# Discounted price process - martingale

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?

• 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 Nov 10 '20 at 18:57
• 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 $m$? Nov 11 '20 at 11:05
• @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? Nov 12 '20 at 12:11
• @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? Nov 12 '20 at 18:45

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 $$\mathbb{E}^\mathbb{Q}[S_t]=S_0e^{(r-q)t}$$.

• Thanks! You lost $\ln$ in $\omega$ :) Nov 15 '20 at 15:04
• 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 $-i$? Mar 20 at 20:57
• @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. Mar 21 at 10:24
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