Show Python code
import numpy as np
import matplotlib.pyplot as plt
np.random.seed(42)Seminar 5: Expectation and Variance of Discrete Random Variables
In previous lectures we introduced discrete random variables and their distributions. We now discuss the most important numerical characteristics of a random variable: its expectation and variance.
Roughly speaking, the expectation describes the center of a distribution, while the variance measures the spread of the values around that center.
1 Expectation of a Discrete Random Variable
Let \(\xi\) be a discrete random variable taking values
\[ X_\xi=\{x_1,x_2,\ldots\}. \]
The expectation (or mean) of \(\xi\) is defined by
\[ \mathbb E[\xi] = \sum_i x_i\,\mathbb P(\xi=x_i), \]
provided that the series \(\sum_i |x_i|\,\mathbb P(\xi=x_i)\) converges.
1.1 Properties of Expectation
The expectation enjoys several important properties.
1.1.1 Linearity
For any constants \(a,b\in\mathbb R\),
\[ \mathbb E[a\xi+\eta b] = a\,\mathbb E[\xi]+b\,\mathbb E[\eta]. \]
A remarkable fact is that independence is not required for this property.
1.1.2 Expectation of a Function
Let \(\varphi:\mathbb R\to\mathbb R\) be a function. Then
\[ \mathbb E[\varphi(\xi)] = \sum_i \varphi(x_i)\, \mathbb P(\xi=x_i). \]
This formula allows us to compute expectations of transformed random variables directly from the distribution of \(\xi\).
1.1.3 Monotonicity
If \(\xi \le \eta\) almost surely, then \(\mathbb E[\xi] \le \mathbb E[\eta]\).
1.1.4 Absolute Value Inequality
For every random variable,
\[ \bigl|\mathbb E[\xi]\bigr| \le \mathbb E[|\xi|]. \]
This inequality is often useful when estimating expectations.
1.1.5 Product of Independent Random Variables
If \(\xi\) and \(\eta\) are independent, then
\[ \mathbb E[\xi\eta] = \mathbb E[\xi]\, \mathbb E[\eta]. \]
This property generally fails if the random variables are not independent.
1.2 Simulations on Expectation
In this part of the seminar, we use simulations to understand expectation empirically. The goal is not only to compute formulas, but to see what these quantities mean when we generate data.
Throughout the notebook, we use the empirical mean
\[ \overline X_n=\frac{1}{n}\sum_{i=1}^n X_i \]
as an approximation to the theoretical expectation \(\mathbb E[X]\).
1.2.1 Simulation 1: Expectation of a Fair Die
Let \(\xi\) be the result of rolling a fair die. Then \(\xi\) takes values \(1,2,3,4,5,6\) with equal probabilities. Therefore,
\[ \mathbb E[\xi]=\frac{1+2+3+4+5+6}{6}=3.5. \]
Notice that \(3.5\) is not a possible outcome of a die roll. This is an important point: the expectation is not necessarily the most likely value, nor even a possible value. It is a long-run average.
Show Python code
n = 1_000
rolls = np.random.randint(1, 7, size=n)
running_average = np.cumsum(rolls) / np.arange(1, n + 1)
plt.figure(figsize=(10, 5))
plt.plot(running_average, label="Running average")
plt.axhline(3.5, linestyle="--", label="Theoretical expectation = 3.5")
plt.xlabel("Number of rolls")
plt.ylabel("Average value")
plt.title("Running Average of Fair Die Rolls")
plt.legend()
plt.grid(True)
plt.show()
1.2.2 Simulation 2: Rare Event Bernoulli Random Variable
Let \(\xi\sim\operatorname{Ber}(0.05)\). We may interpret \(\xi=1\) as a user clicking an advertisement and \(\xi=0\) as not clicking it. Then
\[ \mathbb E[\xi]=0.05. \]
Most observations are equal to \(0\), but the long-run average still approaches \(0.05\).
Show Python code
n = 20_000
p = 0.05
x = np.random.binomial(1, p, size=n)
running_average = np.cumsum(x) / np.arange(1, n + 1)
plt.figure(figsize=(10, 5))
plt.plot(running_average, label="Running average")
plt.axhline(p, linestyle="--", label="Theoretical expectation = 0.05")
plt.xlabel("Number of observations")
plt.ylabel("Average value")
plt.title("Running Average for Bernoulli(0.05)")
plt.legend()
plt.grid(True)
plt.show()
print("Empirical mean:", np.mean(x))
print("Theoretical mean:", p)
Empirical mean: 0.051
Theoretical mean: 0.052 Variance
While the expectation tells us where a distribution is centered, it does not tell us how spread out the values are.
