# Mean correcting martingale argument

Let $X=(X_t,t\in\mathbb{N})$ be a pure jump Lévy process (i.e $X_t$ does not have a diffusion component). Let further $Y=(Y_t,t\in\mathbb{N})$ be a process that is used to time-change the process $X$ (there are several examples in Schoutens (2003) of these type of processes, such as the integrated CIR model). E.g, the time-changed Lévy process is given by $X_Y=(X_{Y_t},t\in \mathbb{N})$.

In Schoutens (2003), it is argued that for a price process $S=(S_t,t\in\mathbb{N})$ given by

$S_t=S_0 \frac{\exp((r-q)t)}{\mathbb{E}(\exp{X_{Y_t}}\vert y_0)}\exp{X_{Y_t}}$

where $q$ is the dividend yield, $r$ the interest rate, and $y_0$ the initial value of the underlying process for which $Y$ is based on, puts us into the risk-neutral world by mean-correcting arguments (because of the factor $\frac{\exp({r-q}t)}{\mathbb{E}(\exp{X_{Y_t}}\vert y_0)}$).

Later on Schouten suggests to use the process $S$ to price options directly in the risk neutral world. It seems that this is all based on arbitrage-free principles.

So now to my question: In what way does the mean correcting argument in this case yield an Equivalent Martingale measure? They don't even define the Physical measure? Even though the discounted price process is a martingale(?) how can it be sure that it is also equivalent to some physical measure?

Schoutens (2003) Lévy processes in finance, Wiley.

• I did not have a deep look into this but doesn't it simply amount to making sure that the risk neutral expectation of the forward asset price is the forward price i.e. $\Bbb{E} [S_t \vert \mathcal{F}_0] = F(0,T)$, which indeed should be the case under the risk-neutral measure (you can equivalently show that $F(t,T)$ is then a $\Bbb{Q}$ martingale as it should by definition) – Quantuple Jul 3 '17 at 7:23
• As for your side question for a given dynamics under the risk neutral measure there could be infinitely many different associated dynamics under the physical measure depending on your working modelling assumptions (notably model completeness). This transpires through the "parametrisation" of the RN derivative. – Quantuple Jul 3 '17 at 7:28
• @Quantuple I think I partly follow. But what still bothers me is how the discounted process $S_t e^{-(r-q)t}$ still could be a martingale. $\mathbb{E}(S_t e^{-(r-q)t} \vert \mathcal{F}_0)=S_0$ is easily shown, but how to show it with respect to the filtration $\mathcal{F}_s$, $0<s\le t$, that is, how to show that $\mathbb{E}(S_t e^{-(r-q)t} \vert \mathcal{F}_s)=S_s e^{-(r-q)s}$? – noidea Jul 28 '17 at 20:17