Why do I not get a standard normal random variable after applying central limit theorem?

Question

Suppose that $S_n = \sum_{i=1}^n X_i$, where $X_i$ has $ P(X_i = 1) = P(X_i = -1 ) = 1/2$.

I am trying to use central limit theorem to calculate $$\lim_{n \to \infty} P\left(S_n < \frac{1}{3} \sqrt{n}\right)$$

My attempt is that: $E(S_n) = 0$ and $\operatorname{Var}(S_n) = n$, hence $$\lim_{n \to \infty} P\left(S_n < \frac{1}{3} \sqrt{n}\right) = \lim_{n \to \infty} P\left(\frac{S_n - n\mu}{\sigma\sqrt{n}}< \frac{\frac{1}{3} \sqrt{n} - n\mu}{\sigma \sqrt{n}}\right) = \lim_{n \to \infty} P\left(\frac{S_n}{n} < \frac{\sqrt{n}}{3n}\right) = 0.5$$

Do I use CLT correctly? Why is $\frac{S_n}{n}$ not a standard normal random variable?

Answer 1

I see what the problem is. In the formula of central limit theorem, $\mu$ and $\sigma$ are $X_i$'s mean and variance but not $S_n$'s mean and variance.

Answer 2

It is obvious that $\operatorname{E}[S_n] = 0$. For the variance, since the $X_i$ are independent,

$$\begin{align} \operatorname{Var}[S_n] &\overset{\text{ind}}{=} \sum_{i=1}^n \operatorname{Var}[X_i] \\ &= n \operatorname{Var}[X_1] \\ &= n \left((1 - 0)^2 \Pr[X_1 = 1] + (-1-0)^2 \Pr[X_1 = -1]\right) \\ &= n. \end{align}$$ Then $$\begin{align} \Pr\left[S_n \le \frac{\sqrt{n}}{3}\right] &= \Pr\left[\frac{S_n - \operatorname{E}[S_n]}{\sqrt{\operatorname{Var}[S_n]}} \le \frac{\sqrt{n}/3 - 0}{\sqrt{n}} \right] = \Pr\left[\frac{S_n - \operatorname{E}[S_n]}{\sqrt{\operatorname{Var}[S_n]}} \le \frac{1}{3} \right]. \end{align}$$ As $n \to \infty$, $$\frac{S_n - \operatorname{E}[S_n]}{\sqrt{\operatorname{Var}[S_n]}} \to Z$$ where $Z \sim \operatorname{Normal}(0,1)$ is a standard normal random variable; hence $$\lim_{n \to \infty} \Pr\left[S_n \le \frac{\sqrt{n}}{3}\right] = \Pr[Z \le 1/3].$$

If you don't like the application of the CLT in this manner, you can also do it on the sample mean instead of the sample total; i.e., define $$\bar X_n = \frac{S_n}{n}$$ and we have

$$\Pr\left[S_N \le \frac{\sqrt{n}}{3}\right] = \Pr\left[\bar X_n \le \frac{1}{3\sqrt{n}}\right] = \Pr\left[\frac{\bar X_n - \operatorname{E}[\bar X_n]}{\sqrt{\operatorname{Var}[\bar X_n]}} \le \frac{\frac{1}{3 \sqrt{n}} - 0}{\sqrt{\frac{1}{n}}}\right],$$

since the variance of $\bar X_n$ is simply $\frac{1}{n^2} \operatorname{Var}[S_n] = \frac{1}{n}$.