# Renewal for Levy Processes

Suppose $$X(t)$$ is a Levy process with almost surely positive increments (for all $$t_1 < t_2$$ $$P(X(t_1) < X(t_2)) = 1$$)

Define

$$\nu X(t) := \sup \{\tau \in \mathbb{R_+}| X(\tau) < t\}$$

It is not hard to see, that $$\nu X$$ is also a stochastic process with almost surely positive increments.

My question is:

Is it a Levy process too?

It is not hard to see, that $$\nu X (0) = 0$$ and for all $$t_1 < t_2$$ $$\nu X(t_1) - \nu X(t_2) = \mu\{\tau \in \mathbb{R}| X(\tau) \in (t_1; t_2]\}$$, which means, that the conditions (1), (3) and (4) are satisfied. However, I do not know, whether the increments of $$\nu X(t)$$ are independent or not. They are indeed uncorrelated, but uncorrelatedness $$\neq$$ independence.

One can also notice, that $$\nu \nu X(t) = X(t)$$ almost surely. However, it does not seem to be very helpful.

• To clarify: (i) $X(t)$ is usually only defined on $t\geq 0$, and so the sup over $\tau \in \mathbb{R}$ is weird and I'm not sure how it is defined. (ii) (related) How do you get $vX(0)=0$? I wonder if you may want to use an alternative definition $$Y(t)=\inf\{\tau \geq 0 | X(\tau) \leq t\} \quad \forall t \geq 0$$ This indeed satisfies $Y(0)=0$ (with prob 1). – Michael Jun 21 at 19:03
• An intuitive argument against stationary increments: If we define $X(t) = \sum_{i=1}^{N(t)} V_i$ where $N(t)$ is a Poisson process of rate $\lambda>0$ and $\{V_i\}$ are independent and i.i.d. positive random variables (not exponentially distributed), then $vX(t)$ is constant over intervals that span durations of $V_i$. So intuitively, if $vX(t)$ has been constant for a while, it is likely because the associated $V_i$ variable is large, and the non-memoryless property of the distribution of $V_i$ suggests that the remaining time is not independent of the elapsed time. – Michael Jun 21 at 19:14

No. For a counter-example just take $$X(t)$$ as a Poisson process.
For $$t \geq 0$$, let $$X(t)$$ count the number of arrivals in a Poisson process of rate $$\lambda>0$$. Note that $$X(t)$$ is a non-decreasing staircase function that is right-continuous. It has i.i.d. exponential inter-arrival times $$\{T_i\}_{i=1}^{\infty}$$ and has jumps $$\{J_i\}_{i=1}^{\infty}$$ of height $$1$$ (so $$J_i=1$$ for all $$i$$).
Now the process $$Z(t) = vX(t)$$ simply exchanges interarrival times and jumps: $$Z(t)$$ has inter-arrival times of 1, and jumps of heights $$\{T_i\}_{i=1}^{\infty}$$. Since $$Z(t)$$ has arrivals at periodic times, it does not have stationary increments (it is not a Levy process).
Also, your edit in reaction to my first comment does not fix the problem of $$t=0$$:
$$Z(0)=vX(0)= \sup\{\tau \in [0, \infty): X(\tau)<0\} = \sup \phi \neq 0$$ This is because there is no time $$\tau \geq 0$$ for which $$X(\tau)<0$$. So your definition of $$vX(0)$$ is taking a supremum over the empty set.
• It also seems that with your current definition, the process $v X(t)$ is left-continuous but not right-continuous. – Michael Jun 22 at 19:31