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Let $X(t)$ be a Lévy process defined on $t\geq 0$. Suppose that, with probability 1, $X(t)\geq 0$ for all $t \geq 0$. That is: $$ P\left[\left(\inf_{t\geq 0} X(t)\right) \geq 0\right] =1$$ Does that imply that $X(t)$ is nondecreasing with probability 1?

The definition of "Lévy process" I am using is given in the 4 properties from the Wikipedia link below ($X(0)=0$ almost surely; $X(t)$ has stationary and independent increments; $X(t)$ has continuity in probability): https://en.wikipedia.org/wiki/L%C3%A9vy_process

Context:

1) I thought of this question while working on this problem: Wald Martingale for Levy Processes

2) Given any increasing sequence of nonnegative times $\{t_i\}_{i=1}^{\infty}$ (chosen before looking at the stochastic process realization) I can prove that with prob 1 $X(t)$ is nondecreasing over that countable sequence. Indeed for any $i\in \{1, 2, 3, ...\}$ we get by stationary increments: $$ P[X(t_{i+1})-X(t_i)\geq 0] = P[X(t_{i+1}-t_i)\geq 0]=1$$ Thus $$ P\left[\cap_{i=1}^{\infty} \{X(t_{i+1})\geq X(t_i)\}\right]=1$$ The difficulty of proving that $X(t)$ is nondecreasing in general (via the stationary increments property) is that we need to look at an uncountably infinite number of times.

TheBridge
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Michael
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  • I'm not sure I see this even in the countable case. I see that $P[X(t_{i+1}) - X(t_i) \geq 0] = P[X(t_{i+1}-t_i) \geq 0]$, I just don't see that that probability is $1$ just because it happened to be true for this instance of $X(t)$. I see this as analogous to a good bet: If I flip a coin and either lose a dime or win a dollar on the flip, that doesn't mean that the process is necessarily non-decreasing, just because a particular process with those dynamics never dropped below $0$, does it? Maybe I'm misunderstanding the question. – Brian Tung Jun 21 '19 at 18:01
  • @BrianTung : I think you have a misunderstanding: We are told that with probability 1, $X(t)\geq 0$ for all $t \geq 0$. In other words $P[\left(\inf_{t\geq 0} X(t)\right) \geq 0]=1$. Thus $P[X(t_{i+1}-t_i)\geq 0]=1$. – Michael Jun 21 '19 at 18:04
  • Ahh, I did miss that (the "with probability $1$"). The perils of reading too fast. Thanks. – Brian Tung Jun 21 '19 at 18:05
  • I will edit teh question to clarify . – Michael Jun 21 '19 at 18:05
  • @BrianTung : Today I resolved the question by realizing (i) The proof I give in the question can be repeated on the set of all pairs of rationals; (ii) The answer to my question is either "yes" or "no" depending on if the definition of "Levy process" also requires right-continuity. (I found a counter-example where, with positive probability, the sample path of the nonnegative process $X(t)$ is not nondecreasing.) – Michael Jun 22 '19 at 17:00

1 Answers1

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I just realized that I can do a similar proof with the countably infinite set of pairs of rationals!

Let $C$ be the set of all pairs of nonnegative rationals $(x,y)$ such that $x<y$. For each particular $(x,y) \in C$ we have (by the same argument I give in the question itself): \begin{align} P[X(x) \leq X(y)] &= P[X(y)-X(x)\geq 0] \\ &\overset{(a)}{=} P[X(y-x)-X(0)\geq 0] \\ &\overset{(b)}{=} P[X(y-x)\geq 0] \\ &= 1 \end{align} where (a) holds by the stationary increment property; (b) holds because $X(0)=0$ with probability 1.

Thus, because $C$ is a countably infinite set, we have: $$ P[\cap_{(x,y)\in C} \{X(x)\leq X(y)\}] = 1$$ That is, with probability 1, the sample path $X(t)$ has no instance of rationals $x,y$ such that $0\leq x<y$ and $X(x)>X(y)$.