The variance of a random variable \(\xi\) is defined by
\[ \operatorname{Var}(\xi) = \mathbb E\!\left[(\xi-\mathbb E[\xi])^2\right]. \]
The quantity
\[ \sigma_\xi = \sqrt{\operatorname{Var}(\xi)} \]
is called the standard deviation.
2.1 Alternative Formula for Variance
Expanding the square gives
\[ \operatorname{Var}(\xi) = \mathbb E[\xi^2] - \bigl(\mathbb E[\xi]\bigr)^2. \]
This formula is often much easier to use in computations.
2.2 Properties of Variance
\(\operatorname{Var}(\xi)\ge 0\). Moreover, \(\operatorname{Var}(\xi)=0 \iff \mathbb P(\xi=c)=1\), for some constant \(c\).
For any constants \(a,b\in\mathbb R\),
\[ \operatorname{Var}(a\xi+b) = a^2\operatorname{Var}(\xi). \] 3. If \(\xi\) and \(\eta\) are independent, then \[\operatorname{Var}(\xi+\eta)=\operatorname{Var}(\xi)+\operatorname{Var}(\eta).\]
3 Important Examples
The following expectations and variances are useful enough to remember.
| Distribution | Expectation | Variance |
|---|---|---|
| \(\operatorname{Ber}(p)\) | \(p\) | \(p(1-p)\) |
| \(\operatorname{Bin}(n,p)\) | \(np\) | \(np(1-p)\) |
| \(\operatorname{Geom}(p)\) | \(\frac{1}{p}\) | \(\frac{1-p}{p^2}\) |
| \(\operatorname{Pois}(\lambda)\) | \(\lambda\) | \(\lambda\) |
| \(U(\{1,\ldots,n\})\) | \(\frac{n+1}{2}\) | \(\frac{n^2-1}{12}\) |
3.1 Comparing Different Distributions
The following figure compares a Binomial, Poisson, and Geometric distribution.
For each distribution:
- the red dashed line shows the expectation \(\mu\);
- the blue dotted lines show \(\mu\pm\sigma\);
- the green dotted lines show \(\mu\pm2\sigma\).
Although expectation and variance provide useful summaries of a distribution, they do not completely describe its shape. Notice how the three distributions differ in symmetry, spread, and skewness despite having well-defined means and variances.
Throughout the notebook, we use the empirical variance
\[ s_n^2 = \frac{1}{n} \sum_{i=1}^{n} \left(X_i-\overline X_n\right)^2 \]
as an approximation to the theoretical variance \(\operatorname{Var}(X)\).
Show Python code
import numpy as np
import matplotlib.pyplot as plt
np.random.seed(42)
n_sim = 20_000
fig, axes = plt.subplots(1, 3, figsize=(18, 5))
# =====================================================
# Binomial
# =====================================================
n = 20
p = 0.3
x = np.random.binomial(n, p, size=n_sim)
mu = n * p
sigma = np.sqrt(n * p * (1 - p))
ax = axes[0]
ax.hist(
x,
bins=np.arange(-0.5, n + 1.5, 1),
density=True,
edgecolor="black",
alpha=0.7
)
ax.axvline(mu, color="red", linestyle="--", linewidth=3)
ax.axvline(mu - sigma, color="blue", linestyle=":", linewidth=2)
ax.axvline(mu + sigma, color="blue", linestyle=":", linewidth=2)
ax.axvline(mu - 2*sigma, color="green", linestyle=":", linewidth=2)
ax.axvline(mu + 2*sigma, color="green", linestyle=":", linewidth=2)
ax.axvspan(mu-sigma, mu+sigma, alpha=0.15)
ax.axvspan(mu-2*sigma, mu+2*sigma, alpha=0.08)
ax.set_title(
rf"Binomial$(20,0.3)$" "\n"
rf"$\mu={mu:.1f},\ \sigma={sigma:.2f}$"
)
ax.set_xlabel("Value")
ax.set_ylabel("Density")
ax.grid(alpha=0.3)
# =====================================================
# Poisson
# =====================================================
lam = 5
