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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");

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 \(p(0.5 \le Y \le 2\;\text{or}\;3\le Y \le 4)\).