# Limit of a Symmetric Random Walk

I'm given a probability space of ($$\Omega$$, $$\mathcal{F}$$, $$\mathbb{P}$$) and am asked to look into a symmetric random walk with its n-step defined as

$$X_k = \Bigg\{ \begin{matrix} 1 & \text{with probability 1/2} \\ -1 & \text{with probability 1/2} \end{matrix}$$

where $$X_n$$ and $$X_m$$ are independent for all $$n \ne m$$.

$$S_{n} = \sum_{k=1}^{n}X_k, \text{ i = 1, 2, ..., t}$$

where $$S_0 = 0$$. Defining

$$W_t^{(n)}=\frac{S_{\lfloor nt \rfloor}}{\sqrt{n}} = \frac{1}{\sqrt{n}} \sum_{i=1}^{\lfloor nt \rfloor}X_k$$

I want to show that for fixed time t

$$\lim_{n \to \infty}W_t^{(n)} = \frac{S_{\lfloor nt \rfloor}}{\sqrt{n}} = \lim_{n \to \infty} \frac{1}{\sqrt{n}} \sum_{i=1}^{\lfloor nt \rfloor}X_k \overset{D}{\to} \mathcal{N}(0,t)$$

I know by the central limit theorem that $$\frac{X_N}{\sqrt{N}} \overset{D}{\to} \mathcal{N}(0,1)$$ as $$N \to \infty$$ and I have found a lecture that states

$$\lim_{n \to \infty}\frac{S_{\lfloor nt \rfloor}}{\sqrt{n}} = \lim_{n \to \infty}\frac{S_{\lfloor nt \rfloor}}{\sqrt{\lfloor nt \rfloor}}\frac{\sqrt{\lfloor nt \rfloor}}{\sqrt{n}} \to \mathcal{N}(0,1)\sqrt{t} \overset{D}{=} \mathcal{N}(0,t)$$

However, I am confused as to how each step in this equation leads to the other. I'd greatly appreciate if anyone could provide some insight on this. Thanks.

If $$Y_n \to Y$$ in distribution then $$Y_{n_k} \to Y$$in distribution for any subsequence $$\{n_k\}$$. Here $$\{ {\lfloor {nt} \rfloor}\}$$ is a subsequence of $$\{1,2,\cdots\}$$. Next, $$Y_n \to Y$$ in distribution and $$a_n \to a>0$$ implies $$a_nY_n \to aY$$ in distribution. Finally $$\{\sqrt {\lfloor {nt} \rfloor}\}/\sqrt n \to \sqrt t$$ as $$n \to \infty$$.