x = np.random.poisson(lam, size=n_sim)
mu = lam
sigma = np.sqrt(lam)
ax = axes[1]
ax.hist(
x,
bins=np.arange(-0.5, np.max(x)+1.5, 1),
density=True,
edgecolor="black",
alpha=0.7
)
ax.axvline(mu, color="red", linestyle="--", linewidth=3)
ax.axvline(mu - sigma, color="blue", linestyle=":", linewidth=2)
ax.axvline(mu + sigma, color="blue", linestyle=":", linewidth=2)
ax.axvline(mu - 2*sigma, color="green", linestyle=":", linewidth=2)
ax.axvline(mu + 2*sigma, color="green", linestyle=":", linewidth=2)
ax.axvspan(mu-sigma, mu+sigma, alpha=0.15)
ax.axvspan(mu-2*sigma, mu+2*sigma, alpha=0.08)
ax.set_title(
rf"Poisson$(5)$" "\n"
rf"$\mu={mu:.1f},\ \sigma={sigma:.2f}$"
)
ax.set_xlabel("Value")
ax.grid(alpha=0.3)
# =====================================================
# Geometric
# =====================================================
p = 0.2
x = np.random.geometric(p, size=n_sim)
mu = 1/p
sigma = np.sqrt((1-p)/p**2)
ax = axes[2]
ax.hist(
x,
bins=np.arange(0.5, 26.5, 1),
density=True,
edgecolor="black",
alpha=0.7
)
ax.axvline(mu, color="red", linestyle="--", linewidth=3)
ax.axvline(mu - sigma, color="blue", linestyle=":", linewidth=2)
ax.axvline(mu + sigma, color="blue", linestyle=":", linewidth=2)
ax.axvline(mu - 2*sigma, color="green", linestyle=":", linewidth=2)
ax.axvline(mu + 2*sigma, color="green", linestyle=":", linewidth=2)
ax.axvspan(mu-sigma, mu+sigma, alpha=0.15)
ax.axvspan(mu-2*sigma, mu+2*sigma, alpha=0.08)
ax.set_title(
rf"Geometric$(0.2)$" "\n"
rf"$\mu={mu:.1f},\ \sigma={sigma:.2f}$"
)
ax.set_xlabel("Value")
ax.grid(alpha=0.3)
plt.suptitle(
"Expectation (red) and Standard-Deviation Regions",
fontsize=16
)
plt.tight_layout()
plt.show()
3.2 Example: Sample Mean Becomes Less Variable
Let \(X_1,\ldots,X_n\) be independent observations with expectation \(\mu\) and variance \(\sigma^2\). The sample mean is
\[ \overline X_n=\frac{1}{n}\sum_{i=1}^n X_i. \]
Its variance is
\[ \operatorname{Var}(\overline X_n)=\frac{\sigma^2}{n}. \]
Thus, as \(n\) grows, the sample mean becomes more stable.
Show Python code
n_sim = 20_000
sample_sizes = [1, 2, 5, 10, 30, 100]
plt.figure(figsize=(10, 5))
for n in sample_sizes:
samples = np.random.exponential(scale=1, size=(n_sim, n))
sample_means = samples.mean(axis=1)
plt.hist(sample_means, bins=60, density=True, alpha=0.35, label=f"n={n}")
plt.xlabel("Sample mean")
plt.ylabel("Density")
plt.title("Distribution of the Sample Mean for Different Sample Sizes")
plt.legend()
plt.grid(True)
plt.show()
3.3 Example: Apples and Oranges
A basket contains \(4\) apples and \(6\) oranges. We randomly choose \(3\) fruits without replacement. Let \(\xi\) be the number of apples among the selected fruits. Let’s compute the expectation and variance of \(\xi\).
Then \(\xi\) can take the values \(0,1,2,3\). Since we are sampling without replacement, \(\xi\) has a hypergeometric distribution:
\[ \xi\sim\operatorname{Hypergeom}(N=10,K=4,n=3), \]
where \(N=10\) is the total number of fruits, \(K=4\) is the number of apples, and \(n=3\) is the number of selected fruits.
In the simulation below, we repeat the experiment many times and compare the empirical distribution with the theoretical one.
