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Let us suppose we have an infinite number if Bernoulli variables that can take one of two values: $-1,1$, with equal probability. Then their sum converges to a normal probability distribution.

What about variables that can take three values, $-1,0,1$ with equal probability? Does the probability distribution of their sum also converge to the normal distribution?

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  • $\begingroup$ Bernoulli variables are $0$ or $1$ but this does not affect your question $\endgroup$
    – Henry
    Commented Sep 26, 2023 at 0:37

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The short answer is yes.

Firstly, you said when you have Bernoulli random variables that take values $-1, 1,$ then their sum converges to a normal distribution (presumably you mean by the Central Limit Theorem). This is slightly false; actually, the normalized sum will converge. In particular, if your variables are $X_1, X_2, \dots,$ then $$\frac{1}{\sqrt{n}} \sum_{i=1}^n X_i \stackrel{d}{\to} \mathcal{N}(0, \text{Var}(X)).$$

The Central Limit Theorem in general states that if you have a sequence of independent random variables $X_1, X_2, \dots$ that have the same distribution, so that $\sigma^2 := \text{Var}(X_i) < \infty$ and $\mu := \mathbb{E}[X_i]$ is the expected value, then $$\frac{1}{\sqrt{n}} \sum_{i=1}^n (X_i - \mu) \stackrel{d}{\to} \mathcal{N}(0, \sigma^2).$$

So indeed, this still works for the "3-valued Bernoulli variables" you mentioned. To know the variance of the normal distribution that the normalized sum converges to, you just need to compute the variables of your "3-valued Bernoulli variables."

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