^2}{2})$$ enclose a prior mass $$X\propto r^C$$, where $$r$$ is a radius in C dimension. Therefore the posterior $$\text{d}P\propto L \text{d}X=L X\text{d}\log X$$ induce a variance on $$\log X$$:

$$\begin{equation} \langle \delta (\log X)^2\rangle\propto C/2 \end{equation}$$

So the posterior is broadly distributed over a range $$-H \pm \sqrt{C}$$ in $$\log X$$, where $$H$$ is the information $$H=\int\log(\frac{dP}{dX})dP$$

I cannot wrap my head around the estimate for the variance of $$\log X$$, and the subsequent posterior bulk range. Have someone else gone through this derivation?

EDIT: Proof attempt

The quantity I need to compute is the variance of $$\log X$$ under the posterior distribution $$dP$$ $$\begin{eqnarray} \int (\log X - \langle\log X\rangle )^2 dP & = & \int_0^1 (\log X - \langle\log X\rangle )^2 L dX\\ &=& \int_{-\infty}^{0} (Y - \langle Y\rangle )^2 \exp({-(\exp{Y})^{\frac{2}{C}})}\exp{Y}dY \end{eqnarray}$$ however I cannot find the path down to the estimate $$C/2$$