Show Python code
import numpy as np
import matplotlib.pyplot as plt
from math import comb
np.random.seed(42)
# =====================================================
# Parameters
# =====================================================
n_sim = 100_000
N = 10 # total number of fruits
K = 4 # number of apples
M = 6 # number of oranges
n = 3 # number of selected fruits
# Basket: 1 = apple, 0 = orange
basket = np.array([1] * K + [0] * M)
# =====================================================
# Simulation
# =====================================================
simulated_xi = np.array([
np.sum(np.random.choice(basket, size=n, replace=False))
for _ in range(n_sim)
])
k_values = np.arange(0, n + 1)
empirical_probs = np.array([
np.mean(simulated_xi == k)
for k in k_values
])
theoretical_probs = np.array([
comb(K, k) * comb(M, n - k) / comb(N, n)
for k in k_values
])
# =====================================================
# Theoretical values
# =====================================================
mu = n * K / N
var = (
n
* (K / N)
* (1 - K / N)
* (N - n)
/ (N - 1)
)
# =====================================================
# Empirical values
# =====================================================
empirical_mean = np.mean(simulated_xi)
empirical_var = np.var(simulated_xi)
empirical_prob_gt_1 = np.mean(simulated_xi > 1)
theoretical_prob_gt_1 = np.sum(theoretical_probs[k_values > 1])
# =====================================================
# Plot
# =====================================================
fig, axes = plt.subplots(
1, 2,
figsize=(17, 6),
gridspec_kw={"width_ratios": [1.05, 1.55]}
)
fig.suptitle(
r"Sampling 3 Fruits Without Replacement",
fontsize=18,
weight="bold",
y=1.03
)
# =====================================================
# Left panel: basket drawing
# =====================================================
ax = axes[0]
apple_positions = [
(0.9, 1.25), (1.55, 1.25), (2.2, 1.25), (1.25, 0.6)
]
orange_positions = [
(3.05, 1.25), (3.7, 1.25), (4.35, 1.25),
(3.35, 0.6), (4.0, 0.6), (4.65, 0.6)
]
# Basket shadow
shadow = plt.Circle(
(2.75, 0.8),
2.55,
color="gray",
alpha=0.08,
zorder=0
)
ax.add_patch(shadow)
# Basket outline
basket_outline = plt.Circle(
(2.75, 0.85),
2.50,
fill=False,
linewidth=3,
color="saddlebrown",
alpha=0.8,
zorder=1
)
ax.add_patch(basket_outline)
# Apples
for x_pos, y_pos in apple_positions:
ax.scatter(
x_pos, y_pos,
s=1000,
color="tomato",
edgecolor="darkred",
linewidth=2,
zorder=3
)
ax.text(
x_pos, y_pos,
"A",
color="white",
fontsize=15,
ha="center",
va="center",
weight="bold",
zorder=4
)
# Oranges
for x_pos, y_pos in orange_positions:
ax.scatter(
x_pos, y_pos,
s=1000,
color="orange",
edgecolor="darkorange",
linewidth=2,
zorder=3
)
ax.text(
x_pos, y_pos,
"O",
color="white",
fontsize=15,
ha="center",
va="center",
weight="bold",
zorder=4
)
ax.text(
2.75, 2.25,
"Basket: 4 apples and 6 oranges",
ha="center",
fontsize=14,
weight="bold"
)
ax.text(
2.75, -0.35,
r"Choose $3$ fruits without replacement",
ha="center",
fontsize=12
)
ax.text(
2.75, -0.75,
r"$\xi =$ number of apples selected",
ha="center",
fontsize=13,
color="darkblue",
bbox=dict(
boxstyle="round,pad=0.35",
facecolor="aliceblue",
edgecolor="steelblue"
)
)
ax.set_xlim(0, 5.5)
