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


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

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.

  • $\begingroup$ 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). $\endgroup$ – Michael Jun 21 at 19:03
  • $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ It also seems that with your current definition, the process $v X(t)$ is left-continuous but not right-continuous. $\endgroup$ – Michael Jun 22 at 19:31

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