Finite generator master equations and uniformization — Examples

The matrix $Q$ in the Example has off-diagonal entries $\alpha ,\beta \ge 0$ and both columns sum to zero, so it satisfies [F1]; its exit rates are $\alpha$ and $\beta$.

Suppose $s>0$ and choose $\nu =s$. Then $P=I+Q\mathrm{/}s=\left(\begin{array}{cc}\beta \mathrm{/}s& \beta \mathrm{/}s\\ \alpha \mathrm{/}s& \alpha \mathrm{/}s\end{array}\right)$. Both columns equal the probability vector $v=(\beta \mathrm{/}s,\alpha \mathrm{/}s{)}^T$. Therefore $P{e}_{\mathrm{o}\mathrm{f}\mathrm{f}}=v$ and $Pv=v$, which by induction gives $P^k{e}_{\mathrm{o}\mathrm{f}\mathrm{f}}=v$ for every $k\ge 1$.

By [F2], the on-coordinate at time $t$ is ${\sum }_{k\ge 1}{e}^{-st}(st{)}^k(\alpha \mathrm{/}s)\mathrm{/}k!$. By [F3] the sum of the masses over $k\ge 1$ is $1-e^{-st}$, so $p_{\mathrm{o}\mathrm{n}}(t)=\frac{\alpha }{s}(1-e^{-st})$. The off-coordinate is $1-p_{\mathrm{o}\mathrm{n}}(t)$ because the solution is a probability vector.

If $s=0$, nonnegativity of $\alpha ,\beta$ implies $\alpha =\beta =0$, so $Q=0$ and the master equation keeps $e_{\mathrm{o}\mathrm{f}\mathrm{f}}$ fixed. If $s>0$ and $\alpha =0$, the formula gives $p_{\mathrm{o}\mathrm{n}}=0$; if $\beta =0$, it gives $p_{\mathrm{o}\mathrm{n}}=1-e^{-\alpha t}$. The endpoint cases agree with the stated rates.

The off-diagonal entries of $Q$ are nonnegative and its columns sum to zero; its exit rates are $1,2,2$. Thus [F1] holds, $\nu =2$ dominates every exit rate, and $P=I+Q\mathrm{/}2=\left(\begin{array}{ccc}1\mathrm{/}2& 1\mathrm{/}2& 0\\ 1\mathrm{/}2& 0& 1\\ 0& 1\mathrm{/}2& 0\end{array}\right)$.

Starting from $e_0$, successive multiplication by this $P$ gives $P{e}_0=(1\mathrm{/}2,1\mathrm{/}2,0{)}^T$, $P^2{e}_0=(1\mathrm{/}2,1\mathrm{/}4,1\mathrm{/}4{)}^T$, $P^3{e}_0=(3\mathrm{/}8,1\mathrm{/}2,1\mathrm{/}8{)}^T$, and $P^4{e}_0=(7\mathrm{/}16,5\mathrm{/}16,1\mathrm{/}4{)}^T$. Each vector is obtained by multiplying the preceding vector by the displayed matrix.

By [F4], the first five Poisson weights at parameter $1$ are $e^{-1}(1,1,1\mathrm{/}2,1\mathrm{/}6,1\mathrm{/}24)$. Weighting the five vectors from $e_0$ through $P^4{e}_0$ and adding coordinatewise gives $u_4=e^{-1}(703\mathrm{/}384,277\mathrm{/}384,5\mathrm{/}32{)}^T$ and total retained mass $65\mathrm{/}(24e)$. By normalization [F4], the exact omitted mass is $1-65\mathrm{/}(24e)=\mathbb{P}(\mathrm{Pois}(1)>4)$. This subtraction is an exact identity, not a floating-point tail-estimation rule.

Applying [F3] with $\mu =1$ and $K=4$ gives $\mathbb{P}(\mathrm{Pois}(1)>4)\le e^{-1}\mathrm{/}(5!(1-1\mathrm{/}6))=1\mathrm{/}(100e)$. This upper bound is computed from a positive geometric majorant and does not use a floating-point subtraction.

The system has two channels, increments $+1$ and $-1$, and propensities $a_{+}(n)=1$ and $a_{-}(n)=n$; the death channel is inactive at $n=0$, so every positive-propensity poststate remains in ${\mathbb{N}}_0$ and the total rate $1+n$ is finite. Thus [F1] applies.

