# convergence to SDE

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 $\sum\limits_j(B_{t_{j+1}}-B_{t_j})$ ?

• 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". Jun 29, 2018 at 16:34
• I'm not sure if my edits has made it any clearer... Jun 29, 2018 at 16:43
• So you are asking if the discretized solution converges to the exact solution? Jun 29, 2018 at 17:01
• 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. Jun 29, 2018 at 17:13

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.