Quantiles of the standard Normal#
Quantiles are a great way to summarize a random variable with a few numbers. Let’s start with the standard Normal. Take:
The definition of is this:
The \(q\) quantile of \(Z\) is the value \(z_q\) such that the probability of \(Z\) being less that \(z_q\) is \(q\).
Mathematically, you want to find a value \(z_q\)
The median of the standard Normal#
For example, the \(0.5\) quantile \(z_{0.5}\) satisfies the property:
This is known as the median of \(Z\). In words, 50% of the probability of \(Z\) to the left of the median. For the standard normal, we have because of the symmetry of the PDF about zero that:
Of course, scipy.stats knows about the median:
Show code cell source
MAKE_BOOK_FIGURES=False
import matplotlib as mpl
import matplotlib.pyplot as plt
%matplotlib inline
import matplotlib_inline
matplotlib_inline.backend_inline.set_matplotlib_formats('svg')
import seaborn as sns
sns.set_context("paper")
sns.set_style("ticks")
def set_book_style():
plt.style.use('seaborn-v0_8-white')
sns.set_style("ticks")
sns.set_palette("deep")
mpl.rcParams.update({
# Font settings
'font.family': 'serif', # For academic publishing
'font.size': 8, # As requested, 10pt font
'axes.labelsize': 8,
'axes.titlesize': 8,
'xtick.labelsize': 7, # Slightly smaller for better readability
'ytick.labelsize': 7,
'legend.fontsize': 7,
# Line and marker settings for consistency
'axes.linewidth': 0.5,
'grid.linewidth': 0.5,
'lines.linewidth': 1.0,
'lines.markersize': 4,
# Layout to prevent clipped labels
'figure.constrained_layout.use': True,
# Default DPI (will override when saving)
'figure.dpi': 600,
'savefig.dpi': 600,
# Despine - remove top and right spines
'axes.spines.top': False,
'axes.spines.right': False,
# Remove legend frame
'legend.frameon': False,
# Additional trim settings
'figure.autolayout': True, # Alternative to constrained_layout
'savefig.bbox': 'tight', # Trim when saving
'savefig.pad_inches': 0.1 # Small padding to ensure nothing gets cut off
})
def save_for_book(fig, filename, is_vector=True, **kwargs):
"""
Save a figure with book-optimized settings.
Parameters:
-----------
fig : matplotlib figure
The figure to save
filename : str
Filename without extension
is_vector : bool
If True, saves as vector at 1000 dpi. If False, saves as raster at 600 dpi.
**kwargs : dict
Additional kwargs to pass to savefig
"""
# Set appropriate DPI and format based on figure type
if is_vector:
dpi = 1000
ext = '.pdf'
else:
dpi = 600
ext = '.tif'
# Save the figure with book settings
fig.savefig(f"{filename}{ext}", dpi=dpi, **kwargs)
def make_full_width_fig():
return plt.subplots(figsize=(4.7, 2.9), constrained_layout=True)
def make_half_width_fig():
return plt.subplots(figsize=(2.35, 1.45), constrained_layout=True)
if MAKE_BOOK_FIGURES:
set_book_style()
make_full_width_fig = make_full_width_fig if MAKE_BOOK_FIGURES else lambda: plt.subplots()
make_half_width_fig = make_half_width_fig if MAKE_BOOK_FIGURES else lambda: plt.subplots()
import numpy as np
import scipy.stats as st
Z = st.norm()
Z.median()
0.0
Other quantiles of the standard Normal#
Another interesting quantile is \(z_{0.025}\). So, \(z_{0.025}\) marks the point below which \(Z\) lies with probability \(2.5\)%. This is not trivial to find though. You really need to solve the nonlinear equation:
But scipy.stats can do this for you using the function Z.ppf():
z_025 = Z.ppf(0.025)
print(f'z_025 = {z_025:1.2f}')
z_025 = -1.96
Let’s verify that this is indeed giving me the \(0.025\) quantile. If I plug it in the CDF I should get \(0.025\):
print(f'Phi(z_025) = {Z.cdf(z_025):1.3f}')
Phi(z_025) = 0.025
Okay, it looks good!
Let’s also find \(z_{0.975}\):
z_975 = Z.ppf(0.975)
print(f'z_975 = {z_975:1.2f}')
z_975 = 1.96
Nice! This is just \(-z_{0.025}\). We could have guessed it!
Credible intervals#
Alright, these two quantiles are particularly important. Why, because the probability that \(Z\) is between them is 95%! We we could write:
\(Z\) is between -1.96 and +1.96 with probability 95%.
This is a very nice summary of the uncertainty in \(Z\). This is called the 95% (central) credible interval of \(Z\).
Now if you are like me, you would simplify this a bit more by writing:
\(Z\) is between -2 and +2 with probability (approximately) 95%.
Who wants to remember that 1.96…
Let’s visualize the 95% (central) credible interval by shaded the PDF:
Show code cell source
fig, ax = make_full_width_fig()
zs = np.linspace(-6.0, 6.0, 200)
Phis = Z.pdf(zs)
ax.plot(zs, Phis)
idx = (zs >= -2) & (zs <= 2)
ax.fill_between(zs[idx], 0.0, Phis[idx], color='r', alpha=0.5, label='95% credible interval')
ax.set_xlabel('$z$')
ax.set_ylabel(r'$\phi(z)$')
plt.legend(loc='best')
save_for_book(fig, 'ch12.fig6')
Let’s end by finding the 99.9% credible interval of \(Z\). We need the following quantiles:
\(z_{0.001}\):
z_001 = Z.ppf(0.001)
print(f'z_001 = {z_001:1.2f}')
z_001 = -3.09
\(z_{0.999}\):
z_999 = Z.ppf(0.999)
print(f'z_999 = {z_999:1.2f}')
z_999 = 3.09
So, we can now write:
\(Z\) is between -3.09 and 3.09 with probability 99.8%.
Or the more practical:
\(Z\) is between -3 and 3 with probability (about) 99.8%.
How can I think about this intuitively? Well, if you sample many many times from \(Z\) approximately 2 out of a 1000 samples will be outside of the interval \([-3, 3]\). Let’s test this computationally:
# Take 1,000,000 samples
zs = Z.rvs(size=1_000_000)
# Count the number of zs that are outside the range
idx = (zs < -3) | (zs > 3)
# How many samples out of 1,000?
zs[idx].size / 1_000_000 * 1_000
2.626
Questions#
Modify the code above to find the 99.99% central credible interval of \(Z\).