Let $X = \{ X_t \in {\bf R}, t \geq 0 \}$ be a 1-dimensional (real) Levy process. Suppose further that the distribution of $X_t$ is not concentrated on a grid. (This forces the distribution of $X_t$ to have a Lebesgue density).

For a fixed $p \in (0,1)$, let $Q_t(p)$ be the quantile function of $X_t$, i.e $$ {\bf P}(X_t \leq Q_t(p)) = p. $$

If $X$ is a symmetric process and $Y_t = X_t + \alpha t$, then $Q_t(1/2)$ is a linear function of $t$. Indeed, $Q_t(1/2) = \alpha t$.

Do Levy process have this property in general? In other words, for a general Levy process $X_t$, can one find a $p$ that makes $Q_t(p)$ a linear function of time?


1 Answer 1


First of all, $X$ being non-lattice is not enough for $X_t$ being absolutely continuous. A simple counter-example is $$X_t = \sum_{n=1}^\infty \frac{N_t^{(n)}}{n!} \, ,$$ where $N_t^{(n)}$ are independent Poisson processes. If $X$ is non-lattice and not a compound Poisson process, then the distribution of $X_t$ is continuous. However, the definition of the quantile $Q_t(p)$ does not depend on continuity $X_t$; what is needed here is that the support of $Q_t(p)$ is an interval. This is the case when $X$ has a Gaussian component or when the closed semigroup generated by the support of Lévy measure of $X$ is an interval. (A detailed discussion of distributional properties of a Lévy process can be found, for example, in Sato's book Lévy processes and infinitely divisible distributions.)

It is quite simple to see that $Q_t(\tfrac12)$ need not be linear. Consider first the usual Poisson process $X_t$. Then $Q_t(\tfrac12) = 0$ for $t < \log 2$, and $Q_t(\tfrac12) = 1$ when $\log 2 < t < T$, where $T$ satisfies $e^{-T} (1 + T) = \tfrac12$. In particular, $Q_t$ is not linear in this case.

If one insists on a Lévy process $X_t$ with a continuous distribution, then it is enough to add an "epsilon" of Brownian motion to the above example. This is intuitively clear, I hope, but a detailed argument would require a somewhat technical estimate.


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