# Quantiles of a Levy process

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?

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.