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I'm having difficulty with this old qualifying exam problem. Suppose we have a sequence of independent R.V's $\{X_n\}_{n\in\mathbb{N}}$ satisfying,

$$ \mathbb{P}(X_n = \pm n^2) = \frac{1}{12n^2}, \;\;\;\; \mathbb{P}(X_n = \pm n) = \frac{1}{12}, \;\;\;\; \mathbb{P}(X_n = 0) = 1 - \frac{1}{6} - \frac{1}{6n^2}$$

I managed to show that the Lindeberg condition does not hold. However, the problem states that the sequence $\frac{S_n}{b_n}$ still converges in distribution to a standard normal, where,

$$ S_n = \sum\limits_{k=1}^nX_k, \;\;\; b_n^2 = \frac{n(n+1)(2n+1)}{18} $$

Is there any cheeky way to show this? I tried using characteristic functions, but I feel like this method would be too inefficient on a timed exam.

EDIT: I consulted with the graduate director and he told me that there was a typo in the question. The question was actually taken from Chung's "A Course in Probability Theory" text (Section 7.2, Exercise 10). The correct problem states:

A CLT may well hold with a different sequence of constants $b_n$. Prove that Lindberg's condition is not satisfied. Nonetheless, if we take $b_n^2 = n^3/18$, then $S_n/b_n$ converges in dist. to the standard normal. The point is that abnormally large values may not count! Hint: Truncate out the abnormal value

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I think there is somthing off with the problem. Let's recall the following well-known statement:

Let $(X_n)_{n \in \mathbb{N}}$ be a sequence of real-valued independent random variables such that $\mathbb{E}(X_k)=0$ and $\sigma_k^2 = \mathbb{E}(X_k^2)>0$. Set $s_n^2 := \sum_{k=1}^n \sigma_k^2$. If $(X_n)_{n \in \mathbb{N}}$ satisfies the Feller condition $$\lim_{n \to \infty} \max_{1 \leq k \leq n} \frac{\sigma_k^2}{s_n^2} = 0, \tag{F}$$ then the following two statements are equivalent:

  1. $\frac{1}{s_n} (X_1+\ldots+X_n)$ converges in distribution to a standard Gaussian random variable.
  2. The Lindeberg condition holds: $$\forall \epsilon>0: \quad \lim_{n \to \infty} \frac{1}{s_n^2} \sum_{k=1}^n \int_{|X_k|>\epsilon s_k} |X_k|^2 \, d\mathbb{P}=0. \tag{L}$$

In the given example, we have

$$\sigma_k^2 = \mathbb{E}(X_k^2) = k^4 \mathbb{P}(|X_k| = k^2) + k^2 \mathbb{P}(|X_k| = k) = \frac{1}{3} k^2$$

and so

$$s_n^2 = \sum_{k=1}^n \sigma_k^2 = \frac{1}{18} n (n+1) (2n+1).$$

In particular, we find

$$\max_{k \leq n} \frac{\sigma_k^2}{s_n^2} \leq 6 \frac{n^2}{n (n+1)(2n+1)} \xrightarrow[]{n \to \infty} 0,$$

i.e. the Feller condition (F) holds. By the above theorem, this means that the CLT holds if, and only if, the Lindeberg condition (L) is satisfied. However, $s_k \approx k^{3/2}$ for large $k$, and therefore

$$\int_{|X_k|>\epsilon s_k} |X_k|^2 \, d\mathbb{P} = \int_{|X_k| = k^2} k^4 \, d\mathbb{P} = \frac{1}{6}k^2$$

for large $k$ implying that

$$\frac{1}{s_n^2} \sum_{k=1}^n \int_{|X_k|>\epsilon s_k} |X_k|^2 \, d\mathbb{P}\approx \frac{3}{n(n+1) (2n+1)} \sum_{k=1}^n k^2 = \frac{1}{18}$$

does not converge to $0$ as $n \to \infty$; this shows that (L) does not hold, and consequently (CLT) cannot hold true.

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  • $\begingroup$ Thank you. Apparently there was a problem with the question. The CLT holds for the sequence $n^3/18$ and not $n(n+1)(2n+1)/18$. $\endgroup$ – Flowsnake Jan 18 '18 at 22:00
  • $\begingroup$ If we replace $X_k$ by $\sigma X_k$, then wouldn't the $\sigma$ cancel out in the ratio $\frac{1}{s_n}\sum\limits_{k=1}^nX_k$? Then we would be back to square one. I'm still not sure why the ratio cannot converge to any $N(0, \sigma^2)$. $\endgroup$ – Flowsnake Jan 19 '18 at 20:55
  • $\begingroup$ @Flowsnake Ah you are right; it would cancel out. Something still feel off to me, but it might well be that there is an error in my logic. I'll think about it $\endgroup$ – saz Jan 19 '18 at 21:11
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Adding to what saz said, I found similar problems from a different approach. I took your idea of characteristic functions (although by simplicity reduced to moment generating functions). For each $X_n$ we have $$M_{X_n}(t)=\frac56-\frac1{6n^2}+(e^{nt}+e^{-nt})\frac1{12}+(e^{n^2t}+e^{-n^2t})\frac1{12n^2}.$$

