# Justification of interchanging Expectation and Limit in Ito Integral Approximation

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!

• What are your assumptions on $X$...? You will need some integrability and measurability condition. – saz Apr 5 at 5:12
• I'll edit them in – Slade Apr 5 at 5:14
• 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 – 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 $$\Pi_n = \{0=t_0^{(n)}<\ldots with mesh size tending to zero. In this case, $$(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.

• 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. – Slade Apr 5 at 17:40
• Just to confirm, $L^2$ convergence implies $\lim_{n \to \infty} \|\phi_n-f\|_{L^2(\lambda_T \otimes \mathbb{P})} = 0$, and not $\|\lim_{n \to \infty} \phi_n-f\|_{L^2(\lambda_T \otimes \mathbb{P})} = 0$? – Slade Apr 5 at 18:06
• @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 $$\lim_{n \to \infty} \|\phi_n-f\|_{L^2(\mu)} = 0.$$ – saz Apr 5 at 19:34