I have done this exercise but I have done something wrong because I don't get the correct result for the next part of this exercise (this is part B).
I posted something earlier that is related to another part of this exercise, but I will put all of the information in this post again. I would really appreciate it if you could help me since I have done something wrong - I will comment at the end on what I think I did wrong. Thanks a lot!
QUESTION:
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 $X(t):=W(t)\mathbb{1}_{[0,1]}(t),t\ge 0$.
D)
(i) Evaluate $\mathbb{E}\left(\left[W\left(\frac{k}{n}\right)-W(t)\right]^2\right)$ for all $t\in\left(\frac{k}{n},\frac{k+1}{n}\right]$, using the definition of Brownian motion.
(ii) Calculate the distance of $X_n(t)$ from $X(t)$ in H, i.e. $$d_{\mathcal{H}}(X_n,X):=\mathbb{E}\left(\int_0^{\infty}(X_n(t)-X(t))^2dt\right),$$ for all $n\ge 1$.
(iii) Show that $d_{\mathcal{H}}(X_n,X)\to 0,\quad n\to\infty$
ATTEMPT:
In parts B and C I proved that $X_n\in\mathcal{H}$ and $X\in\mathcal{H}$.
Someone on stackexchange helped me with part D(i) and the answer for this and my attempt at part D(ii) are:
D(i) $$\mathbb{E}\left(\left[W\left(\frac{k}{n}\right)-W(t)\right]^2\right)=\mathbb{E}\left(\left[W\left(\frac{k}{n}-t\right)\right]^2\right)=\frac{k}{n}-t$$
D(ii) We have $$d_{\mathcal{H}}(X_n,X)=\mathbb{E}\left(\int_{0}^{\infty}(X_n(t)-X(t))^2dt\right)=\sum_{k=0}^{n-1}\int_{\frac{k}{n}}^{\frac{k+1}{n}}\left(\frac{k}{n}-t\right)dt=\sum_{k=0}^{n-1}\left[\frac{k}{n}t-\frac{t^2}{2}\right]_{\frac{k}{n}}^{\frac{k+1}{n}}=\sum_{k=0}^{n-1}-\frac{1}{2n^2}=-\frac{1}{2n}$$ But how can the distance be negative? Although this satisfies $d_{\mathcal{H}}(X_n,X)\to 0$ as $n\to\infty$, the same method I used does not satisfy part E so it must be wrong.
What did I do wrong? My guess is the limits of the integration? The limits for $X(k/n)$ are $t\in\left(\frac{k}{n},\frac{k+1}{n}\right)$ and for $X(t)$ it is $t\in[0,1]$
The sign in the first part of your solution is wrong. Note that you consider the case $\frac{k}{n} < t$, so the expectation is equal to $t - \frac{k}{n}$.