# $CLT$ and $LLN$ give different results

I tried to solve a problem two different ways and I got different results.

Let $( X_i )_{i \in \mathbb{N}}$ be a series of independent, identically distributed random variables, with $\mathbb{E}[X_i] = 1$ and $\mathbb{V}[X_i] = 1$

Determine

$$\lim_{n \to \infty} \mathbb{P}\left(\frac{1}{\sqrt{n}} \sum_{i=1}^n X_i \leq \sqrt{n}\right)$$

Here are the two approaches that I tried.

Central limit theorem \begin{align*} & \lim_{n \to \infty} \mathbb{P}\left(\frac{1}{\sqrt{n}} \sum_{i=1}^n X_i \leq \sqrt{n}\right) \\ = {} & \lim_{n \to \infty} \mathbb{P}\left(\frac{1}{\sqrt{n}} \sum_{i=1}^n (X_i - 1) \leq \sqrt{n} - \frac{n}{\sqrt{n}}\right) \\ = {} & \lim_{n \to \infty} \mathbb{P}\left(\frac{1}{\sqrt{n}} \sum_{i=1}^n (X_i - 1) \leq 0\right) \\ = {} & \Phi_{0,1}(0) = \frac{1}{2} \end{align*}

Law of large numbers

\begin{align*} & \lim_{n \to \infty} \mathbb{P}\left(\frac{1}{\sqrt{n}} \sum_{i=1}^n X_i \leq \sqrt{n}\right) \\ = {} & \lim_{n \to \infty} \mathbb{P}\left(\frac{1}{n} \sum_{i=1}^n X_i \leq 1 \right) \\ \geq {} & \lim_{n \to \infty} \mathbb{P}\left(\frac{1}{n} \sum_{i=1}^n X_i = 1\right) \\ = {} & 1 \end{align*} according to the strong law of large numbers. This then means that $$\lim_{n \to \infty} \mathbb{P}\left(\frac{1}{\sqrt{n}} \sum_{i=1}^n X_i \leq \sqrt{n}\right) = 1$$

What am I doing wrong here? My understanding is that the CLT solution is correct, but I don't see what I did wrong with applying the law of large numbers either.

If e.g. $X_1$ has continuous distribution then so has $\sum_{i=1}^nX_i$ for every $n$.
Consequence: $$\mathsf P(\frac1n\sum_{i=1}^nX_i=1)=0\text{ for every }n$$ so that also: $$\lim_{n\to\infty}\mathsf P(\frac1n\sum_{i=1}^nX_i=1)=0\neq1$$
• That law says here that $P(\lim_{n\to\infty}\frac1n\sum_{i=1}^nX_i=1)=1$. That is essentially different from $\lim_{n\to\infty}P(\frac1n\sum_{i=1}^nX_i=1)=1$. – drhab Jun 24 '18 at 13:07
What the (strong) law of large numbers says: \require{cancel} \begin{align} \text{right: } & \Pr\left( \lim_{n\to\infty} \frac 1 n \sum_{i=1}^n X_i = 1\right) = 1. \\ \\ \text{wrong: } & \xcancel{\lim_{n\to\infty} \Pr\left( \frac 1 n \sum_{i=1}^n X_i = 1\right) = 1.} \end{align}