This is essentially a replication of a question I found over at math.se, which didn't get the answers I hoped for.
Let $\{ X_i \}_{i \in \mathbb{N}}$ be a sequence of independent, identically distributed random variables, with $\mathbb{E}[X_i] = 1$ and $\mathbb{V}[X_i] = 1$.
Consider the evaluation of
$$
\lim_{n \to \infty} \mathbb{P}\left(\frac{1}{\sqrt{n}} \sum_{i=1}^n X_i \leq \sqrt{n}\right)
$$
This expression has to be manipulated since, as is, both sides of the inequality event tend to infinity.
A) TRY SUBTRACTION
Before considering the limiting statement, subtract $\sqrt{n}$ from both sides:
$$\lim_{n \to \infty} \mathbb{P}\left(\frac{1}{\sqrt{n}} \sum_{i=1}^n X_i -\sqrt{n} \leq \sqrt{n}-\sqrt{n} \right) = \lim_{n \to \infty} \mathbb{P}\left(\frac{1}{\sqrt{n}} \sum_{i=1}^n (X_i – 1) \leq 0\right) \\
= \Phi(0) = \frac{1}{2}$$
the last equality by the CLT, where $\Phi()$ is the standard normal distribution function.
B) TRY MULTIPLICATION
Multiply both sides by $1/\sqrt{n}$
$$\lim_{n \to \infty} \mathbb{P}\left(\frac {1}{\sqrt{n}}\cdot \frac{1}{\sqrt{n}} \sum_{i=1}^n X_i \leq \frac {1}{\sqrt{n}}\cdot\sqrt{n} \right) = \lim_{n \to \infty} \mathbb{P}\left(\frac{1}{n} \sum_{i=1}^n X_i \leq 1\right) $$
$$= \lim_{n \to \infty} \mathbb{P}\left(\bar X_n \leq 1\right) = \lim_{n \to \infty}F_{\bar X_n}(1) = 1$$
where $F_{\bar X_n}()$ is the distribution function of the sample mean $\bar X_n$, which by the LLN converges in probability (and so also in distribution) to the constant $1$, hence the last equality.
So we get conflicting results.
Which is the right one? And why the other is wrong?
Best Answer
The error here is likely in the following fact: convergence in distribution implicitly assumes that $F_n(x)$ converges to $F(x)$ at points of continuity of $F(x)$. As the limit distribution is of a constant random variable, it has a jump discontinuity at $x=1$, hence it is incorrect to conclude that the CDF converges to $F(x)=1$.