Could you help me with the following:
I have that $$T(x):=\frac{X(nx)-E[X(nx)]}{\sqrt{n}} \xrightarrow{d} N(0, \frac{x^k}{k})$$ for each fixed $x>0$, where we also have that $\frac{X(nx)}{t}$ is the empirical Distribution function of a Distribution, where $\frac{E[X(nx)]}{t}$ is the Distribution function.
Then, it is meantioned that this proves that $T(x) \xrightarrow{d} W(\frac{x^k}{k})$, where W is a Standard brownian Motion.
Why? Is there any reference for that? As reference, it is mentioned the following Theorem, which does not help me: Suppose the $\xi_n$ are Independent and have a common Distribution function $F(t)$. Then, with $$Y_n(t,\omega)=\sqrt{n} (F_n(t,\omega)-F(t)),$$ $Y_n \xrightarrow{d} Y$, where $Y$ is the Gaussian random element of $D$ specified by $E[Y(t)]=0$, $E[Y(s)Y(t)]=F(s)(1-F(t))$ for $s\leq t$.
Thus, what really is my question: What has a Standard brownian Motion to do with a Gaussian random element? Could you explain this to me? I would really appreciate any help!!
(DEFINITION STANDARD BROWNIAN MOTION: $W(0)=0$ and $t \mapsto W(t)$ is continuous with probability 1 and $W(t)$ has stationary, jointly Independent increments and the increment $W(t_1+t_2)-W(t_1)$ is normal-distributed $N(0,t_2)$.)
For every nonnegative $u$, the distribution of $W(u)$ is $N(0,u)$ hence, for any parameter $u(x)$, the assertion that $T(x)\stackrel{d}{\to}N(0,u(x))$ and the assertion that $T(x)\stackrel{d}{\to}W(u(x))$ are logically equivalent.