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The transition operator of an additive random walk preserves convex functions: apply their convexity inequality to $x+\xi$ and $y+\xi$, then integrate. The pointwise maximum of convex functions preserves convexity too, so induction proves convexity of the Snell envelope Bellman functions. Expectations are allowed to be extended-valued until finiteness is justified. Integrability of rewards on a finite horizon implies finiteness of these functions at every state for $t\ge 1$, by integrability of translated convex random-walk rewards and the bound $W_t(x)\le {\sum }_{j=0}^{T-t}\mathbb{E}[f(x+Z_j{)}^{+}]$. At time zero only $X_0=0$ is observed; if $W_0$ is infinite elsewhere, its finite value at zero has a constant convex extension representing the same Snell envelope. Full statewise Bellman recursion at time zero requires an additional translated-integrability hypothesis.

For a Markov process with reward $f$, the value starting at time $t$ in state $s$ is the supremum of expected rewards over admissible stopping times. In finite discrete time it obeys $V(T,s)=f(s)$ and $V(t,s)=\max \{f(s),PV(t+1,\cdot )(s)\}$ for the transition operator $P$, wherever expectations are well-defined. Statewise absolute integrability of every remaining-horizon reward makes these values finite.

Let $\nu$ be the common increment law. The random walk has the Markov property, since ${\xi }_{t+1}$ is independent of ${\mathcal{F}}_t$ and $X_{t+1}=X_t+{\xi }_{t+1}$. Its transition operator is $Ph(x)=\int h(x+z)\nu (dz)$ wherever that integral exists.

Set $V(T,x)=f(x)$. For $1\le t<T$, suppose the Borel function $V(t+1,\cdot )$ has been constructed and represents $U_{t+1}$. On the Borel setdefine the optimal stopping value function recursively byOutside $A_t$, assign the finite Borel value $f(x)$. Measurability of integrals against a fixed probability law proves measurability of both the set and the function. Integrability of $U_{t+1}=V(t+1,X_t+{\xi }_{t+1})$ and independence imply $X_t\in A_t$ almost surely andThe Snell recursion thus proves $U_t=V(t,X_t)$ by induction.

At time zero the generated sigma-algebra is trivial, so $U_0$ is deterministic. Since $X_0=0$, define $V(0,x)=U_0$ for all $x$, withwhen $T\ge 1$. This supplies a globally finite deterministic representation without assuming integrability of rewards from every unvisited starting state. HenceWith the additional statewise integrability normally used to define a value function for arbitrary starting states, one can instead use the Bellman maximum at time zero too. For $T=0$, the representation is simply $U_0=f(0)$.

The key convexity calculation is that the random-walk transition operator preserves convex functions. If $h$ is convex, then for $0<\theta <1$,The pointwise maximum of convex functions is convex as well: the convexity inequality holds for each candidate and is bounded above by the convex combination of the two endpoint maxima. Thus the Bellman recursion proves convexity of a random-walk Snell value function by backward induction from $V(T,\cdot )=f$.

To ensure that the functions in this argument are genuinely finite, rather than silently imposing a growth bound on $f$, use integrability of translated convex random-walk rewards. Write $Z_j$ for a sum of $j$ independent increments with the same law. For $j\le T-1$, the assumptions give integrability of $f(Z_j)$ and $f(Z_{j+1})$. The negative part of a finite convex function is Lipschitz, so translating $Z_j$ preserves integrability of the negative part. For the positive part and $x>0$, if the increment law is unbounded above, take an independent increment $\eta$ and $p=\mathbb{P}(\eta \ge x)>0$. On that event $Z_j+x$ lies between $Z_j$ and $Z_j+\eta$, so convexity givesIf increments are bounded above by $M$, then $Z_j\le jM$ and the same convexity argument between $Z_j$ and $jM+x$ bounds the positive part by $f(Z_j{)}^{+}+f(jM+x{)}^{+}$. For $x<0$, use the corresponding lower-tail argument. Thus every translated reward $f(x+Z_j)$ is integrable for $j\le T-1$.