ax.set_ylim(-1.1, 2.65)
ax.set_aspect("equal")
ax.axis("off")
# =====================================================
# Right panel: empirical vs theoretical distribution
# =====================================================
ax = axes[1]
width = 0.36
bars_emp = ax.bar(
k_values - width / 2,
empirical_probs,
width=width,
label="Simulation",
edgecolor="black",
linewidth=1.2,
alpha=0.75
)
bars_th = ax.bar(
k_values + width / 2,
theoretical_probs,
width=width,
label="Theory",
edgecolor="black",
linewidth=1.2,
alpha=0.75
)
# Labels above bars
for bar in bars_emp:
height = bar.get_height()
ax.text(
bar.get_x() + bar.get_width() / 2,
height + 0.015,
f"{height:.3f}",
ha="center",
va="bottom",
fontsize=10
)
for bar in bars_th:
height = bar.get_height()
ax.text(
bar.get_x() + bar.get_width() / 2,
height + 0.015,
f"{height:.3f}",
ha="center",
va="bottom",
fontsize=10
)
# Expectation line
ax.axvline(
mu,
color="red",
linestyle="--",
linewidth=2.5,
label=fr"$\mathbb{{E}}\xi={mu:.2f}$"
)
ax.set_xticks(k_values)
ax.set_xlabel(r"Number of apples selected, $\xi$", fontsize=12)
ax.set_ylabel("Probability", fontsize=12)
ax.set_title(
r"Distribution of $\xi$",
fontsize=14,
weight="bold"
)
ax.set_ylim(
0,
max(max(empirical_probs), max(theoretical_probs)) + 0.22
)
ax.grid(axis="y", alpha=0.25)
ax.spines["top"].set_visible(False)
ax.spines["right"].set_visible(False)
# Legend outside the plot
ax.legend(
loc="upper left",
bbox_to_anchor=(1.02, 1.00),
frameon=True
)
# Summary box outside the plot
summary_text = (
rf"$\mathbb{{E}}\xi$: simulation {empirical_mean:.3f}, theory {mu:.3f}" "\n"
rf"$\operatorname{{Var}}\xi$: simulation {empirical_var:.3f}, theory {var:.3f}" "\n"
rf"$\mathbb{{P}}(\xi>1)$: simulation {empirical_prob_gt_1:.3f}, theory {theoretical_prob_gt_1:.3f}"
)
ax.text(
1.02,
0.70,
summary_text,
transform=ax.transAxes,
ha="left",
va="top",
fontsize=11,
bbox=dict(
boxstyle="round,pad=0.5",
facecolor="white",
edgecolor="gray",
alpha=0.95
)
)
plt.tight_layout(rect=[0, 0, 0.84, 1])
plt.show()
# =====================================================
# Numerical output
# =====================================================
print("Theoretical probabilities:")
for k, prob in zip(k_values, theoretical_probs):
print(f"P(xi = {k}) = {prob:.4f}")
print()
print(f"Empirical mean = {empirical_mean:.4f}")
print(f"Theoretical mean = {mu:.4f}")
print()
print(f"Empirical variance = {empirical_var:.4f}")
print(f"Theoretical variance = {var:.4f}")
print()
print(f"Empirical P(xi > 1) = {empirical_prob_gt_1:.4f}")
print(f"Theoretical P(xi > 1) = {theoretical_prob_gt_1:.4f}")
Theoretical probabilities:
P(xi = 0) = 0.1667
P(xi = 1) = 0.5000
P(xi = 2) = 0.3000
P(xi = 3) = 0.0333
Empirical mean = 1.1985
Theoretical mean = 1.2000
Empirical variance = 0.5591
Theoretical variance = 0.5600
Empirical P(xi > 1) = 0.3311
Theoretical P(xi > 1) = 0.33333.4 Example: The Hat-Check Problem
A group of \(n\) students throw their hats at graduation. After that, each student randomly picks one hat from the floor. Let \(\xi\) be the number of students who get their own hat back.
In the simulation below, we randomly permute the hats many times and compare the empirical mean and variance with the theoretical values.
