Separate stationary solve residual from generator error

Let Q be the true finite column generator, let $\nu >0$ satisfy $\nu \ge {\max }_j{r}_j$, and put $P=I+Q\mathrm{/}\nu$. Suppose $q=\delta (P^m)<1$ for some $m\ge 1$. Let $\tilde{Q}$ be any computed matrix model on the same finite state set, let $y\in {\mathbb{R}}^n$, and let x be a column probability vector with $c=\mathrm{\parallel }y-x{\mathrm{\parallel }}_1$. Then, for the unique stationary probability ${\pi }_Q$ of Q,

$\mathrm{\parallel }y-{\pi }_Q{\mathrm{\parallel }}_1\le c+\frac{m}{\nu (1-q)}(\mathrm{\parallel }\tilde{Q}x{\mathrm{\parallel }}_1+\mathrm{\parallel }(Q-\tilde{Q})x{\mathrm{\parallel }}_1)\mathrm{.}$

The first residual is the solve residual for the computed model; the second is the model perturbation evaluated at the chosen candidate. If $\tilde{Q}x=0$, the solve-residual term vanishes. Every term must be supplied as an exact value or a certified outward upper bound before the displayed expression is reported as a numerical certificate.

In particular, if P and y are rational and the exact branch of Correct and certify a finite-chain stationary candidate returns certified, its output x and c satisfy these conditions.