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An increasing integrable process $A_t$ is natural if $E\int_0^t m_s dA_s = E\int_0^t m_{s-} dA_s$ for every bounded right-continuous martingale.

If both the Reimann-Stieltjes integrals $\int_0^t m_s dA_s, \int_0^t m_{s-} dA_s$ exist, then I think it can be shown that they should be equal almost surely. Therefore, the definition of a natural process seems 'un-natural', as it requires only equality of expectations. I am not sure what I am missing.

jpv
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1 Answers1

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I think it can be shown that they should be equal almost surely.

Not at all. Let $A_t = \mathbf{1}_{[1,+\infty)}(t)$ and $m$ be a martingale that always jumps at $1$, e.g. $m_t = \xi \mathbf{1}_{[1,+\infty)}$, where $\xi=\pm 1$ with probability $1/2$. Then $0=\int_0^2 m_{s-} dA_s \neq \int_0^2 m_{s} dA_s = \xi$.

However, this process $A$ is natural (what can be more "natural"?) e.g. $\mathbb{E}\int_0^t (m_s-m_{s-}) dA_s = \mathbb{E}\xi = 0$.

Another useful fact: a process is natural iff it is predictable.

zhoraster
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  • zhoraster: Thanks for the reply. I think the integrals do not exist in the Reimann-Stieltjes sense since $m$ and $A$ share a point of discontinuity. – jpv Feb 19 '17 at 16:07
  • I understand that the integrals depend on which point is chosen, right or left. If the integrals are understood in this sense then it seems correct. Using standard definition for the integrals creates difficulties. Probably, this created my confusion. – jpv Feb 19 '17 at 17:24
  • @jpv, I understand your confusion. But in stochastic analysis one rarely defines integral as Riemann-Stieltjes. – zhoraster Feb 19 '17 at 17:34
  • @jpv Could you elaborate on what you mean by choosing left and right point? Also, in the above example, since A_{t} is a right continuous step function, the Riemann-Stieltjes integral exists iff m_{t} is left continuous. So isn't the equality trivially satisfied? – Sudheesh Surendranath Apr 23 '20 at 09:28