Then for $S_n$, since the $X_1,\ldots,X_n$ are independent $$M_{S_n}(t)=\prod_{k=1}^n M_{X_k}(t)=\prod_{k=1}^n \left(\frac56-\frac1{6k^2}+(e^{kt}+e^{-kt})\frac1{12}+(e^{k^2t}+e^{-k^2t})\frac1{12k^2} \right).$$

Finally, if we define $Z_n=\tfrac{S_n}{b_n}$, then

$$M_{Z_n}(t)=M_{S_n}\left(\frac{t}{b_n}\right)=\prod_{k=1}^n \left(\frac56-\frac1{6k^2}+(e^{kt/b_n}+e^{-kt/b_n})\frac1{12}+(e^{k^2t/b_n}+e^{-k^2t/b_n})\frac1{12k^2} \right).$$

Now, since MGF of $Z_n$ are well defined in a neighborhood of $t=0$, it's a theorem that $$Z_n \to N(0,1) \, (\mathcal D) \quad \iff \quad M_{Z_n}(t) \to e^{\tfrac{t^2}2},$$ or more generally $$Z_n \to Y \, (\mathcal D) \quad \iff \quad M_{Z_n}(t) \to M_Y(t).$$

Since the expression of $Z_n$ is not even close to "friendly", I've estimated it's values and found out that (apparently) $$M_{Z_n}(t)\to \infty, \, t\neq 0.$$

This implies that $Z_n$ does not converge in distribution to a distribution with a well defined MGF, and certainly, that it does not converge to a normal distribution.

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  • $\begingroup$ Thank you. Apparently there was a problem with the question. The CLT holds for the sequence $n^3/18$ and not $n(n+1)(2n+1)/18$. $\endgroup$ – Flowsnake Jan 18 '18 at 22:01
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I believe I've gotten to a solution. Following the hint, define the truncated R.V's, $Y_j = X_jI\{|X_j| \leq j\}$. Then,

$$ \sum\limits_{j=1}^\infty\mathbb{P}(X_j \neq Y_j) = \sum\limits_{j=1}^\infty\mathbb{P}(X_j = \pm j^2) = \sum\limits_{j=1}^\infty\frac{1}{6j^2} < \infty $$

By Borel-Cantelli, $\mathbb{P}(X_j \neq Y_j \;\; i.o.) = 0$, i.e. $\{X_j \neq Y_j\}_{j\in\mathbb{N}}$ can only occur finitely many times. Thus, $S_n^* = \sum\limits_{j=1}^nY_j$ and $S_n = \sum\limits_{j=1}^nX_j$ have the same asymptotic distribution. One can show that the Lindeberg condition applied to $\{Y_j\}_{j\in\mathbb{N}}$ does hold. Thus,

$$ S_n/b_n \sim S_n^*/b_n \xrightarrow{n\rightarrow\infty} N(0,1) $$

where $b_n^2 = \sum\limits_{j=1}^n\mathbb{E}[Y_j^2] = \sum\limits_{j=1}^n\frac{j^2}{6} \sim \frac{n^3}{18}$. The notation $\sim$ is taken to mean "asymptotically equal".

EDIT: To justify why $S_n/b_n$ and $S_n^*/b_n$ have the same asymptotic distribution, I used the following results from Chung's "A Course in Probability Theory".

DEFINITION. Two sequences of R.V.'s $\{X_n\}$ and $\{Y_n\}$ are said to be equivalent iff $$\sum\limits_n\mathbb{P}(X_n \neq Y_n) < \infty$$ Theorem. If $\{X_n\}$ and $\{Y_n\}$ are equivalent, then, $$ \sum\limits_{n}(X_n - Y_n) $$ converges almosut surely. Moreover if $a_n$ is a monotonic sequence diverging to $+\infty$, then, $$ \frac{1}{a_n}\sum\limits_{j=1}^n(X_j - Y_j) \xrightarrow{n\rightarrow\infty} 0 $$ almost surely.

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  • $\begingroup$ How exactly do you justify that $$S_n/b_n \to N(0,1) \iff S_n^*/b_n \to N(0,1)$$ ...? $\endgroup$ – saz Jan 19 '18 at 7:58
  • $\begingroup$ Please see my edit. I referred to the textbook that the problem was taken from. $\endgroup$ – Flowsnake Jan 19 '18 at 20:52
  • $\begingroup$ I'm well aware of the result about the equivalence of sequences. What was bothering me is that you are mixing almost sure convergence and convergence in distribution... but after I have thought once more about it, I think that your reasoning is fine. $\endgroup$ – saz Jan 20 '18 at 9:49

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