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MAKE_BOOK_FIGURES=Trueimport numpy as npimport scipy.stats as stimport matplotlib as mplimport matplotlib.pyplot as plt%matplotlib inlineimport matplotlib_inlinematplotlib_inline.backend_inline.set_matplotlib_formats('svg')import seaborn as snssns.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 set_notebook_style():    plt.style.use('seaborn-v0_8-white')    sns.set_style("ticks")    sns.set_palette("deep")    mpl.rcParams.update({        # Font settings - using default sizes        'font.family': 'serif',        'axes.labelsize': 10,        'axes.titlesize': 10,        'xtick.labelsize': 9,        'ytick.labelsize': 9,        'legend.fontsize': 9,                # Line and marker settings        'axes.linewidth': 0.5,        'grid.linewidth': 0.5,        'lines.linewidth': 1.0,        'lines.markersize': 4,                # Layout settings        'figure.constrained_layout.use': True,                # Remove only top and right spines        'axes.spines.top': False,        'axes.spines.right': False,                # Remove legend frame        'legend.frameon': False,                # Additional settings        'figure.autolayout': True,        'savefig.bbox': 'tight',        'savefig.pad_inches': 0.1    })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()else:    set_notebook_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()

Sampling Estimates of the Cumulative Distribution Function

Let \(Y\) be a random variable. Suppose that \(F(y)\) is the cumulative distribution function of \(Y\). We want to estimate \(F(y)\) using samples of \(Y\). The key is to observe that it can be written as an expectation using the indicator function:

\[ F(y) = p(Y\le y) = \int_{-\infty}^y p(y)dy = \int_{-\infty}^y 1_{[-\infty,y]}(y)p(y)dy = \mathbb{E}[1_{[-\infty,y]}(Y)]. \]

So, if we had a sample \(Y_1,Y_2,\dots,Y_N\) of \(Y\), we could estimate \(F(y)\) as:

\[ \bar{F}_N(y) = \frac{1}{N}\sum_{i=1}^N1_{[-\infty,y]}(Y_i) = \frac{\text{number of }Y_i \le y}{N}. \]

This is known as the empirical cumulative distribution function (CDF). We saw this estimate in a previous lecture. Now you know where it comes from!

When solving the uncertainty propagation problem, we have a random variable \(X\) and a function \(g\). We want to estimate the cumulative distribution function of \(Y=g(X)\): Then, we can write:

\[ F(y) = \mathbb{E}[1_{[-\infty,y]}(Y)] = \mathbb{E}[1_{[-\infty,y]}(g(X))]. \]

So, if we have independent samples \(X_1,X_2,\dots,X_N\) of \(X\), we can estimate \(F(y)\) as:

\[ \bar{F}_N(y) = \frac{1}{N}\sum_{i=1}^N1_{[-\infty,y]}(g(X_i)) = \frac{\text{number of }g(X_i) \le y}{N}. \]

Example: 1D CDF

We will continue using the 1D test function of Example 3.4 of [Robert and Casella, 2004]. Assume that \(X\sim U([0,1])\) and pick:

\[ g(x) = \left(\cos(50x) + \sin(20x)\right)^2. \]

import numpy as np

# define the function here:
g = lambda x: (np.cos(50 * x) + np.sin(20 * x)) ** 2

# We will not write code for the empirical CDF as it is already in here:
# https://www.statsmodels.org/stable/generated/statsmodels.distributions.empirical_distribution.ECDF.html
from statsmodels.distributions.empirical_distribution import ECDF

# Maximum number of samples to take
max_n = 10000 
# Generate samples from X
x_samples = np.random.rand(max_n)
# Get the corresponding Y's
y_samples = g(x_samples)

# Build ECDF with 10 samples
ecdf_10 = ECDF(y_samples[:10])

# Build ECDF with 50 samples
ecdf_100 = ECDF(y_samples[:100])

# Build ECDF with all samples
ecdf_all = ECDF(y_samples)

# Make the plot
fig, ax = plt.subplots()
# Points on which to evaluate the CDF's
ys = np.linspace(y_samples.min(), y_samples.max(), 100)
ax.plot(ys, ecdf_10(ys), ".-", label=r"$N=10$")
ax.plot(ys, ecdf_100(ys), "--", label=r"$N=100$")
ax.plot(ys, ecdf_all(ys), label=f"$N={max_n:d}$")
ax.set_xlabel("$y$")
ax.set_ylabel(r"$\bar{F}_N(y)$")
plt.legend(loc="best", frameon=False)
sns.despine(trim=True);

Let’s use the empirical CDF to find the probability that \(Y\) takes specific values. For example, we calculate the probability that \(Y\) is between \(1\) and \(3\). We have:

\[ p(1\le Y\le 3) = F(3) - F(1) \approx \bar{F}_N(3) - \bar{F}_N(1). \]

Let’s calculate this numerically for various choices of \(N\):

# Estimate of the probability with 10 samples:
p_Y_in_set_10 = ecdf_10(3.0) - ecdf_10(1.0)
print(f"N = 10:\t\tp(1 <= Y <= 3) ~= {p_Y_in_set_10:.2f}")
# Estimate of the probability with 100 samples:
p_Y_in_set_100 = ecdf_100(3.0) - ecdf_100(1.0)
print(f"N = 100:\tp(1 <= Y <= 3) ~= {p_Y_in_set_100:.2f}")
# Estimate of the probability with all 10000 samples:
p_Y_in_set_all = ecdf_all(3.0) - ecdf_all(1.0)
print(f"N = 1000:\tp(1 <= Y <= 3) ~= {p_Y_in_set_all:.2f}")

N = 10:		p(1 <= Y <= 3) ~= 0.40
N = 100:	p(1 <= Y <= 3) ~= 0.31
N = 1000:	p(1 <= Y <= 3) ~= 0.33

Questions

• Why is the empirical CDF for small \(N\) discontinuous?

• How do you know how many samples you need? For now, think about it on your own. We will answer in Lecture 10.

• Use the best empirical CDF we have constructed so far to find the probability of that \(Y\) is in \([0.5, 2]\) or \([3,4]\), i.e., see \(p(0.5 \le Y \le 2\;\text{or}\;3\le Y \le 4)\).