A nonlinear observation makes a symmetric two-member EnKF miss posterior spread

The predicted observations of both ensemble members equal $1$, so their empirical observation variance is $0$. Their empirical state-observation cross-covariance is also $0$, because the observation deviations are both zero. By [F5] and $\gamma >0$, the gain is zero; the update is therefore $x^{a,1}=-1$, $x^{a,2}=1$ for every perturbation realization. The updated empirical variance remains $1$.

By [A1], [F1], and [F3], the prior and Gaussian likelihood have densities. The likelihood is positive and bounded, so its prior integral is finite and positive; Bayes' formula [F4] makes the posterior density proportional to the even function $e^{-x^2\mathrm{/}2}w(x^2)$, where $w(t)=e^{-t^2\mathrm{/}(2\gamma )}$. The posterior mean is therefore zero by symmetry. Put $T=X^2$. Since $\mathbb{E}T=1$ by [F2], $\mathbb{E}[(T-1)w(T)]=\mathbb{E}[(T-1)(w(T)-w(1))]<0.$ The equality uses $\mathbb{E}(T-1)=0$. The inequality holds because $w$ is strictly decreasing on $[0,\mathrm{\infty })$, so the integrand is negative whenever $T\mathrm{\ne }1$; the standard normal density gives $\mathbb{P}(T=1)=0$. The integrand is integrable since $0<w\le 1$ and $\mathbb{E}T<\mathrm{\infty }$. Thus the posterior variance is $\mathrm{Var}(X\mid Y=0)=\frac{\mathbb{E}[Tw(T)]}{\mathbb{E}[w(T)]}<1.$ The finite EnKF ensemble variance from step 1.1 is $1$, so this admissible nonlinear finite-ensemble update does not equal the exact posterior law.