reformat display math, add mystnb figure tags

Author: mmcky

lectures/heavy_tails.md

@@ -61,10 +61,10 @@ To explain this concept, let's look first at examples.
 The classic example is the [normal distribution](https://en.wikipedia.org/wiki/Normal_distribution), which has density
 
 $$ 
-    f(x) = \frac{1}{\sqrt{2\pi}\sigma} 
-    \exp\left( -\frac{(x-\mu)^2}{2 \sigma^2} \right)
-    \qquad
-    (-\infty < x < \infty)
+f(x) = \frac{1}{\sqrt{2\pi}\sigma} 
+\exp\left( -\frac{(x-\mu)^2}{2 \sigma^2} \right)
+\qquad
+(-\infty < x < \infty)
 $$
 
 
@@ -78,6 +78,12 @@ We can see this when we plot the density and show a histogram of observations,
 as with the following code (which assumes $\mu=0$ and $\sigma=1$).
 
 ```{code-cell} ipython3
+---
+mystnb:
+  figure:
+    caption: Histogram of observations
+    name: hist-obs
+---
 fig, ax = plt.subplots()
 X = norm.rvs(size=1_000_000)
 ax.hist(X, bins=40, alpha=0.4, label='histogram', density=True)
@@ -101,6 +107,12 @@ X.min(), X.max()
 Here's another view of draws from the same distribution:
 
 ```{code-cell} ipython3
+---
+mystnb:
+  figure:
+    caption: Histogram of observations
+    name: hist-obs2
+---
 n = 2000
 fig, ax = plt.subplots()
 data = norm.rvs(size=n)
@@ -174,6 +186,12 @@ data = yf.download('AMZN', '2015-1-1', '2022-7-1')
 ```
 
 ```{code-cell} ipython3
+---
+mystnb:
+  figure:
+    caption: Daily Amazon returns
+    name: dailyreturns-amzn
+---
 s = data['Adj Close']
 r = s.pct_change()
 
@@ -194,7 +212,18 @@ Several of observations are quite extreme.
 We get a similar picture if we look at other assets, such as Bitcoin
 
 ```{code-cell} ipython3
-s = yf.download('BTC-USD', '2015-1-1', '2022-7-1')['Adj Close']
+:tags: [hide-output]
+data = yf.download('BTC-USD', '2015-1-1', '2022-7-1')
+```
+
+```{code-cell} ipython3
+---
+mystnb:
+  figure:
+    caption: Daily Bitcoin returns
+    name: dailyreturns-btc
+---
+s = data['Adj Close']
 r = s.pct_change()
 
 fig, ax = plt.subplots()
@@ -211,6 +240,12 @@ The histogram also looks different to the histogram of the normal
 distribution:
 
 ```{code-cell} ipython3
+---
+mystnb:
+  figure:
+    caption: Histogram (Normal vs Bitcoin returns)
+    name: hist-normal-btc
+---
 r = np.random.standard_t(df=5, size=1000)
 
 fig, ax = plt.subplots()
@@ -274,10 +309,6 @@ like
 We return to these points [below](https://intro.quantecon.org/heavy_tails.html#why-do-heavy-tails-matter).
 
 
-
-
-
-
 ## Visual comparisons
 In this section, we will introduce important concepts such as the Pareto distribution, Counter CDFs, and Power laws, which aid in recognizing heavy-tailed distributions.
 
@@ -300,6 +331,12 @@ distribution](https://en.wikipedia.org/wiki/Cauchy_distribution), which is
 heavy-tailed.
 
 ```{code-cell} ipython3
+---
+mystnb:
+  figure:
+    caption: Histogram of Cauchy distribution
+    name: hist-cauchy
+---
 n = 120
 np.random.seed(11)
 
@@ -353,6 +390,12 @@ The exponential distribution is a light-tailed distribution.
 Here are some draws from the exponential distribution.
 
 ```{code-cell} ipython3
+---
+mystnb:
+  figure:
+    caption: Histogram of Exponential distribution
+    name: hist-exponential
+---
 n = 120
 np.random.seed(11)
 
@@ -394,14 +437,22 @@ exponential random variable.
 
