2
$\begingroup$

I have a doubt about the central limit theorem.

Let $(X_n)$ be an IID sequence, each $X_n$ distributed as $X$ where

$\mathbb{E}[X]=0$, $\sigma^2 = Var(X) < \infty$.

Define $S_n = X_1 + \dots + X_n$, and set

$G_n = \frac{S_n}{\sigma\sqrt{n}}$

Now, the theorem tells us that

$\lim_{n \to \infty}\mathbb{P}(G_n \le x) = \Phi(x)$ where $\Phi(x)$ is the distribution function of the standard normal distribution.

Now is it FORMALLY correct to say, for example, that

$\lim_{n \to \infty}\mathbb{P}(G_n \le 1+\frac{1}{n}) = \Phi(\lim_{n \to \infty}(1+\frac{1}{n})) = \Phi(1) \simeq 0.8413$?

Thanks to who will solve my (perhaps trivial) doubt.

$\endgroup$
5
  • $\begingroup$ No: $x$ is a number, not a function of $n$. $\endgroup$ Sep 1, 2017 at 20:15
  • $\begingroup$ In this case, how can I estimate that probability? $\endgroup$
    – LJG
    Sep 1, 2017 at 20:17
  • $\begingroup$ The CLT is merely telling you how to compute $\lim_{n \to \infty} P(G_n \leq x)$ for fixed $x$ not depending on $n$. You might hope by some interchange of limits result or similar that you can replace $x$ by a sequence $x_n$ converging to $x$ but that's not the same result. $\endgroup$
    – Ian
    Sep 1, 2017 at 20:18
  • $\begingroup$ @LordSharktheUnknown Do you mean "not a function of $n$" instead - $x$ certainly is a function of $x$ ;) $\endgroup$
    – B. Mehta
    Sep 1, 2017 at 20:19
  • $\begingroup$ LJG: You accepted instantly the answer below. As a consequence your question basically disappeared from the list of those to answer, which made that you are left with the unique answer below. Too bad... $\endgroup$
    – Did
    Sep 3, 2017 at 11:10

1 Answer 1

2
$\begingroup$

No, it is not formally correct because $1+1/n$ is a sequence of numbers which depends on $n$, whereas the CLT as stated only holds for fixed real numbers $x$.

However, it is still true that $\lim_{n \to \infty} P(G_n \leq 1+1/n) = \Phi(1)$. In order to prove this, one only needs to use the fact that the functions $F_n(x):=P(G_n \leq x)$ are increasing, together with continuity of the limit $\Phi$. Indeed, for $N \leq n$, one has that $F_n(1+1/n) \leq F_n(1+1/N)$, and thus for any fixed $N$ $$\limsup_{n\to \infty}F_n(1+1/n) \leq \limsup_{n \to \infty} F_n(1+1/N) = \Phi(1+1/N)$$ Next, we may let $N \to \infty$ on both sides, and since the LHS does not depend on $N$, one sees by continuity of $\Phi$ that $$\limsup_{n \to \infty}F_n(1+1/n) \leq \lim_{N\to \infty} \Phi(1+1/N) = \Phi(1)$$ On the other hand, one clearly has that $F_n(1+1/n) \geq F_n(1)$ fo every $n$, and therefore we see that $$\liminf_{N \to \infty} F_n(1+1/n) \geq \liminf_{n \to \infty} F_n(1) = \Phi(1)$$ which (together with the preceding expression) gives the desired result.

In fact, because of this question, it is actually true that one has uniform convergence of the cdf's $F_n$ to $\Phi$, which means that $F_n(x_n) \to \Phi(x)$ whenever $x_n \to x$ (and the proof uses the same ideas).

$\endgroup$
12
  • $\begingroup$ Thank you for the reply. Could you give me some advice to estimate in a formal way that probability? $\endgroup$
    – LJG
    Sep 1, 2017 at 20:33
  • $\begingroup$ We did not study Berry–Esseen theorem. Anyway, for the theorem you talked about before, can I write safely that $\lim_{n \to \infty}\mathbb{P}(G_n \le 1+\frac{1}{n}) = \Phi(1)$? $\endgroup$
    – LJG
    Sep 1, 2017 at 20:50
  • $\begingroup$ @LJG Yes, see the linked question $\endgroup$
    – shalop
    Sep 1, 2017 at 20:51
  • $\begingroup$ Excellent. Thank you very much. $\endgroup$
    – LJG
    Sep 1, 2017 at 20:53
  • 1
    $\begingroup$ @Shalop I've seen the changes and thank you for the further clarification. $\endgroup$
    – LJG
    Sep 3, 2017 at 7:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .