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I have a stochastic process given as $X_t(\omega)=\sum_{m=1}^\infty a_m 1_{\{0<b_m\leq t\}}(\omega) $, where $(a_m,b_m)\in(\mathbb{R}^d,\mathbb{R}_{+})$.

I know that that the sum converge absolutely for all $k\in\mathbb{N}$ inserted instead of $t\in\mathbb{R}_+$, and from this I should be able to conclude that the process is càdlàg (at least the author of the notes conclude without further explaination).

I know that $X_{t-}(\omega)=\sum_{m=1}^\infty a_m 1_{\{0<b_m< t\}}(\omega)$, so the jump at time t must be $\Delta X_t = a_m 1_{\{t\}}$, and the left-hand limit exists (is this right?). But how am I supposed to conclude that the process is càdlàg with what I know?

Additionally, I am confused about the role of the $k\in\mathbb{N}$, since the sum would converge for all $t>0$.

saz
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Limbo
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  • Sorry but what exactly is stochastic here? You could also (or is it, equivalently?) explain the notation $1_{{0<b_m\leq t}}(\omega) $. – Did Mar 20 '15 at 21:38
  • Sorry for not making it clear. Both $a_m$ and $b_m$ are stochastic, so I could just as well write $1_{(0,t]}(b_m(\omega))$. – Limbo Mar 21 '15 at 09:40

1 Answers1

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For each fixed $M \in \mathbb{N}$,

$$X_t^M := \sum_{m=0}^M a_m 1_{\{t \geq b_m\}}$$

defines a càdlàg process. Moreover, for fixed $T>0$, we have

$$\sup_{t \in [0,T]} \left| X_t^M - X_t \right| \leq \sum_{m=M+1}^{\infty} |a_m| 1_{\{T \geq b_m\}}.$$

Since the series on the right-hand side converges by assumption, we find that the left-hand side converges uniformly on $[0,T]$ to $0$ as $M \to \infty$. This means that

$$X_t = \sum_{m=0}^M a_m 1_{\{t \geq b_m\}}, \qquad t \in [0,T]$$

is the uniform limit of càdlàg functions; hence, càdlàg. As $T>0$ is arbitrary, the claim follows.

saz
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  • Thank you for the answer! Could you elaborate a little on why $X_t^M$ is a càdlàg process? And why does the author in the present case use that the sum converge absolutely for all $k\in\mathbb{N}$ and not for all $t>0$? – Limbo Mar 21 '15 at 09:35
  • @Limbo Well, an process of the form $$Y_t := a 1_{{t \geq b}}$$ is certainly cadlag, isn't it? Since $X_t^M$ is a (finite) linear combination of such processes, it is itself cadlag. Concerning your second question: If the sum converges absolutely for all $k$, then this already implies that it converges absolutely for all $t$. – saz Mar 21 '15 at 14:10