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spell check and example admonition (#536)
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longye-tian authored Jul 25, 2024
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Expand Up @@ -51,6 +51,9 @@ will converge to their population means.

Let's see an example of the LLN in action before we go further.

```{prf:example}
:label: lln_ex_ber
Consider a [Bernoulli random variable](https://en.wikipedia.org/wiki/Bernoulli_distribution) $X$ with parameter $p$.
This means that $X$ takes values in $\{0,1\}$ and $\mathbb P\{X=1\} = p$.
Expand All @@ -68,6 +71,7 @@ $$
\mathbb E X
= 0 \cdot \mathbb P\{X=0\} + 1 \cdot \mathbb P\{X=1\} = \mathbb P\{X=1\} = p
$$
```

We can generate a draw of $X$ with `scipy.stats` (imported as `st`) as follows:

Expand Down Expand Up @@ -369,7 +373,8 @@ The LLN fails to hold here because the assumption $\mathbb E|X| < \infty$ is vio

The LLN can also fail to hold when the IID assumption is violated.

For example, suppose that
```{prf:example}
:label: lln_ex_fail
$$
X_0 \sim N(0,1)
Expand All @@ -384,6 +389,7 @@ $$
$$
Therefore, the distribution of $\bar X_n$ is $N(0,1)$ for all $n$!
```

Does this contradict the LLN, which says that the distribution of $\bar X_n$
collapses to the single point $\mu$?
Expand Down Expand Up @@ -439,9 +445,9 @@ n \to \infty
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 $\sigma$.


The striking implication of the CLT is that for **any** distribution with
The striking implication of the CLT is that for any distribution with
finite [second moment](https://en.wikipedia.org/wiki/Moment_(mathematics)), the simple operation of adding independent
copies **always** leads to a Gaussian(Normal) curve.
copies always leads to a Gaussian(Normal) curve.



Expand Down Expand Up @@ -599,7 +605,7 @@ $$
$$
where $\alpha, \beta, \sigma$ are constants and $\epsilon_1, \epsilon_2,
\ldots$ is IID and standard norma.
\ldots$ are IID and standard normal.
Suppose that
Expand Down

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