Last week, George Lowther provided a rather sophisticated counter-example of a continuous process $\{W(t):t \geq 0\}$ with $W(0)=0$ and $W(t)-W(s) \sim {\rm N}(0,t-s)$ for all $0 \leq s < t$, yet not a Brownian motion; see link text. Apparently, his approach relied heavily on special properties of BM.

Now, what about the analogue for a Poisson process: Can you find an example of a càdlàg (right-continuous with left limits) process $\{X(t):t \geq 0\}$ with $X(0)=0$ and $X(t)-X(s) \sim {\rm Poi}(t-s)$ for all $0 \leq s < t$, yet not being a Poisson process?

Bonus question: If the last question turns out to be too easy to answer (etc.), then what about the general Lévy process case? That is, given a Lévy process $X$ (defined below) with law $\mu_t$ at time $t>0$, does there exist a càdlàg process $\tilde X$ with $\tilde X(0) = 0$ and $\tilde X(t)-\tilde X(s) \sim \mu_{t-s}$ for all $0 \leq s < t$, which is yet not identical in law to $X$ (hence not a Lévy process)? [Here, assume that $X$ is non-deterministic, equivalently, $\mu_t$ is not a $\delta$-distribution.]

Definition: A stochastic process $X=\{X(t):t \geq 0\}$ is a Lévy process (say, real-valued) if the following conditions are satisfied: (1) $X(0)=0$ a.s.; (2) $X$ has independent increments; (3) $X$ has stationary increments; (4) $X$ is stochastically continuous; (5) Almost surely, the function $t \mapsto X(t)$ is right-continuous (for $t \geq 0$) and has left limits (for $t>0$). [In fact, condition (4) is implied by (1), (3), and (5).]

PS: you are still welcome to try and find a simpler counter-example for the Brownian motion case.

Last week, George Lowther provided a rather sophisticated counter-example of a continuous process $\{W(t):t \geq 0\}$ with $W(0)=0$ and $W(t)-W(s) \sim {\rm N}(0,t-s)$ for all $0 \leq s < t$, yet not a Brownian motion; see link text. Apparently, his approach relied heavily on special properties of BM.

Now, what about the analogue for a Poisson process: Can you find an example of a càdlàg (right-continuous with left limits) process $\{X(t):t \geq 0\}$ with $X(0)=0$ and $X(t)-X(s) \sim {\rm Poi}(t-s)$ for all $0 \leq s < t$, yet not being a Poisson process?

Bonus question: If the last question turns out to be too easy to answer (etc.), then what about the general Lévy process case? That is, given a Lévy process $X$ (defined below) with law $\mu_t$ at time $t>0$, does there exist a càdlàg process $\tilde X$ with $\tilde X(0) = 0$ and $\tilde X(t)-\tilde X(s) \sim \mu_{t-s}$ for all $0 \leq s < t$, which is yet not identical in law to $X$ (hence not a Lévy process)? [Here, assume that $X$ is non-deterministic, equivalently, $\mu_t$ is not a $\delta$-distribution.]

Definition: A stochastic process $X=\{X(t):t \geq 0\}$ is a Lévy process (say, real-valued) if the following conditions are satisfied: (1) $X(0)=0$ a.s.; (2) $X$ has independent increments; (3) $X$ has stationary increments; (4) $X$ is stochastically continuous; (5) Almost surely, the function $t \mapsto X(t)$ is right-continuous (for $t \geq 0$) and has left limits (for $t>0$). [In fact, condition (4) is implied by (1), (3), and (5).]

PS: you are still welcome to try and find a simpler counter-example for the Brownian motion case.

Last week, George Lowther provided a rather sophisticated counter-example of a continuous process $\{W(t):t \geq 0\}$ with $W(0)=0$ and $W(t)-W(s) \sim {\rm N}(0,t-s)$ for all $0 \leq s < t$, yet not a Brownian motion; see link text. Apparently, his approach relied heavily on special properties of BM.

Now, what about the analogue for a Poisson process: Can you find an example of a c`adl`ag (right-continuous with left limits) process $\{X(t):t \geq 0\}$ with $X(0)=0$ and $X(t)-X(s) \sim {\rm Poi}(t-s)$ for all $0 \leq s < t$, yet not being a Poisson process?

``Bonus question'': If the last question turns out to be too easy to answer (etc.), then what about the general L'evy process case? That is, given a L'evy process $X$ (defined below) with law $\mu_t$ at time $t>0$, does there exist a c`adl`ag process $\tilde X$ with $\tilde X(0) = 0$ and $\tilde X(t)-\tilde X(s) \sim \mu_{t-s}$ for all $0 \leq s < t$, which is yet not identical in law to $X$ (hence not a L'evy process)? [Here, assume that $X$ is non-deterministic, equivalently, $\mu_t$ is not a $\delta$-distribution.]

Definition: A stochastic process $X=\{X(t):t \geq 0\}$ is a {\it L'evy process} (say, real-valued) if the following conditions are satisfied: (1) $X(0)=0$ a.s.; (2) $X$ has independent increments; (3) $X$ has stationary increments; (4) $X$ is stochastically continuous; (5) Almost surely, the function $t \mapsto X(t)$ is right-continuous (for $t \geq 0$) and has left limits (for $t>0$). [In fact, condition (4) is implied by (1), (3), and (5).]

PS: you are still welcome to try and find a simpler counter-example for the Brownian motion case.

Last week, George Lowther provided a rather sophisticated counter-example of a continuous process $\{W(t):t \geq 0\}$ with $W(0)=0$ and $W(t)-W(s) \sim {\rm N}(0,t-s)$ for all $0 \leq s < t$, yet not a Brownian motion; see link text. Apparently, his approach relied heavily on special properties of BM.

Now, what about the analogue for a Poisson process: Can you find an example of a càdlàg (right-continuous with left limits) process $\{X(t):t \geq 0\}$ with $X(0)=0$ and $X(t)-X(s) \sim {\rm Poi}(t-s)$ for all $0 \leq s < t$, yet not being a Poisson process?

Bonus question: If the last question turns out to be too easy to answer (etc.), then what about the general Lévy process case? That is, given a Lévy process $X$ (defined below) with law $\mu_t$ at time $t>0$, does there exist a càdlàg process $\tilde X$ with $\tilde X(0) = 0$ and $\tilde X(t)-\tilde X(s) \sim \mu_{t-s}$ for all $0 \leq s < t$, which is yet not identical in law to $X$ (hence not a Lévy process)? [Here, assume that $X$ is non-deterministic, equivalently, $\mu_t$ is not a $\delta$-distribution.]

Definition: A stochastic process $X=\{X(t):t \geq 0\}$ is a Lévy process (say, real-valued) if the following conditions are satisfied: (1) $X(0)=0$ a.s.; (2) $X$ has independent increments; (3) $X$ has stationary increments; (4) $X$ is stochastically continuous; (5) Almost surely, the function $t \mapsto X(t)$ is right-continuous (for $t \geq 0$) and has left limits (for $t>0$). [In fact, condition (4) is implied by (1), (3), and (5).]

PS: you are still welcome to try and find a simpler counter-example for the Brownian motion case.