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Suppose $X=(X_t)$ is a stochastic process. I have a question about the notation $X_0$. Am I right, that $X_0(\omega)$ is not a constant. It depends on $\omega$ and can therefore have different values for different $\omega$? When do we know that $X_0$ is always a constant? Are there assumptions on the filtration? Suppose I know that $M_t:=X_t-X_0$ is a local martingale. Can I conclude that $X_t$ is a local martingale too? Obviously if $X_0$ is a constant, this is true. This is the motivation for my question.

Thank you for your help


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$X_0$ is a snapshot of the stochastic process at time $t_0$ and hence is random variable – jay-sun Oct 30 '12 at 18:19
Phil Protter says your process is a local martingale, with stopping like $T = |X_0| > n$ or ...., but not everyone agrees. I think the definition in Revuz & Yor rules that sort of stopping time out. – mike Oct 30 '12 at 20:42
up vote 0 down vote accepted

By definition, if $(X_t)_{t\geq 0}$ is a stochastic process, then $X_t$ is a random variable for every $t\geq 0$. So indeed $X_0$ is a random variable (i.e. it is $\mathcal{F}$-$\mathcal{B}(\mathbb{R})$ measurable). If moreover $X_0$ is $\mathcal{F}_0$-measurable and integrable, then the "constant" process $(Y_t)_{t\geq 0}$ is a martingale, where $Y_t=X_0$ for every $t\geq 0$. This is just because $$ E[Y_t\mid \mathcal{F}_s]=E[X_0\mid \mathcal{F}_s]=X_0 \quad \text{a.s.},\quad 0\leq s<t, $$ as $X_0$ is $\mathcal{F}_0$-measurable and $\mathcal{F}_0\subseteq \mathcal{F}_s$. So if $(M_t)_{t\geq 0}$ is a martingale, then $(M_t+Y_t)_{t\geq 0}=(X_t)_{t\geq 0}$ is also a martingale.

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