Edit 1 : Additionally assuming right-continuity

Now suppose that The Levy process is also required to be right-continuous everywhere (definitions of Levy processes vary, this is not on the 4 properties of the wikipedia link but it is in other definitions):

Assuming right-continuity, if there are times $u,v$ such that $0\leq u<v<\infty$ and $X(u)>X(v)$, it would mean that there are rational numbers $u+\epsilon$ and $v+\delta$ (with $\epsilon\geq 0, \delta\geq 0$) such that $u+\epsilon < v+\delta$ and $X(u+\epsilon)>X(v+\delta)$. But we already know that existence of any two rational numbers $x<y$ for which $X(x)>X(y)$ is an event that has probability 0.

So, indeed, if the Levy process is nonnegative with probability 1, and if it is also required to be right-continuous everywhere, then it is also nondecreasing with probability 1.

Edit 2: Counter-example

Here is an example process $X(t)$ that satisfies the 4 properties on wikipedia, is surely nonnegative, but is surely not right continuous, and with positive probability it is not nondecreasing:

For $t\geq 0$, let $N(t)$ count the number of arrivals in a Poisson process with rate $\lambda=1$. Notice that $N(t)$ surely takes integer values. (Perhaps it is better to say that $N(t)$ "almost surely" takes integer values if we are worried about the possibility of an infinite number of arrivals in finite time. Of course, we can remove all such instances from the sample space without affecting distributions, and if we view $N(t)$ as the result of such a "trimmed" sample space then we get back the ability to say "surely").

Let $U$ be an independent random variable that is uniform over $(0,1)$. For $t\geq 0$ define $$Z(t) = \left\{ \begin{array}{ll} 0 &\mbox{ if $t\neq U$} \\ 77.5 & \mbox{ if $t=U$} \end{array} \right.$$ Define $X(t) = N(t)+Z(t)$. It is easy to see that $X(t)$ satisfies the 4 properties on wikipedia because: $X(0)=0$ almost surely; $X(t)$ has stationary and independent increments (indeed, given any finite or countably infinite list of (deterministic) times $\{t_1, t_2, t_3, ...\}$, with probability 1 $U$ does not lie on those times hence the $Z(t)$ process does not affect increments during those times); it has the "continuity in probability" property.

But $X(t)$ is not right-continuous: Surely, $X(t)$ is integer-valued for all $t \neq U$, while $X(U)=N(U)+77.5$, which is not an integer. So, surely, $X(t)$ is not right-continuous at $t=U$ (a sequence of integers cannot converge to the noninteger value $N(U)+77.5$).

And $X(t)$ is not nondecreasing: We show that $P[X(U)>X(U+1)]>0$ and so with positive probability the sample path $X(t)$ is not nondecreasing: \begin{align} P[X(U)> X(U+1)] &=P[77.5+N(U)>0+N(U+1)] \\ &\overset{(a)}{\geq} P[77.5 > N(U+1)]\\ &\overset{(b)}{\geq} P[77.5 > N(2)]\\ &>0 \end{align} where (a) holds because $N(U)\geq 0$ surely; (b) holds because $N(t)$ is surely nondecreasing and $U+1\leq 2$ so $N(U+1)\leq N(2)$.

Michael
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  • Notice that the wikipedia link (mysteriously) gives a side comment that says the 4 properties imply that a "version" of $X(t)$ is right-continuous. This is not explained, but after a web search I find that $Y(t)$ is considered to be a "version" of $X(t)$ if for all $t\geq 0$ we have $P[X(t)=Y(t)]=1$. [This is not the same as saying that $P[X(t)=Y(t) \quad \forall t \geq 0] = 1$.] So in the counter-example above it is indeed true that $N(t)$ is a right-continuous "version" of $N(t)+Z(t)$. – Michael Jun 22 '19 at 16:56
  • I have asked a question about clarifying the definition of Levy process here: https://math.stackexchange.com/questions/3270884/what-is-the-true-definition-of-a-levy-process – Michael Jun 22 '19 at 17:37