Show that $\mathbb{E}\left(\int_0^1X_n(t)dW(t)-\int_0^1X(t)dW(t)\right)^2\to 0$

Question

I tried to shove this question in another one of my posts as a follow-up question, but I deleted my comment and will post it instead. I would really appreciate if someone could help me spot out and correct the error.

This past exam problem consists of several parts. The part that keeps on giving me a result that makes no sense is part E. Part E is supposed to be the easiest one along with part B.

EXERCISE:

Let $W(t)$, $t\in\mathbb{R}_+$ be a Brownian motion with its natural filtration $\mathcal{F}_t, t\in\mathbb{R}_+$. Let $$\mathcal{H}:=\{h(t):h(t)\text{ is an adapted process, }\mathbb{E}\left[\int_0^{\infty}h^2(t)dt\right]<\infty\}$$ denote the set of general integrands with respect to $W(t)$.

Consider the stochastic processes $$X_n(t):=\sum_{k=0}^{n-1}W\left(\frac{k}{n}\right)\mathbb{1}_{\left(\frac{k}{n},\frac{k+1}{n}\right]}(t), t\ge 0,$$ for $n\ge 1$ and define $X(t):=W(t)\mathbb{1}_{[0,1]}(t),t\ge 0$.

(A) $X_n\in\mathcal{H}$ for all $n\ge 1$.

(B) $X\in\mathcal{H}$.

(C) $\mathbb{E} \left( \left[ W \left( \frac{k}{n} \right)-W(t) \right]^2 \right) = \mathbb{E}\left( \left[ W \left( t-\frac{k}{n} \right) \right]^2 \right) = t-\frac{k}{n}$ for all $t\in\left(\frac{k}{n},\frac{k+1}{n}\right]$.

(D) $d_{\mathcal{H}}(X_n,X):=\mathbb{E}\left(\int_0^{\infty}(X_n(t)-X(t))^2dt\right)=\frac{1}{2n}$ for all $n\ge 1$ and $d_{\mathcal{H}}(X_n,X)\to 0,\quad n\to\infty$.

(E) Conclude that $\mathbb{E}\left(\int_0^1X_n(t)dW(t)-\int_0^1X(t)dW(t)\right)^2\to 0$ as $n\to\infty$.

ATTEMPT number 1 - (E): $$\mathbb{E}\left(\int_0^1X_n(t)dW(t)-\int_0^1X(t)dW(t)\right)^2=\mathbb{E}\left(\int_0^{1}(X_n(t)-X(t))^2dt\right)\quad=\sum_{k=0}^{n-1}\int_0^1 \left(t-\frac{k}{n}\right) dt =\sum_{k=0}^{n-1}\left[\frac{t^2}{2}-\frac{k}{n}t\right]_0^1=\sum_{k=0}^{n-1}\left(\frac{1}{2}-\frac{k}{n}\right)=\frac{1}{2},$$ so the result is wrong.

Answer 1

I think it is obvious from the phrasing of the question that you conclude E from part D without an additional calculation.

I cannot be bothered to figure out how you made the calculation in your attempt and it makes absolutely no sense to me whatsoever, but

Isn't

$$0 \leq \mathbb{E}\left(\int_0^{1}(X_n(t)-X(t))^2dt\right)\leq\mathbb{E}\left(\int_0^{1}(X_n(t)-X(t))^2dt\right),$$

since the integrand is positive?

Then since the right hand converging like $1/n$, by sandwich rule, so is the integral in the middle? Then use the first equality from your attempt.

Answer 2

I think the issue is that the indicator functions are not considered. Let $a_k = k/n$, and $A_k = [ a_k, a_{k+1} ], k=0,...,n-1.$ Then, \begin{align*} \mathbf{E} \int_0^1 (X_n(t) - W(t) )^2dt &= \mathbf{E} \int_0^1 \left(\sum_{k=0}^{n-1} W(a_k)^21_{A_k}(t) + W(t)^2 - 2 \sum_{k=0}^{n-1} W(a_k)W(t)1_{A_k}(t) \right) dt \\ &= \sum_{k=0}^{n-1} a_k /n + \int_0^1 tdt - 2 \sum_{k=0}^{n-1} a_k/n, \end{align*} because $\int_0^1 1_{A_k}(t)dt = 1/n $. From here, you can see that part (d) is true and hence (e) follows.