Data assimilation: 3DVAR, 4DVAR, and ensemble filters — Examples

For $N=2$, $\bar{X}=(X_1+X_2)\mathrm{/}2$, so $X_1-\bar{X}=(X_1-X_2)\mathrm{/}2$ and $X_2-\bar{X}=-(X_1-X_2)\mathrm{/}2$. By [F3], $C_2=\frac{1}{2}((X_1-\bar{X}{)}^2+(X_2-\bar{X}{)}^2)=\frac{(X_1-X_2{)}^2}{4}\mathrm{.}$

The event $A=\{-0.1<X_1<0.1,3<X_2<3.1\}$ has positive probability: each interval has positive probability by [F1], and the variables are independent. On $A$, $X_2-X_1>2.9$, so step 1.1 gives $C_2>{2.9}^2\mathrm{/}4=2.1025>1$. The true variance of $N(0,1)$ is $1$ by [F2]; hence the finite-ensemble covariance differs from it with positive probability.

Since $u_1=\mathrm{\Psi }(u_0)=u_0$, the one-variable objective from [F1] is $J(u)=\frac{1}{2}{u}^2+\frac{1}{2}(1-u^2{)}^2=\frac{1}{2}-\frac{1}{2}{u}^2+\frac{1}{2}{u}^4\mathrm{.}$

Differentiating gives $J^{\mathrm{\prime }}(u)=u(2u^2-1)$. Its only stationary points are $0$ and $\pm 1\mathrm{/}\sqrt{2}$. The derivative changes from negative to positive at each of $\pm 1\mathrm{/}\sqrt{2}$, so both are strict local minima. The objective is even, $J(\pm 1\mathrm{/}\sqrt{2})=3\mathrm{/}8<J(0)=1\mathrm{/}2$, and $J(u)\to \mathrm{\infty }$ as $\mathrm{\mid }u\mathrm{\mid }\to \mathrm{\infty }$. Every global minimum is therefore attained at a stationary point, and these two distinct points are both global minimizers. Thus the minimizer is not unique, so an iterative routine's finite definition gives no guarantee that it returns a designated unique solution.

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.

Here $S=HB{H}^T+\mathrm{\Gamma }=2$ and $K=B{H}^T{S}^{-1}=1\mathrm{/}2$. Thus the analysis is $m^a=0+(1\mathrm{/}2)(2-0)=1$.

The independent Gaussian prior and observation-error model in the Example is the one-step model specified by [F1]. Completing the square gives $J(u)=\frac{1}{2}{u}^2+\frac{1}{2}(2-u{)}^2=(u-1{)}^2+1$, so its unique minimizer is $1$, agreeing with step 1.1 and [F3]. The Kalman covariance is $P^a=1-(1\mathrm{/}2)2(1\mathrm{/}2)=1\mathrm{/}2$ by [F4]; [F5] therefore identifies the exact conditional law as $N(1,1\mathrm{/}2)$. The 3DVAR output is its MAP point and supplies no covariance output.