Editing Central limit theorem (section)

===Calculating the variance===
Let {{mvar|S<sub>n</sub>}} be the sum of {{mvar|n}} random variables. Many central limit theorems provide conditions such that {{math|{{mvar|S<sub>n</sub>}}/{{sqrt|Var({{mvar|S<sub>n</sub>}})}}}} converges in distribution to {{math|''N''(0,1)}} (the normal distribution with mean 0, variance 1) as {{math|{{mvar|n}} → ∞}}. In some cases, it is possible to find a constant {{math|''σ''<sup>2</sup>}} and function {{mvar|f(n)}} such that {{math|{{mvar|S<sub>n</sub>}}/(σ{{sqrt|{{mvar|n⋅f}}({{mvar|n}})}})}} converges in distribution to {{math|''N''(0,1)}} as {{math|{{mvar|n}}→ ∞}}.

{{math theorem | name = Lemma<ref>{{cite journal|last1=Hew|first1=Patrick Chisan|title=Asymptotic distribution of rewards accumulated by alternating renewal processes|journal=Statistics and Probability Letters|date=2017|volume=129 |pages=355–359 |doi=10.1016/j.spl.2017.06.027}}</ref> | math_statement = Suppose <math>X_1, X_2, \dots</math> is a sequence of real-valued and strictly stationary random variables with <math>\operatorname E(X_i) = 0</math> for all {{nowrap|<math>i</math>,}} {{nowrap|<math>g : [0,1] \to \R</math>,}} and {{nowrap|<math>S_n = \sum_{i=1}^{n} g\left(\tfrac{i}{n}\right) X_i</math>.}} Construct

<math display="block">\sigma^2 = \operatorname E(X_1^2) + 2\sum_{i=1}^\infty \operatorname E(X_1 X_{1+i})</math>

# If <math>\sum_{i=1}^\infty \operatorname E(X_1 X_{1+i})</math> is absolutely convergent, <math>\left| \int_0^1 g(x)g'(x) \, dx\right| < \infty</math>, and <math>0 < \int_0^1 (g(x))^2 dx < \infty</math> then <math>\mathrm{Var}(S_n)/(n \gamma_n) \to \sigma^2</math> as <math>n \to \infty</math> where {{nowrap|<math>\gamma_n = \frac{1}{n}\sum_{i=1}^{n} \left(g\left(\tfrac{i}{n}\right)\right)^2</math>.}}
# If in addition <math>\sigma > 0</math> and <math>S_n/\sqrt{\mathrm{Var}(S_n)}</math> converges in distribution to <math>\mathcal{N}(0,1)</math> as <math>n \to \infty</math> then <math>S_n/(\sigma\sqrt{n \gamma_n})</math> also converges in distribution to <math>\mathcal{N}(0,1)</math> as {{nowrap|<math>n \to \infty</math>.}}
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