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We can generate a draw of $X$ with `scipy.stats` (imported as `st`) as follows:
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@@ -369,7 +373,8 @@ The LLN fails to hold here because the assumption $\mathbb E|X| < \infty$ is vio
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The LLN can also fail to hold when the IID assumption is violated.
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For example, suppose that
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```{prf:example}
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:label: lln_ex_fail
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$$
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X_0 \sim N(0,1)
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$$
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Therefore, the distribution of $\bar X_n$ is $N(0,1)$ for all $n$!
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```
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Does this contradict the LLN, which says that the distribution of $\bar X_n$
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collapses to the single point $\mu$?
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Here $\stackrel { d } {\to} N(0, \sigma^2)$ indicates [convergence in distribution](https://en.wikipedia.org/wiki/Convergence_of_random_variables#Convergence_in_distribution) to a centered (i.e., zero mean) normal with standard deviation $\sigma$.
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The striking implication of the CLT is that for **any** distribution with
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The striking implication of the CLT is that for any distribution with
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finite [second moment](https://en.wikipedia.org/wiki/Moment_(mathematics)), the simple operation of adding independent
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copies **always** leads to a Gaussian(Normal) curve.
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copies always leads to a Gaussian(Normal) curve.
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@@ -599,7 +605,7 @@ $$
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$$
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where $\alpha, \beta, \sigma$ are constants and $\epsilon_1, \epsilon_2,
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