Let $V(n)=1+n$. Its sublevel set $\{n:V(n)\le m\}$ is finite for each integer $m\ge 1$, so $V$ is proper. For $n\ge 1$, $QV(n)=1(V(n+1)-V(n))+n(V(n-1)-V(n))=1-n\le 1+n=V(n)$; at $n=0$ the same value $QV(0)=1$ satisfies this bound. Hence [F2] with $c=1$ and [A1] proves the immigration-death chain is nonexplosive, as required by [F3].

On $F=\{0,1\}$, state $0$ jumps to $1$ at rate $1$; state $1$ jumps to $0$ at rate $1$ and exits to $2\mathrm{\notin }F$ at rate $1$. These off-diagonal rates are nonnegative, and the cemetery construction [F4] gives the displayed ${\tilde{Q}}_F$ with zero column sums. Multiplying by the lifted probability vector gives $p_0^{\mathrm{\prime }}=-p_0+p_1$, $p_1^{\mathrm{\prime }}=p_0-2p_1$, and $c_F^{\mathrm{\prime }}=p_1$, with initial vector $(1,0,0{)}^T$.

From $p_1^{\mathrm{\prime }}=p_0-2p_1$ we have $p_0=p_1^{\mathrm{\prime }}+2p_1$; differentiating and using $p_0^{\mathrm{\prime }}=-p_0+p_1$ yields $p_1^{\mathrm{\prime }\mathrm{\prime }}+3p_1^{\mathrm{\prime }}+p_1=0$, with $p_1(0)=0$ and $p_1^{\mathrm{\prime }}(0)=1$. The characteristic roots are $r_{\pm }=(-3\pm \sqrt{5})\mathrm{/}2$ and $r_{+}-r_{-}=\sqrt{5}$, so the unique solution is $p_1(t)=(e^{r_{+}t}-e^{r_{-}t})\mathrm{/}\sqrt{5}$.

Since $c_F^{\mathrm{\prime }}=p_1$ and $c_F(0)=0$, integrating the expression in step 2.1 gives the displayed formula for $c_F(t)$. By [F6], this coordinate is $\mathbb{P}({\tau }_F\le t)$ and equals the embedded killed-vector ${\mathrm{\ell }}^1$ error.

Fix $t>0$ and choose ordered times $0<s_1<s_2<s_3<t$. The joint density for jumps $0\to 1$ at $s_1$, $1\to 2$ at $s_2$, and $2\to 1$ at $s_3$, followed by no further jump through $t$, is $e^{-s_1}\cdot 1\cdot e^{-2(s_2-s_1)}\cdot 1\cdot e^{-3(s_3-s_2)}\cdot 2\cdot e^{-2(t-s_3)}$, which is strictly positive; integrating this positive density over the nonempty ordered-time simplex gives positive probability. Such paths contribute to the full marginal at state $1$ but have already been killed at their first exit from $F$.

$Q_A=0$ is a finite column generator, and the constant vector $p_A(t)=e_1$ solves $p_A^{\mathrm{\prime }}=Q_A{p}_A$ with the stated initial value. By [F2] it is the unique law, and its mean is $0\cdot 0+1\cdot 1+2\cdot 0=1$.

The off-diagonal entries of $Q_B$ are nonnegative and its columns sum to zero, so [F1] applies. Writing $p_B=(p_0,p_1,p_2{)}^T$, its master equation is $p_0^{\mathrm{\prime }}=p_1$, $p_1^{\mathrm{\prime }}=-2p_1$, $p_2^{\mathrm{\prime }}=p_1$ with initial values $(0,1,0)$. Solving gives $p_1=e^{-2t}$ and $p_0=p_2=(1-e^{-2t})\mathrm{/}2$; direct differentiation verifies these formulas and their initial data, and [F2] makes this the unique law.

The mean under $p_B(t)$ is $0\cdot p_0+1\cdot p_1+2\cdot p_2=e^{-2t}+1-e^{-2t}=1$, the same as the mean under $p_A(t)$, while $p_B$ has positive mass at $0$ for $t>0$ and $p_A$ has none. Thus the same mean trajectory corresponds to two different full distributions.