I am trying to do a past exam paper to practice, but I don't know if I have answered this question properly... I would really appreciate it if someone could double check it. Thanks a lot!
QUESTION: 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.
ANSWER: I have that $$\mathbb{E}\left(\left[W\left(\frac{k}{n}\right)-W(t)\right]^2\right)=\mathbb{E}\left(W^2\left(\frac{k}{n}\right)\right)+\mathbb{E}\left(W^2(t)\right)-\mathbb{E}\left(2W\left(\frac{k}{n}\right)W(t)\right)=\frac{k}{n}+t$$
No, it is not correct; the identity $$\mathbb{E}(W(k/n) \cdot W(t))=0$$ (which you used in your calculation) does not hold true.
Hint: A Brownian motion has stationary increments, i.e. $W_t-W_{k/n}\stackrel{d}{=} W_{t-k/n}$. So, $$\mathbb{E} \left( \left[ W \left( \frac{k}{n} \right)-W(t) \right]^2 \right) = \mathbb{E}\left( \left[ W \left( t-\frac{k}{n} \right) \right]^2 \right) = \dots$$