Square root covariance update error and stability — Examples

The exact Gram matrix is $Z{Z}^T=\left[\begin{array}{cc}2& 1\\ 1& 1\end{array}\right]\mathrm{.}$ Its innovation variance is $2$, its cross covariance is $1$, and its exact posterior covariance is $1-1\mathrm{/}2=1\mathrm{/}2$.

The perturbed array and its Gram matrix are $\widehat{Z}=\left[\begin{array}{cc}1& 11\mathrm{/}10\\ 0& 1\end{array}\right],\widehat{Z}{\widehat{Z}}^T=\left[\begin{array}{cc}221\mathrm{/}100& 11\mathrm{/}10\\ 11\mathrm{/}10& 1\end{array}\right]\mathrm{.}$ Here $\mathrm{\parallel }E{\mathrm{\parallel }}_F=1\mathrm{/}10$, $\mathrm{\parallel }Z{\mathrm{\parallel }}_F=\sqrt{3}$, and the Gram difference is $\left[\begin{array}{cc}21\mathrm{/}100& 1\mathrm{/}10\\ 1\mathrm{/}10& 0\end{array}\right]$, whose Frobenius norm is $\sqrt{641}\mathrm{/}100$; it is bounded by [F1]. The perturbed innovation remains positive, and its Schur complement is $1-\frac{(11\mathrm{/}10{)}^2}{221\mathrm{/}100}=1-\frac{121}{221}=\frac{100}{221}>0.$ This differs from $1\mathrm{/}2$ by $-21\mathrm{/}442$.

By [F2], the perturbed Schur complement is exactly the represented factor product $L_{22}{L}_{22}^T$, so it is positive semidefinite; here it is the positive scalar $100\mathrm{/}221$. The nonzero prearray perturbation changes the posterior covariance, while the factor representation retains positive semidefiniteness.

Since $0<c<1$, for $(x_1,x_2)\mathrm{\ne }(0,0)$, $(x_1,x_2)P^{-}(x_1,x_2{)}^T=(x_1+c{x}_2{)}^2+(1-c^2)x_2^2>0.$ Thus $P^{-}$ is positive definite. Also $1-c^2=2^{-23}-2^{-48}>0$. The exact innovation is $S=H{P}^{-}{H}^T+R=4-2^{-23}+2^{-22}=4+2^{-23}>0.$

In precision $p=24$, the exact sum $1+c=2-2^{-24}$ is halfway between the representable predecessor $2-2^{-23}$ and $2$; ties-to-even rounds it to $2$. Hence the computed $B$ is $(2,2{)}^T$. The next exact addition is $4+2^{-22}$, halfway between $4$ and its next representable successor $4+2^{-21}$; ties-to-even rounds it to $S_{\mathrm{f}\mathrm{l}}=4$. All these operands and results are normal.

The computed gain is $K=(1\mathrm{/}2,1\mathrm{/}2{)}^T$. Multiplication gives $(KS)K^T=\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]$ exactly, so the rounded subtractive update is $P_{\mathrm{s}\mathrm{u}\mathrm{b}}=\left[\begin{array}{cc}0& -2^{-24}\\ -2^{-24}& 0\end{array}\right]\mathrm{.}$ Its eigenvalues are $2^{-24}$ and $-2^{-24}$, so it is indefinite. This shows cancellation can lose positive semidefiniteness even with an exact positive-definite prior and positive measurement variance. By contrast, when the factor procedure succeeds, [F3] gives a represented covariance that is an exact factor product; this example makes no claim that rounding a later materialization of that product preserves the property.

The prediction array is $[F{L}_0,L_Q]=[0,0]$. Its QR upper factor is zero, so $L^{-}=0$ and $P^{-}=0$. The predicted mean is $m^{-}=1\cdot 2=2$. The innovation is $v=7-1\cdot 2=5$, and its covariance is $S=H{P}^{-}{H}^T+R=1$. Thus the canonical innovation is positive definite even though $P_0$ and $Q$ are singular.

The measurement prearray is $Z=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\mathrm{.}$ Its exact QR gives $L_{11}=1$, $L_{21}=0$, $L_{22}=0$. The solve gives $z=5$, so $m^{+}=m^{-}+L_{21}z=2$, while the represented posterior covariance is $L_{22}{L}_{22}^T=0$.

The canonical covariance update is $P^{+}=0-0\cdot 1^{-1}\cdot 0=0$ and the canonical mean update is $2+0\cdot 1^{-1}\cdot 5=2$. They agree with the factor output in step 2.1. The zero posterior factor is admissible by [F3]; no Cholesky factor of a positive-definite state covariance is required.

The signal part of the observation variance is $H{P}^{-}{H}^T={10}^{-6}$, so $R\mathrm{/}(H{P}^{-}{H}^T)={10}^{-6}$ and the observation noise is small relative to the predicted signal variance. The innovation variance and cross covariance are $S={10}^{-6}+{10}^{-12}=\frac{1000001}{{10}^{12}},B={10}^{-3}\mathrm{.}$ Thus the exact gain is $K=\frac{B}{S}=\frac{{10}^9}{1000001}\mathrm{.}$

In the scalar block QR, $L_{11}^2=S$ and $L_{21}{L}_{11}=B$, so $L_{21}\mathrm{/}{L}_{11}=K$. If $L_{11}\widehat{z}-v=\rho ={10}^{-9}$, then $\widehat{z}=(v+\rho )\mathrm{/}{L}_{11}$. The induced mean-update error is therefore $L_{21}\widehat{z}-Kv=K\rho =\frac{1}{1000001}\approx 9.99999000001\times {10}^{-7}\mathrm{.}$ An absolute solve residual of ${10}^{-9}$ has produced an update error about ${10}^{-6}$ because the gain is about ${10}^3$. The $K^{\mathrm{\#}}\rho$ term in [F2] provides a certified upper-bound counterpart when only a gain norm bound is available.