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For a piecewise-deterministic Markov process, conditional on no jump before $u$, the state is $z(u)=\phi (s,x,u)$ and the survival probability is $S(u)=\exp [-{\int }_s^u\alpha (z(r))dr]$. The first jump has probability density function $\alpha (z(u))S(u)$; conditioning on its time and its destination gives $P_{s,t}f(x)=S(t)f(z(t))+{\int }_s^t\alpha (z(u))S(u)\int P_{u,t}f(y)K_u(z(u),dy)du$. This proves the identity for bounded Borel measurable functions under nonexplosion.

The no-jump part of a piecewise-deterministic Markov process. The chain rule gives ${\partial }_t{P}_{s,t}^{\alpha }f=P_{s,t}^{\alpha }{L}_t^{\alpha }f$, where $L_t^{\alpha }f=a(t,\cdot )f^{\prime }-\alpha f$. The time subscript is essential for a time-dependent vector field.

The displayed identity requires a correction to the process construction in the source. In a piecewise-deterministic Markov process, the instantaneous hazard function is evaluated at the current flowing state $X_{u-}$, and the destination is sampled from $K_u(X_{u-},dy)$. Keeping the jump times and destinations of the original Markov jump process $Y$ instead evaluates both quantities at its last reset state $Y_{u-}$, which generally differs from $X_{u-}$. The source PDF confirms that these original jump times and destinations were retained. Consequently its definition does not in general produce the asserted transition kernel, or even a Markov process in the state variable $X$ alone.

For an explicit counterexample, take $a(t,x)=1$, $\alpha (x)=2+\tanh x$, $y_0=0$, and let every $K_t(x,dy)$ be the standard normal distribution, independently of $t,x$. These are smooth coefficients with bounded positive jump rate, so nonexplosion holds. In the literal construction, the first holding time has the exponential distribution of rate $2$. Thus the probability distribution of $X_t$ has an atom of mass $e^{-2t}$ at $t$. Paths with at least one jump contribute no atom there, because the reset has a normal density. The asserted identity instead assigns to this no-jump atom the masswhich differs for every $t>0$.

Here is the full derivation for the intended piecewise-deterministic Markov process. Assume nonexplosion; smoothness alone does not guarantee it. Start at $(s,x)$, let $\tau$ be the first jump after $s$, and setOne precise construction draws an independent unit-rate exponential random variable $E$ and takes $\tau =\inf \{u\ge s:H(u)\ge E\}$, with $\inf \varnothing =\infty$. This implements a hazard function along the flow of a vector field, rather than a fixed exponential holding time at the last reset. Its survival probability and jump-time probability density function areEquivalently, $S^{\prime }=-\alpha (z)S$ with $S(s)=1$. On $\{\tau >t\}$, the terminal state is $z(t)$. Conditional on $\tau =u$, draw the new state $y$ from $K_u(z(u),dy)$ and restart the same construction from $(u,y)$. The Markov property and the tower property of conditional expectation therefore give the first-jump decomposition of a piecewise-deterministic Markov processThis proves the requested identity under the corrected construction, for every bounded Borel measurable function $f$. The two contributions partition according to whether the first jump occurs before $t$; later jumps are already contained in $P_{u,t}$.

For comparison, the literal construction does have a different renewal identity when started at a reset with $X_s=Y_s=x$. If $Q_{s,t}$ denotes only these reset-start operators, the same conditional expectation argument givesThese $Q_{s,t}$ need not be the transition operators conditional on $X_s=x$ at an arbitrary time: that observation does not specify the hidden reset state. This distinction explains exactly why replacing only the between-jump paths is insufficient.

Use the corrected piecewise-deterministic Markov process from part (b). For a smooth test function $f$, its time-dependent infinitesimal generator isIndeed, over an interval of length $h$, deterministic motion contributes $ha(t,x)f^{\prime }(x)$, while a jump has probability $h\alpha (x)+o(h)$ and changes the conditional mean of $f$ by $K_{t}f(x)-f(x)$. Terms involving both drift and a jump contribute only to the remainder. Thus $P_{t,t+h}f=f+h{L}_{t}f+o(h)$.

The Dynkin formula, or this infinitesimal expansion integrated against the current probability distribution, gives the weak evolution identityTake $f\in C_c^{\infty }(\mathbb{R})$, so integration by parts has no boundary contribution. Assuming the displayed jump integrals are finite, Fubini's theorem givesConsequently the forward equation of a piecewise-deterministic Markov process isHere $A_t=L_t^{\ast }$ is the formal adjoint operator acting on probability density functions. Its terms are transport, loss from $x$, and incoming mass from every $y$. In particular, the incoming rate is $\alpha (y)$, evaluated at the departure state. The operator denoted $A$ in the source is generally time-dependent. Because $\int m_t(y,x)dx=1$, the gain and loss integrals cancel; with vanishing transport flux at infinity, $\frac{d}{dt}\int p_t=0$.

The deterministic initial state in part (b) does not actually yield an everywhere smooth full density. For finite times the no-jump contribution is the positive measurewhere ${\delta }_z$ is the Dirac measure at $z$. Smooth jump destination probability density functions cannot remove this positive atom. The density equation above is therefore conditional on the extra density hypothesis, for example with a suitable random initial probability density function. For the specified fixed initial state, write ${\nu }_t$ for the whole probability measure and use insteadThis weak measure evolution includes both the atom and any continuous component. It becomes the displayed partial differential equation whenever ${\nu }_t(dx)=p_t(x)dx$ with the assumed regularity. It does not describe the literal hidden-reset construction without the correction in part (b).