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 $n \to \infty$.


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