My reason for asking this question is because I can't seem to justify extending the results from the Ito Integral of elementary functions to the continuous form after taking the limit. For example, if I prove that the Ito integral of a simple function for finite $n$ is a martingale, I don't understand how to extend this for when we take the limit as $n \to \infty$. So that is my bigger picture question.

I have looked all over for a direct justification of: \begin{align*} \mathbb E\left[\int_{0}^{t}f(W_s,s) \, dW_s \right] &=\mathbb E\left[\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{f(W_{t_{i-1}},{{t}_{i-1}})(W({{t}_{i}})-W({{t}_{i-1}})})\right] \\ &= \underset{n\to \infty }{\mathop{\lim }} \mathbb E\left[\sum\limits_{i=1}^{n}{f(W_{t_{i-1}},{{t}_{i-1}})(W({{t}_{i}})-W({{t}_{i-1}})})\right] \end{align*}

where, $f$ is square integrable, so $\mathbb E [ \int_s^tf^2(\omega, r)dr] \leq \infty$, $f$ is adapted to the natural filtration generated by $W$, and also measurable with respect to the underlying probability space.

where the limit and expectation is interchanged but I haven't been able to find anything precise enough. I have seen mentions of using Dominated Convergence but not to which the bounding random variable is.

I know the sequence converges to the Ito Integral, but the Ito integral isn't necessarily with finite expectation or such that $$\left|\int_{0}^{t}f_s \, dW_s \right| \geq \sum\limits_{i=1}^{n}{f(W_{t_{i-1}},{{t}_{i-1}})(W({{t}_{i}})-W({{t}_{i-1}})})$$ for all $n$.

I tried using the absolute value of the Ito Integral and other types of simple functions but can't seem to pick one that definitely fits the DCT criteria.

Thanks a lot!

  • $\begingroup$ What are your assumptions on $X$...? You will need some integrability and measurability condition. $\endgroup$ – saz Apr 5 at 5:12
  • $\begingroup$ I'll edit them in $\endgroup$ – Slade Apr 5 at 5:14
  • $\begingroup$ Finished. Basically it's the 'usual' conditions on the function, like in Oksendal's book on SDEs $\mathcal V$. I changed it to $f(W_s,s)$ to make it more general as well $\endgroup$ – Slade Apr 5 at 5:22

Interchanging the limit with the expectation is typically justified by choosing an approximating sequence of simple functions $(f_n)_n$ such that the corresponding stochastic integrals $\int_0^t f_n(s) \, dW_s$ converge to $\int_0^t f(s) \, dW_s$ in $L^2(\mathbb{P})$.

If $f: \Omega \times [0,\infty) \to \mathbb{R}$ is a progressively measurable function such that

$$\mathbb{E} \left( \int_0^T f(s)^2 \, ds \right) < \infty$$

for all $T>0$, then there exists a sequence of simple functions $(f_n)_{n \in \mathbb{N}}$ such that

$$\mathbb{E} \left( \int_0^t |f_n(s)-f(s)|^2 \, ds \right) \xrightarrow[]{n \to \infty} 0$$

for all $T>0$. By the very definition of the Itô integral, this implies that

$$\int_0^T f(s) \, dW_s = L^2(\mathbb{P})- \lim_{n \to \infty} \int_0^T f_n(s) \, dW_s \tag{1}$$

for all $T>0$. Note that this implies, in particular, that

$$\mathbb{E} \left( \int_0^T f(s) \, dW_s \right) = \lim_{n \to \infty} \mathbb{E}\left( \int_0^t f_n(s) \, dW_s \right). \tag{2}$$

In general, there is no explicit formula for $f_n$. If $f$ is mean-sequare continuous, i.e.

$$\lim_{s \to t} \mathbb{E}(|f(s)-f(t)|^2) = 0, \qquad t>0,$$

then it can be shown that

$$f_n(s) := \sum_{j=1}^{k(n)} f(t_{j-1}^{(n)}) 1_{[t_{j-1}^{(n)},t_j^{(n)})}(s)$$

does the job for any sequence of partitions $\Pi_n = \{0=t_0^{(n)}<\ldots<t_{k(n)}^{(n)}=T\}$ with mesh size tending to zero. In this case, $(1)$ reads

$$\int_0^t f(s) \, dW_s = L^2(\mathbb{P})-\lim_{n \to \infty} \sum_{j=1}^{k(n)} f(t_{j-1}^{(n)}) (W_{t_j^{(n)}}-W_{t_{j-1}^{(n)}})$$

and $(2)$ becomes

$$\mathbb{E} \left( \int_0^t f(s) \, dW_s \right) = \lim_{n \to \infty} \sum_{j=1}^{k(n)} \mathbb{E} \bigg[ f(t_{j-1}^{(n)}) (W_{t_j^{(n)}}-W_{t_{j-1}^{(n)}}) \bigg]. $$

All the above-mentioned results can be, for instance, found in the monograph Brownian Motion - An Introduction to Stochastic Processes by Schilling & Partzsch.

  • $\begingroup$ Thanks a lot! This is perfect. I have never seen the mean-square continuous condition before so I will check it out and see if it helps with the problems I am having. $\endgroup$ – Slade Apr 5 at 17:40
  • $\begingroup$ Just to confirm, $L^2$ convergence implies $\lim_{n \to \infty} \|\phi_n-f\|_{L^2(\lambda_T \otimes \mathbb{P})} = 0$, and not $ \|\lim_{n \to \infty} \phi_n-f\|_{L^2(\lambda_T \otimes \mathbb{P})} = 0$? $\endgroup$ – Slade Apr 5 at 18:06
  • $\begingroup$ @Slade Yes, the first one is the very definition of $L^2$ convergence: $\phi_n \to f$ in $L^2(\mu)$ if and only if $$\lim_{n \to \infty} \|\phi_n-f\|_{L^2(\mu)} = 0.$$ $\endgroup$ – saz Apr 5 at 19:34
  • $\begingroup$ Okay got it! Thanks so much. Your answers on SE are super helpful $\endgroup$ – Slade Apr 6 at 19:24
  • $\begingroup$ @Slade Glad to hear that; you are welcome. $\endgroup$ – saz Apr 6 at 19:30

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