Let $B_t$ be a Brownian motion. Let $X_t:=e^{-t}B_{e^{2t}}$ and therefore is $N(0,1)$ distributed with $X_0=B_1$. I wanna prove that $\int_0^tX_sds$ is normally distributed too.
I see that $$\lim_{n\to \infty}\frac1n\sum_{k=1}^nX_{kt/n} \to \int_0^tX_sds.$$
Then in the proof that I'm following is written that $\frac1n \sum_{k=1}^nX_{kt/n}$ is a Gaussian process but I'm not able to understand why. I would agree if the we have a sum of independent normally distributed random variable, but in this case it is not. So why it is a gaussian process?
my attempt is the following:$\sum_{k=1}^nX_{kt/n}=\sum_{k=1}^n \big( \sum_{i=0}^tX_{\frac{k(i+1)}{n}}-X_{\frac{k(i)}{n}} \big) +(B_1-B_0)$ where $B_0=0$ and in this way I have a sum of independent normally distributed random variable.
HINT
That $(X_t)_t$ is a Gaussian process does not mean that $X_1$ and $X_2$ are independent, as you claim. But the increments are independent, so I can say that $D_{21} = X_2 - X_1$ is independent of $X_1$, and so $$ X_1 + X_2 = X_1 + \left(X_1 + D_{21}\right) = 2X_1 + D_{21}, $$ which is a sum of zero-mean normally distributed random variables, and hence itself is a zero-mean normal.
UPDATE
In a similar way, note that $$ \begin{split} X_1 + X_2 + X_3 &= X_1 + X_2 + (X_2 + D_{32}) \\ &= X_1 + 2X_2 + D_{32} \\ &= X_1 + 2(X_1 + D_{21}) + D_{32} \\ &= 3X_1 + D_{21} + D_{32}. \end{split} $$