# convergence to standard brownian motion

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.
• thank you for your answer. I think, this is exactly what I am looking for. As the reference I am looking at is much more difficult, I wonder, whether you know a good paper where this is explained relatively easy? This would really help me. As I also have a Problem with the Notation (in my paper), is it true that $W(u) =^d N(0,u)$? I am not sure whether this sign $=^d$ means this. – user146358 Aug 14 '14 at 13:02
• Okay, then I think, this should hold, doesn't it? As we have $W(u+s)-W(s) =^d N(0,u)$, we have with $W(0)=0$ that $W(u)=^d N(0,u)$? Or does this not make sense in context of $W$? I am not sure what $W$ exactly is. Does it have a Distribution, so does it make sense to write $W=^d ...$? – user146358 Aug 14 '14 at 13:07
• Wow, you seem really lost here... Yes, $W(u)\stackrel{d}{=}N(0,u)$ by definition. – Did Aug 14 '14 at 13:09