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 SUBTRACTIONBefore 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 MULTIPLICATIONMultiply 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?

**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 $F(x)=1$.

**Attribution***Source : Link , Question Author : Alecos Papadopoulos , Answer Author : Richard Hardy*