Why is Geometric Brownian motion not a Lévy process?

I'm trying to price an option. The old-style (e.g. Black Scholes 1973) pricing models use the GBM to model the underlying asset, which suffers of some deficiencies wrt volatility smiles and term structures. This is why Heston came up with his stochastic volatility model in 1993, and around the same time jumps were introduced. The modern approach (Cont Tankov 2004, Jackson 2008) proposes to model the underlying asset price as a Lévy process, which allows for more general realistic structure of asset returns. So I'm willing to use this approach.

I'm here having some troubles. I can't exactly get how GBM and Lévy processes are linked. Wikipedia says:

In probability theory, a Lévy process, named after the French mathematician Paul Lévy, is a stochastic process with independent, stationary increments

Now, GBM increments are independent by definition. Does it have the same distribution $\forall t$? I tried to get the answer this way:

Geometric brownian motion: $$X(t) = X(0)\exp{\left( \left(\mu - \frac{\sigma^2}{2}\right)t + \sigma W(t) \right)}$$

First moment of GBM: $$E(X(t)) = X(0)e^{\mu t}$$

Second moment of GBM: $$Var(X(t)) = X(0)^2 e^{2\mu t}(e^{\sigma^2 t} - 1)$$

Can we say that GBM is not stationary because its first and second analytical moments show $t$, and consequently they change in time? Is the non-stationarity characteristic of the GBM the reason why it is not a Lévy process?

Yes, it is not Levy because it doesn't have stationary increments, which you have shown from the moment calculation.

• But is it the same to show the non-stationarity of the stochastic process and the non-stationarity of the increments of the stochastic process? – Puzzle Dec 6 '15 at 18:32
• They are not the same. For example, Brownian motion is non-stationary but has stationary increments. On the other hand, the increments of a GBM are neither stationary nor independent. I haven't looked into the literature that you're citing, but is it possible that the Levy processes there are intended to serve as a replacement for W(t) in the formula for GBM, i.e. that the Levy process models log returns rather than returns? – Brent Kerby Dec 7 '15 at 15:01

As @Brent Kerby said

[..] the increments of a GBM are neither stationary nor independent.

This is the reason why GBM is not a Lévy process. What I instead proved is the non-stationarity of the process itself, which is not taken into account by the definition of Lévy process.

Note

The use made of Lévy processes in modern quantitative finance is the following: instead of using the GBM as the stochastic process followed by the stock prices (as in the Black-Scholes model), different stochastic processes are taken into account to describe the dynamics of the stock prices. Many of this others stochastic processes are Lévy processes.

The main example is the Variance Gamma process, which can be written as a time-changed Brownian Motion ${W_T}_s$ subjected to an independent increasing jump process, a so-called Gamma Lévy process with $T_s \sim Gamma(\alpha s, \beta)$. The process ${W_T}_s$ is then also a Lévy process itself.

The advantage of using Variance Gamma process instead of GBM to model stock prices is that the former takes into account GBM problems such as the Gaussian density decreasing too quickly, absence of variation of the volatility $\sigma$ over time, absence of jumps.