How are these 2 results, derived from the Central Limit Theorem, related?

Question

Unfortunately, it's been a while since I last studied mathematics - so I'm currently fuzzy to say the least. I am now skim reading through lecture notes for a module on Markov Processes and have come across the below that I cannot quite understand.

In my lecture notes, they have derived (from the CLT) that, for a simple random walk

$$ \frac{\frac{X_{n}}{n}-(p-q)}{\sqrt{\frac{4pq}{n}}} \rightarrow N(0, 1) $$

as $n \rightarrow \infty$ (where $p$ is the probability of the process increasing and $q$ is the probability of it decreasing).

I understand the above just fine. However, the notes then go on to say that this is equivalent to

$$ X_{n} \simeq X_{n}' \sim N(n(p-q), 4npq) $$

for large $n$. This is the part I do not understand. Could anyone explain how these expressions are equal, and where the $X_{n}'$ term comes from?

Answer 1

$$\frac{\frac{X_n}{n}-(p-q)}{\sqrt{\frac{4pq}{n}}}=\frac{n}{n}.\frac{\frac{X_n}{n}-(p-q)}{\sqrt{\frac{4pq}{n}}}=\frac{X_n-n(p-q)}{\sqrt{\frac{n^24pq}{n}}}=\frac{X_n-n(p-q)}{\sqrt{n4pq}}$$ I.e. if $X'_n$ is a normal variable with mean $n(p-q)$ and variance $4npq$, then $\frac{X'_n-n(p-q)}{\sqrt{n4pq}} \sim N(0,1)$, so for large n, $X_n$ is $\textbf{approximately}$ equal to $X'_n$, i.e. $X_n\approx X'_n$, not equal.