Suppose $\{U_j\}$ are random variables. Consider the process $$ Y_t = \sum_{k=0}^{\sigma_t}U_k, $$ where $\sigma_t$ is some suitable increasing process taking values in $\mathbb{N}$. The jump times of $Y_t$ are $$ \tau_k = \inf\{t:\sigma_t\geq k\}. $$ To derive the characteristics of $Y$, in Jacod and Shiryaev's book, it is used the following strategy. They consider the set $$ J = \bigcup_{k}[\tau_k] $$ where $[\tau_k]=\{(\omega,\tau_k(\omega))|\omega\in\Omega\}$. The authors say that, by construction $$ Y = 1_{J}\cdot Y\quad(1). $$ I do not understand why identity $(1)$ holds. If $(\omega,t)\notin J$ therefore $1_{J}=0$, but why should be that $Y=0$ as prescribed by $(1)$? In this case, I can only say that $t\neq \tau_k(\omega)$ for all $k$, which means that $t$ is not a jump time for $Y$. Nevertheless it could be that $Y_t(\omega)\neq 0$.
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Could you give the exact location in the book from where you take that? – Mushu Nrek Jan 28 '24 at 21:55
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@MushuNrek sure. You can find it on page 95 of the second edition, the beginning of the proof of Theorem II.3.11 – AlmostSureUser Jan 29 '24 at 10:54
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The key question is what $1_{J} \cdot Y$ actually means. The $\cdot$ does not mean a simple product. The definition used was probably
$$(1_{J} \cdot Y)_t = \sum_{s=0}^t 1_{J}(s)(Y_s-Y_{s-1}),$$
with the convention $Y_{-1} = 0$.
By construction, $\tau_0=0$, so $0 \in J$.
Given $s \ge 1$, to have $Y_s-Y_{s-1} \ne 0$, one must have $\sigma_s>\sigma_{s-1}$, equivalently $s=\tau_k$ for some $k$, i.e. $s \in J$.
Hence $1_{J}(s)(Y_s-Y_{s-1}) = Y_s-Y_{s-1}$ for all $s \ge 0$ Equality (1) follows.
Christophe Leuridan
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You are totally right, thanks. The symbol $\cdot$ indicates in the book, but I presume it is quite customary, stochastic integration. So $1_J\cdot Y$ is the stochastic integral using $Y$ as integrator and $1_{J}$ as integrand. – AlmostSureUser Feb 07 '24 at 10:58