For $t\ge 1$, backward induction also gives the boundIt justifies every convolution in the Bellman recursion at these times, so the arbitrary assignments outside $A_t$ in the previous part are never needed when $f$ is convex. The displayed convexity calculation therefore proves that each $V(t,\cdot )$ for $t\ge 1$ is a finite convex function. The chosen $V(0,\cdot )=U_0$ is constant and hence convex. This proves the requested conclusion for the deterministic representation under the finite-horizon assumptions.

There is a distinction between this representation and a value function for every possible starting point at time zero. Without translated integrability at the full horizon, the latter may take $+\infty$ at unused states. For example, with $T=1$, $f(x)=e^{x^2}$ and $\mathbb{P}({\xi }_1=k)=c{e}^{-k^2-k}$ for $k=1,2,\dots$, the observed reward is integrable, but $\mathbb{E}[f(x+{\xi }_1)]=c{e}^{x^2}{\sum }_{k\ge 1}{e}^{(2x-1)k}$ is infinite for $x\ge 1/2$. The constant time-zero extension avoids claiming a globally finite Bellman function under insufficient assumptions. If integrability is assumed for all times of an infinite random walk, the translated-reward argument with one extra time also makes the full time-zero Bellman function finite and convex.

For the intended random walk model, take $S_0$ to be deterministic, or independent of all the independent and identically distributed random variables $Y_j=S_j-S_{j-1}$. If $\nu$ is their common probability distribution, then $Y_{t+1}$ is independent of ${\mathcal{F}}_t=\sigma (S_0,Y_1,\dots ,Y_t)$. This supplies the Markov property that the backward dynamic programming argument needs. Define the transition operatorand the deterministic optimal stopping value function recursively byAt any state where the conditional law is evaluated, independence givesBackward induction, starting with $U_T=f(S_T)$, proves the intended representationIntegrability along the actual random walk makes the recursion finite at the states used by that process, up to null sets. If finite real-valued value functions on all of $\mathbb{R}$ are intended, a convenient sufficient convention is statewise integrability: $\mathbb{E}|f(s+Y_1+\cdots +Y_k)|<\infty$ for every $s$ and $0\le k\le T$. Indeed each stopping value is bounded in absolute value by the expectation of the sum of the absolute rewards. This convention makes the displayed recursion finite and measurable everywhere; the source only assumes integrability from the given starting process.

The printed independence of the increments alone does not suffice if $S_0$ can reveal future increments. Here is a bounded finite-state counterexample, also with a convex reward. Let $T=2$, let ${\epsilon }_1,{\epsilon }_2$ be independent fair signs, and setThe two increments are exactly ${\epsilon }_1,{\epsilon }_2$, hence independent and identically distributed. But ${\mathcal{F}}_1$ already knows both signs, so $U_1=\max \{S_1^{+},S_2^{+}\}$. On the two positive-probability histories with $S_1=0$, its values are respectively $0$ and $1$. Thus no deterministic $V(1,0)$ can work. Adding independence of $S_0$ from the increments repairs this missing Markov property hypothesis.

Use the random walk interpretation and optimal stopping value function recursion specified in part (b). If $h$ is convex, translating and integrating its convexity inequality gives, for $0<\theta <1$,Thus the transition operator preserves convex functions. Also the pointwise maximum of convex functions is convex: each of two convex functions at an intermediate point is bounded by the same convex combination of their pointwise maximum at the endpoints. Starting with $V(T,\cdot )=f$, backward induction intherefore proves the asserted convexity in the intended model:The statewise integrability convention in part (b) gives finite convex functions on all real states. The same inequality holds for the canonical extended value function wherever the expectations are well-defined. Integrability only along one started process need not make that canonical function finite at unused states: for $T=1$, $S_0=0$, $f(s)=e^{s^2}$ and increment density proportional to $e^{-y^2-|y|}$, $\mathbb{E}f(S_1)<\infty$, but $\mathbb{E}f(s+Y_1)=\infty$ for $s>1/2$. This concerns the canonical recursion away from visited states, rather than the almost-sure identity for $U$. Arbitrary off-state versions of $V$ need not be convex; the recursive version is the one meant here. No zero-mean assumption on the increments was used, and convexity alone does not make $f(S_t)$ a submartingale for an arbitrary drift.

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.