 In particular, if $X$ is exponentially distributed with rate parameter $\alpha$, then
 
-$$   Y = \bar x \exp(X) $$
+$$
+Y = \bar x \exp(X) 
+$$
 
 is Pareto-distributed with minimum $\bar x$ and tail index $\alpha$. 
 
 Here are some draws from the Pareto distribution with tail index $1$ and minimum
 $1$.
 
 ```{code-cell} ipython3
+---
+mystnb:
+  figure:
+    caption: Histogram of Pareto distribution
+    name: hist-pareto
+---
 n = 120
 np.random.seed(11)
 
@@ -425,21 +476,27 @@ light and heavy tails is to look at the
 
 For a random variable $X$ with CDF $F$, the CCDF is the function 
 
-$$ G(x) := 1 - F(x) = \mathbb P\{X > x\} $$
+$$
+G(x) := 1 - F(x) = \mathbb P\{X > x\} 
+$$
 
 (Some authors call $G$ the "survival" function.)
 
 The CCDF shows how fast the upper tail goes to zero as $x \to \infty$.
 
 If $X$ is exponentially distributed with rate parameter $\alpha$, then the CCDF is
 
-$$ G_E(x) = \exp(- \alpha x) $$
+$$
+G_E(x) = \exp(- \alpha x)
+$$
 
 This function goes to zero relatively quickly as $x$ gets large.
 
 The standard Pareto distribution, where $\bar x = 1$, has CCDF
 
-$$ G_P(x) = x^{- \alpha} $$
+$$
+G_P(x) = x^{- \alpha}
+$$
 
 This function goes to zero as $x \to \infty$, but much slower than $G_E$.
 
@@ -505,13 +562,21 @@ The sample counterpart of the CCDF function is the **empirical CCDF**.
 
 Given a sample $x_1, \ldots, x_n$, the empirical CCDF is given by
 
-$$ \hat G(x) = \frac{1}{n} \sum_{i=1}^n \mathbb 1\{x_i > x\} $$
+$$
+\hat G(x) = \frac{1}{n} \sum_{i=1}^n \mathbb 1\{x_i > x\}
+$$
 
 Thus, $\hat G(x)$ shows the fraction of the sample that exceeds $x$.
 
 Here's a figure containing some empirical CCDFs from simulated data.
 
 ```{code-cell} ipython3
+---
+mystnb:
+  figure:
+    caption: Empirical CCDFs
+    name: ccdf-empirics
+---
 def eccdf(x, data):
     "Simple empirical CCDF function."
     return np.mean(data > x)
@@ -690,7 +755,13 @@ def extract_wb(varlist=['NY.GDP.MKTP.CD'],
 Here is a plot of the firm size distribution for the largest 500 firms in 2020 taken from Forbes Global 2000.
 
 ```{code-cell} ipython3
-:tags: [hide-input]
+---
+tags: [hide-input]
+mystnb:
+  figure:
+    caption: Firm size distribution
+    name: firm-size-dist
+--- 
 
 df_fs = pd.read_csv('https://media.githubusercontent.com/media/QuantEcon/high_dim_data/main/cross_section/forbes-global2000.csv')
 df_fs = df_fs[['Country', 'Sales', 'Profits', 'Assets', 'Market Value']]
@@ -711,7 +782,13 @@ Here are plots of the city size distribution for the US and Brazil in 2023 from
 The size is measured by population.
 
