Relating to this question, I have a further one, and hope, someone can help me.
I know that $$\left(X_j - X_{j-1}\right)_{j=1}^t \xrightarrow{d} \left(Y_j\right)_{j=1}^t.$$ Further, we know that $$\left(Y_j\right)_{j=1}^t=^d \left(W(y_j)-W(y_{j-1})\right)_{j=1}^t,$$ where $W$ is a Standard brownian Motion. Now, how do we get that $$\left(X_j\right)_{j=1}^t \xrightarrow{d} \left(W(y_j)\right)_{j=1}^t.$$
I think, it should be rather obvious, but I do not see it.
It follows straight from the definition of convergence in distribution that
$$(X_j^n-X_{j-1}^n)_{j=1}^t \stackrel{d}{\to} (W(y_j)-W(y_{j-1}))_{j=1}^t.$$
Define a mapping $g: \mathbb{R}^t \to \mathbb{R}^t$ by
$$g(x) :=(x_1,x_1+x_2,\ldots,x_1+\ldots+x_t).$$
where $x := (x_1,\ldots,x_t)$. Then $g$ is continuous and
$$g((y_j-y_{j-1})_{j=1}^t) = (y_1,\ldots,y_t) \tag{1}$$
for any $y = (y_1,\ldots,y_t) \in \mathbb{R}^t$. Applying the continuous mapping theorem yields
$$g((X_j^n-X_{j-1}^n)_{j=1}^t) \stackrel{d}{\to} g((W(y_j)-W(y_{j-1}))_{j=1}^t).$$
Hence, by $(1)$, this is equivalent to
$$(X_j^n)_{j=1}^t \stackrel{d}{\to} (W(y_j))_{j=1}^t \qquad \text{as} \,\, n \to \infty.$$