Let $\{X_n\}_{n=1}^\infty$ be an i.i.d sequence with mean $0$ and variance $1$. The CLT tells us that

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

where $\Phi(x), x \in \mathbb{R}$ denotes the probability distribution function of a standard normal variable.

Is it true that $$\lim_{n \to \infty} \mathbb{P}\left\{\frac{\sum_{i=1}^nX_i}{\sqrt{n}} < x\right\} = \Phi(x)$$

as well? (Note that I replaced the inequality by strict inequality).

Thanks in advance.


Yes, because $\Phi$ is continuous. The set $B=(-\infty,x)$ is a continuity set for the Gaussian distribution so $P( S_n/\sqrt n \in B)$ tends to the Gaussian probability of $B$, that is, to $\Phi(x)$. $B$ is a continuity set because its boundary $\partial B=\{x\}$ has Gaussian probability $0$. In other words, because $\Phi$ is continuous at $x$.

Read all about it at the wikipedia articles on the Portmanteau theorem and on continuity sets.

  • $\begingroup$ Thanks! I'll check these things out! $\endgroup$ – user661541 Nov 8 at 15:06

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