A stiff reversible system exposes explicit tau-leap mean instability

Apply [F2] to $v(t,x)=x$ and use [F1]. The exact mean $m(t)=\mathbb{E}{X}_t$ satisfies $m(t)=x_0+{\int }_0^t(c_{2}M-\lambda m(s))ds\mathrm{.}$ This makes $m$ absolutely continuous; differentiating almost everywhere shows that the derivative of $e^{\lambda t}(m(t)-c_{2}M\mathrm{/}\lambda )$ is zero almost everywhere. This product is absolutely continuous, so it is constant; its value at $t=0$ is $x_0-c_{2}M\mathrm{/}\lambda$. Rearranging gives $m(t)=\frac{c_{2}M}{\lambda }+(x_0-\frac{c_{2}M}{\lambda })e^{-\lambda t}\mathrm{.}$

By [F3], one Poisson leap has $\mathbb{E}{Y}_1=(1-\lambda h)x_0+c_{2}Mh$. Its linear mean amplification factor is $1-\lambda h$, whose absolute value exceeds $1$ when $h>2\mathrm{/}\lambda$. For the symmetric case $c_1=c_2=r$ and initial state $x_0=0$, the choice $h=3\mathrm{/}(2r)$ gives $\mathbb{E}{Y}_1=3M\mathrm{/}2$, outside the physical upper bound $M$, whereas the exact expectation from step 1.1 is $\frac{M}{2}(1-e^{-3})<M\mathrm{/}2$. This is a one-step comparison; it does not assume that invalid Poisson leaps remain in the physical state set.