I have a process $S_{t}=S_{0}e^{\left(r-q\right)t+mt+X_{t}}$, where $X_t$ is a Levy process and I want to check for which $m$ the process $e^{-(r-q)t}S_t$ is a martingale. The third condition of a martingale states that for $s\leq t$ $$E(e^{-(r-q)t}S_t|F_s)=e^{-(r-q)s}S_s,$$ where $F_s$ is the filtration generated by the process $S_t$.

Many authors write that this process is a martingale when $E(e^{-(r-q)t}S_t)=S_0$ i.e. when $m=-\frac{1}{t}\ln\left(\phi_{X_{t}}\left(-i\right)\right)$, where $\phi_{X_t}$ is the characteristic function of $X_t$.

Why don't they condition on $F_s$ when they verify that the process is a martingale?

  • $\begingroup$ Yes $X_t$ is a levy process. Why they don't check if $E(e^{-(r-q)t}S_t|F_s)=e^{-(r-q)s}S_s$ but only write that if $m=-\frac{1}{t}\ln\left(\phi_{X_{t}}\left(-i\right)\right)$ then discounted process is a martingale $\endgroup$
    – Math122
    Commented Nov 10, 2020 at 18:57
  • $\begingroup$ Ok but technically they dont show that this is a martingale when $m=-\frac{1}{t}\ln\left(\phi_{X_{t}}\left(-i\right)\right)$, they only check special case when $s=0$. Is it possible to show that this process is a martingale for every $s$ with this $m$? Or for every $s$ we have to choose another $m$? $\endgroup$
    – Math122
    Commented Nov 11, 2020 at 11:05
  • $\begingroup$ @Kevin So in conclusion, there is no such $m\in\mathbb{R}$ that the discounted process is a martingale, but for any fixed moment in time $s$ in which we want to price options, we can find such $m$ and can we value option only for that one time instant? Why it works? How to find general form of $m$ for which this process is a martingale? $\endgroup$
    – Math122
    Commented Nov 12, 2020 at 12:11
  • $\begingroup$ @Kevin So if we price Euopean options we can check only condition $\mathbb{E}^\mathbb{Q}[S_t]=S_0e^{(r-q)t}$ i.e with $s=0$, but for American options we need to have a martingale. Am I right? $\endgroup$
    – Math122
    Commented Nov 12, 2020 at 18:45

1 Answer 1


An exponential Lévy process is typically modelled via $$ S_t = S_0\exp\left(\left(r-q+\omega\right)t+X_t\right),$$ where $X_t$ is a Lévy process with $X_0=0$. A Lévy process includes three model features: a linear drift, diffusive shocks and jumps (which may be large and rare or small and frequent). The number $\omega$ is called martingale correction or Jensen's correction and ensures the martingale property.

For our standard finance theory to work, the reinvested and discounted stock price, $S_te^{-(r-q)t}$, needs to be a martingale under $\mathbb{Q}$ (assuming constant interest rates and dividend yields). Let $(\mathcal{F}_t)$ denote the natural filtration of $X_t$. Then, for any $s\leq t$, \begin{align*} \mathbb{E}^\mathbb{Q}[S_t|\mathcal{F}_s] &= \mathbb{E}^\mathbb{Q}[S_0e^{(r-q+\omega)t+X_s+(X_t-X_s)}|\mathcal{F}_s] \\ &= S_0e^{(r-q+\omega)t} e^{X_s} \mathbb{E}^\mathbb{Q}[e^{X_t-X_s}] \\ &= S_s e^{(r-q+\omega)(t-s)} \mathbb{E}^\mathbb{Q}[e^{X_{t-s}}], \end{align*} where we used that $X_s$ is $\mathcal{F}_s$-measurable, and $X_t-X_s\overset{d}{=} X_{t-s}$ is independent of $\mathcal{F}_s$, see here.

Let $\varphi_{X_t}(u)=\mathbb{E}[e^{iuX_t}]$ be the characteristic function of the Lévy process $X_t$. The Lévy-Khintchine formula states that $\varphi_{X_t}(u)=e^{t\Psi(u)}$ which follows from the infinite divisibility of a Lévy process. The function $\Psi$ is called the characteristic exponent and captures the drift, diffusion and jump components of $X_t$.

Then, \begin{align*} \mathbb{E}^\mathbb{Q}[S_t|\mathcal{F}_s] &= S_s e^{(r-q+\omega)(t-s)} \varphi_{X_{t-s}}(-i) \\ &= S_s e^{(r-q+\omega)(t-s)} e^{(t-s)\Psi(-i)}. \end{align*} Hence, setting $\omega=-\Psi(-i)$ yields \begin{align*} \mathbb{E}^\mathbb{Q}[S_t|\mathcal{F}_s] &= S_s e^{(r-q)(t-s)}, \end{align*} which in turn implies that the discounted reinvested stock price is indeed a $\mathbb{Q}$-martingale.

Note that $$\omega=-\Psi(-i)=-\frac{1}{t}\ln\left(\varphi_{X_t}(-i)\right)$$ is independent of time. Thus, for an exponential Lévy process, the martingale property is ensured to hold if you verify that $\mathbb{E}^\mathbb{Q}[S_t]=S_0e^{(r-q)t}$.

  • $\begingroup$ Thanks! You lost $\ln$ in $\omega$ :) $\endgroup$
    – Math122
    Commented Nov 15, 2020 at 15:04
  • $\begingroup$ I have one question: why can we write that $\mathbb{E}^\mathbb{Q}[e^{X_{t-s}}]= \varphi_{X_{t-s}}(-i)$? The domain of characteristic function is real, so why we can take value at point $-i$? $\endgroup$
    – Math122
    Commented Mar 20, 2021 at 20:57
  • 1
    $\begingroup$ @Math122 Characteristic functions can actually be evaluated in an open strip in the complex plane. So there exists a set $\mathcal{S}=\{z\in\mathbb{C}:a<\text{Im}(z)<b\}$ in which we can evaluate $\varphi$. For a ``sensible stock price model'', we have $a\leq-1$ and $b\geq0$. In these cases, $\varphi_{\ln(S_T)}(-i)=\mathbb{E}[S_T]$ is fine. More details are in Lewis' (2001) seminal paper and Schmelzle's (2010) survey. $\endgroup$
    – Kevin
    Commented Mar 21, 2021 at 10:24

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