 ```{code-cell} ipython3
-:tags: [hide-input]
+---
+tags: [hide-input]
+mystnb:
+  figure:
+    caption: City size distribution
+    name: city-size-dist
+--- 
 
 # import population data of cities in 2023 United States and 2023 Brazil from world population review
 df_cs_us = pd.read_csv('https://media.githubusercontent.com/media/QuantEcon/high_dim_data/main/cross_section/cities_us.csv')
@@ -732,7 +809,13 @@ Here is a plot of the upper tail (top 500) of the wealth distribution.
 The data is from the Forbes Billionaires list in 2020.
 
 ```{code-cell} ipython3
-:tags: [hide-input]
+---
+tags: [hide-input]
+mystnb:
+  figure:
+    caption: Wealth distribution (Forbes Billionaires in 2020)
+    name: wealth-dist
+--- 
 
 df_w = pd.read_csv('https://media.githubusercontent.com/media/QuantEcon/high_dim_data/main/cross_section/forbes-billionaires.csv')
 df_w = df_w[['country', 'realTimeWorth', 'realTimeRank']].dropna()
@@ -782,7 +865,13 @@ df_gdp1.dropna(inplace=True)
 ```
 
 ```{code-cell} ipython3
-:tags: [hide-input]
+---
+tags: [hide-input]
+mystnb:
+  figure:
+    caption: GDP per capita distribution
+    name: gdppc-dist
+--- 
 
 fig, axes = plt.subplots(1, 2, figsize=(8.8, 3.6))
 
@@ -828,6 +917,12 @@ Let's have a look at the behavior of the sample mean in this case, and see
 whether or not the LLN is still valid.
 
 ```{code-cell} ipython3
+---
+mystnb:
+  figure:
+    caption: LLN failure
+    name: fail-lln
+--- 
 from scipy.stats import cauchy
 
 np.random.seed(1234)
@@ -887,7 +982,9 @@ portfolio is $\mu$ and the variance is $\sigma^2$.
 
 If instead the investor puts share $1/n$ of her wealth in each asset, then the portfolio payoff is
 
-$$ Y_n = \sum_{i=1}^n \frac{X_i}{n} = \frac{1}{n} \sum_{i=1}^n X_i. $$  
+$$
+Y_n = \sum_{i=1}^n \frac{X_i}{n} = \frac{1}{n} \sum_{i=1}^n X_i. 
+$$  
 
 Try computing the mean and variance.
 
@@ -918,8 +1015,6 @@ For example, the heaviness of the tail of the income distribution helps
 determine {doc}`how much revenue a given tax policy will raise <mle>`.
 
 
-
-
 (cltail)=
 ## Classifying tail properties
 
@@ -964,7 +1059,9 @@ For example, every random variable with bounded support is light-tailed. (Why?)
 
 As another example, if $X$ has the [exponential distribution](https://en.wikipedia.org/wiki/Exponential_distribution), with cdf $F(x) = 1 - \exp(-\lambda x)$ for some $\lambda > 0$, then its moment generating function is 
 
-$$ m(t) = \frac{\lambda}{\lambda - t} \quad \text{when } t < \lambda $$
+$$
+m(t) = \frac{\lambda}{\lambda - t} \quad \text{when } t < \lambda 
+$$
 
 In particular, $m(t)$ is finite whenever $t < \lambda$, so $X$ is light-tailed.
 
@@ -1023,7 +1120,7 @@ $$
 But then
 
 $$
-    \mathbb E X^r = r \int_0^\infty x^{r-1} \mathbb P\{ X > x \} dx
+\mathbb E X^r = r \int_0^\infty x^{r-1} \mathbb P\{ X > x \} dx
 \geq
 r \int_0^{\bar x} x^{r-1} \mathbb P\{ X > x \} dx
 + r \int_{\bar x}^\infty  x^{r-1} b x^{-\alpha} dx.
@@ -1254,7 +1351,7 @@ assumption leads to a lower mean and greater dispersion.
 The [characteristic function](https://en.wikipedia.org/wiki/Characteristic_function_%28probability_theory%29) of the Cauchy distribution is
 
 $$
-    \phi(t) = \mathbb E e^{itX} = \int e^{i t x} f(x) dx = e^{-|t|}
+\phi(t) = \mathbb E e^{itX} = \int e^{i t x} f(x) dx = e^{-|t|}
 $$ (lln_cch)
 
 Prove that the sample mean $\bar X_n$ of $n$ independent draws $X_1, \ldots,