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How do I show that $\Delta Z_t = \theta \, \Delta t+\Delta B_t\rightarrow dZ_t = \theta \, dt+dB_t$ when $\Delta t\rightarrow0$, where $B_t$ is a standard brownian motion?

Here the sde $dZ_t = \theta \, dt+dB_t$ is short for $Z_t=Z_0+\int_0^t\theta \, ds+\int_0^tdB_s$.

I have always taken it for granted... and can't for the life of me proof this. Is there some other way to approach this? I've tried writing it as $\Delta Z_t=\theta\Delta t+\sqrt{\Delta t}\, \varepsilon$ where $\varepsilon\sim N(0,1)$ but still get nowhere.

I suppose I have to go through the definition of the sde and write it a sum? Something like $\sum\limits_j(B_{t_{j+1}}-B_{t_j})$ ?

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  • $\begingroup$ How do you converge towards a symbol? You should be aware that $θdt+dB_t$ is not a number. If you put yourself into a non-standard-analysis framework where this expression is actually an infinitesimal, please describe this framework, if possible with references. E. Nelson has something like that in his "Radically elementary probability theory". $\endgroup$
    – Lutz Lehmann
    Jun 29, 2018 at 16:34
  • $\begingroup$ I'm not sure if my edits has made it any clearer... $\endgroup$
    – icecream_sandwich07
    Jun 29, 2018 at 16:43
  • $\begingroup$ So you are asking if the discretized solution converges to the exact solution? $\endgroup$
    – Lutz Lehmann
    Jun 29, 2018 at 17:01
  • $\begingroup$ I think I am more confused about why when i discretize it, there is no summation term? If i understand correctly, if i discretize the SDE in the integral form, i should get $\sum_i\theta (t_{i+1}-t_i)+\sum_j (B_{j+1}-B_j)$. Yet I have always understood that when we write $dZ_t = \theta dt +dB_t$, it is correct to say that in a small time interval $\delta t$, the process Z changes according to $\Delta Z_t = \theta \Delta t + \Delta B_t$, and when $\Delta t \rightarrow 0$, we get the SDE. $\endgroup$
    – icecream_sandwich07
    Jun 29, 2018 at 17:13

1 Answer 1

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For this specific case, it is easy to see that if $\theta$ is a nice enough function, by Lebesgue-Stieljes one can approximate $\int \theta dt$ by $\sum\limits_{i=0}^{n-1} \theta(t_i) \Delta t_{i+1}$, where $\Delta t_i := t_i - t_{i-1}$, where $0 = t_0 < t_1 < \dots < t_n := t$ is a partition. For the same partition, one can approximate trivially $B_t = B_t - B_0=\sum\limits_{i=1}^{n} \Delta B_{t_i}$. Similarly, $Z_t - Z_0 = \sum\limits_{i=1}^{n} \Delta Z_{t_i}$. Taking limit for the last two approximation is trivial and converges in every sense to $B_t$ and $Z_t - Z_0$, respectively. This is how we make sense of $dZ_t = \theta dt + dB_t$.

More generally, if both the stochastic and Lebesgue-Stieljes integrals are nontrivial, one has to be careful with which partition to use because stochastic integrals are defined as $L^2$-limits whereas Lebesgue-Stieljes is defined as a.s.-limit. I recommend looking at the proof of Ito's formula as it clearly outlines how to deal with the two.

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