# Does central limit theorem still hold if bounds change?

Let $W_1, W_2, \ldots$ be iid random variables, $S_k = \sum_{i=1}^k W_k$ and $Z\sim N(0,1)$.

Given sequences $\{a_k\}_{k=1}^\infty$ and $\{b_k\}_{k=1}^\infty$ s.t.

$$\lim_{k \to \infty} a_k = a, \lim_{k \to \infty} b_k = b,$$

where $a, b \in [-\infty,\infty]$,

is it true that

$$\lim_{k \to \infty} \mathbb{P}\left( a_k \le \frac{(W_1 + \cdots + W_k)-E[S_k]}{\operatorname{Var}[S_k]} \le b_k \right)$$

$$= \mathbb{P}\left( \lim_{k \to \infty} a_k \le Z \le \lim_{k \to \infty} b_k \right) \text{ ?}$$

According to Larsen and Marx, the central limit theorem states that: If needed (but please state which):

1. assume finite mean, finite variance, integrability, boundedness or whatever

2. assume $a_k, a < 0$.

3. assume $b_k, b > 0$.

It doesn't look like continuity of probability will help.

I tried defining

$$A_k := \left( a_k \le \frac{(W_1 + \cdots + W_k)- \operatorname{E}[S_k]}{\operatorname{Var}[S_k]} \le b_k \right)$$

and

$$A := \left( \lim_{k \to \infty} a_k \le Z \le \lim_{k \to \infty} b_k \right)$$

If the bounds or random variable didn't depend on $k$ or if the random, I think one could apply continuity of probability. I'm not quite sure if

$$\bigcap_{k=1}^{\infty} A_k \text{ or } \bigcup_{k=1}^{\infty} A_k = A$$

• I changed $Z$ ~ $N(0,1)$ to $Z\sim N(0,1)$ and $$\bigcap A_k or \bigcup A_k$$ to $$\bigcap A_k \text{ or } \bigcup A_k$$ and added standard use of \ldots and \cdots and a few other things. ${}\qquad{}$ – Michael Hardy Nov 22 '15 at 20:34
• Thanks @MichaelHardy ^-^ – BCLC Nov 22 '15 at 20:35

This looks to me like a standard limit problem. What I would do is to consider, for each $k$, the difference between the integrals between $a_k$ to $b_k$ and between $a$ and $b$.

For any $\epsilon>0$, the set of $k$ for which the difference exceeds $\epsilon$ is bounded, so the sum for those $k$ can be made arbitrarily small, and the rest of the terms are nice.

• Thanks marty cohen, but what do you mean by 'the sum for those k' ? How do you know such set is bounded? I tried something. Edited OP. Pls check it out – BCLC Nov 22 '15 at 21:47
• For any fixed $\epsilon$, the set of $a_k$ and $b_k$ for which the difference exceeds $\epsilon$ is finite; call it $m$. Then consider the differences for $a_k$ and $b_k$ for $m < k < n$, and let $n$ get large. The terms with $k < m$ will become insignificant and the terms with $m < k < n$, which all have error $< \epsilon$ will dominate, so the overall error will be within, say, $2\epsilon$. Then let $\epsilon \to 0$. As I said, this is a standard technique, where there is a "bad" initial part and a "good" final part which swamps the "bad" part. – marty cohen Nov 23 '15 at 0:27
• Soooo you infer the boundedness of $k_{\epsilon}$ from the fact that 'the set...is finite' ? If so that fact is based on the limits of the sequences existing? Also what exactly is m? Cardinality? Supremum? – BCLC Nov 23 '15 at 2:40
• $m$ is $m(\epsilon)$: if $k > m$ then $|a_k-a|<\epsilon$ and same for $b$. – marty cohen Nov 23 '15 at 4:40
• I am not sure I follow. Is my proof right? I am primarily wondering about the boundedness. Is it related to the fact that a_k and b_k converge? – BCLC Nov 23 '15 at 5:34

Let us show that $\forall \epsilon > 0, \exists K > 0$ s.t.

$$k > K \to |c_k - c| < \epsilon$$

where

$c_k := P(Z_k \in (a_k, b_k))$

$c := P(Z \in (a, b))$

$\forall \epsilon > 0$, $k_{\epsilon}:= \{ \text{all k s.t.} |c_k - c| \ge \epsilon \}$ is bounded (I think). Thus, $\exists M_{\epsilon} > 0$ s.t.

$$|c_{M_{\epsilon}} - c| \ge \epsilon$$

but

$$|c_{M_{\epsilon} + 1} - c| < \epsilon$$

Choose $K = M_{\epsilon}$.