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I'm looking for a sequence of identically distributed random variables $(X_n)_n$ with $\text{var} (X_i) > 0$ so that the law of large numbers (either weak or strong) and the central limit theorem do not apply.

I propose: $\mathbb{P} [X_1 = 0] = \mathbb{P} [X_1 = 1] = \frac{1}{2}$; $X_i = X_1$ for $i > 1$. Thus, $(X_n)_n$ are identically distributed (in fact they are identical) but not independent. $\mu = \mathbb{E} [X_1] = \frac{1}{2}$, $\sigma^2 = \text{var} (X_1) = \frac{1}{4}$.

The weak law (and thus the strong law) doesn't hold for $(X_n)_n$ because $\mathbb{P} [|\frac{1}{n} \sum_{i = 1}^n X_i - \mu| > \varepsilon] = \mathbb{P} [|X_1 - \frac{1}{2}| > \varepsilon]$ for $\varepsilon < \frac{1}{2}$. ($\mathbb{P} [|X_1| = 1] = \mathbb{P} [X_1 = 1] = \frac{1}{2}$.)

The central limit theorem doesn't hold because, for $Z_n = \frac{\sum_{i = 1}^n X_i - n \mu}{\sigma \sqrt{n}} = \sqrt{n} (X_1 - 1)$, $\mathbb{P} [Z_n \le 1] = \mathbb{P} [X_1 \le 1 + \frac{1}{\sqrt{n}}] \underset{n \rightarrow \infty}{\longrightarrow} \mathbb{P} [X_1 \le 1] = 1 \ne \Phi (1)$.

I would appreciate any comment on whether this example is right. Is there any more straightforward example?

Andy
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    How about using the values $1$ and $-1$ instead of $0$ and $1$? – Angina Seng Apr 20 '18 at 18:58
  • Law of large numbers and central limit theorem usually require independence, so what is your point? – herb steinberg Apr 20 '18 at 19:39
  • Thanks for your comment, Lord Shark! That would actually simplify the law of large numbers part. However, in the central limit part, I end up with $Z_n = \sqrt{n} X_1$ and thus $\mathbb{P} [Z_n \le a] = \mathbb{P} [X_1 \le \frac{a}{\sqrt{n}}] \longrightarrow \mathbb{P} [X_1 \le 0] = \frac{1}{2} = \Phi (0)$. While the distributions are different, they are both symmetric, which I'm afraid prevents me from inferring the difference in this case. Good point, though! – Andy Apr 20 '18 at 19:52
  • @herb: That's perfectly right, the clt even requires i.i.d. variables. The problem is, however, stated so that I must provide an example with identically distributed variables, not i.i.d. variables. I'm allowed to use dependent variables.In fact, I think going for dependent variables is the most straightforward (if not only) way to solve the problem. My approach has thus been to search a simple example with dependent variables. – Andy Apr 20 '18 at 19:59
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    Andy - your example is the simplest possible. Almost any sequence of random variables where they are all the same would work. – herb steinberg Apr 21 '18 at 20:55
  • Thank you, herb! Lord Shark, I've also found a way how I can make the point with your example: If $\mathbb{P} [X_1 = 1] = \mathbb{P} [X_1 = -1]$ (and $X_i$ = $X_1$ for $i > 1$), $\sum_{i = 1}^n X_i$ can only take two values: $n$ and $-n$. The standard normal distribution is continuous. (I find this argument is at least as intuitive as mine, which uses the symmetry of the standard normal distribution, and it does simplify the first part of the proof.) – Andy Apr 28 '18 at 12:13

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