Show Python code
import numpy as np
import matplotlib.pyplot as plt
from math import factorial
np.random.seed(42)
# =====================================================
# Parameters
# =====================================================
n_students = 20
n_sim = 100_000
students = np.arange(n_students)
# =====================================================
# Simulation
# =====================================================
fixed_points = []
for _ in range(n_sim):
hats = np.random.permutation(n_students)
fixed_points.append(np.sum(hats == students))
fixed_points = np.array(fixed_points)
# =====================================================
# Empirical and theoretical values
# =====================================================
empirical_mean = np.mean(fixed_points)
empirical_var = np.var(fixed_points)
theoretical_mean = 1
theoretical_var = 1
# =====================================================
# Distribution
# =====================================================
max_k = np.max(fixed_points)
k_values = np.arange(0, max_k + 1)
empirical_probs = np.array([
np.mean(fixed_points == k)
for k in k_values
])
poisson_probs = np.array([
np.exp(-1) / factorial(k)
for k in k_values
])
# =====================================================
# Plot
# =====================================================
fig, ax = plt.subplots(figsize=(11, 6))
width = 0.38
bars_emp = ax.bar(
k_values - width / 2,
empirical_probs,
width=width,
label="Simulation",
edgecolor="black",
linewidth=1.2,
alpha=0.75
)
bars_pois = ax.bar(
k_values + width / 2,
poisson_probs,
width=width,
label=r"Poisson$(1)$ approximation",
edgecolor="black",
linewidth=1.2,
alpha=0.75
)
# Labels only above simulation bars to avoid overlap
for bar in bars_emp:
height = bar.get_height()
if height > 0.02:
ax.text(
bar.get_x() + bar.get_width() / 2,
height + 0.012,
f"{height:.3f}",
ha="center",
va="bottom",
fontsize=10
)
# Expected value line
ax.axvline(
theoretical_mean,
color="red",
linestyle="--",
linewidth=2.5
)
ax.text(
theoretical_mean + 0.08,
0.93 * max(empirical_probs),
r"$\mathbb{E}\xi=1$",
color="red",
fontsize=12,
weight="bold"
)
# Axes and title
ax.set_xticks(k_values)
ax.set_xlabel(
r"Number of students receiving the correct hat, $\xi$",
fontsize=12
)
ax.set_ylabel(
"Probability",
fontsize=12
)
ax.set_title(
rf"Hat-Check Problem ($n={n_students}$ students)",
fontsize=15,
weight="bold"
)
ax.set_ylim(0, max(empirical_probs) + 0.13)
ax.grid(axis="y", alpha=0.25)
ax.spines["top"].set_visible(False)
ax.spines["right"].set_visible(False)
# Legend outside the plot
ax.legend(
loc="upper left",
bbox_to_anchor=(1.02, 1.00),
frameon=True
)
# Summary box outside the plot
summary_text = (
rf"$\mathbb{{E}}\xi$: simulation {empirical_mean:.3f}, theory {theoretical_mean:.3f}" "\n"
rf"$\operatorname{{Var}}\xi$: simulation {empirical_var:.3f}, theory {theoretical_var:.3f}" "\n"
rf"$\mathbb{{P}}(\xi=0)$: simulation {np.mean(fixed_points == 0):.3f}" "\n"
rf"$e^{{-1}} \approx {np.exp(-1):.3f}$"
)
ax.text(
1.02,
0.70,
summary_text,
transform=ax.transAxes,
ha="left",
va="top",
fontsize=11,
bbox=dict(
boxstyle="round,pad=0.5",
facecolor="white",
edgecolor="gray",
alpha=0.95
)
)
plt.tight_layout(rect=[0, 0, 0.80, 1])
plt.show()
# =====================================================
# Numerical summary
# =====================================================
print(f"Number of students = {n_students}")
print()
print(f"Empirical mean = {empirical_mean:.4f}")
print(f"Theoretical mean = {theoretical_mean:.4f}")
print()
print(f"Empirical variance = {empirical_var:.4f}")
print(f"Theoretical variance = {theoretical_var:.4f}")
print()
print(f"Empirical P(xi = 0) = {np.mean(fixed_points == 0):.4f}")
print(f"Poisson approximation = {np.exp(-1):.4f}")
Number of students = 20
Empirical mean = 0.9968
Theoretical mean = 1.0000
Empirical variance = 0.9976
Theoretical variance = 1.0000
Empirical P(xi = 0) = 0.3695
Poisson approximation = 0.36793.5 Example: A Casino Game with Positive Expectation
Consider the following game. In one round, a player wins \(\$10\) with probability \(0.1\) and loses \(\$1\) with probability \(0.9\). Let \(X\) be the profit from one round.
Then
\[ X= \begin{cases} 10, & \text{with probability }0.1,\\ -1, & \text{with probability }0.9. \end{cases} \]
The expected profit is
\[ \mathbb E[X] = 10\cdot 0.1+(-1)\cdot 0.9 = 0.1. \]
So the game has positive expectation. However, this does not mean that a player is guaranteed to win. In the simulation below, we see that even a favorable game can produce losing streaks and negative outcomes over short periods.
Show Python code
import numpy as np
import matplotlib.pyplot as plt
np.random.seed(42)
# =====================================================
# Parameters
# =====================================================
n_rounds = 200
n_players = 100
win_prob = 0.1
win_amount = 10
loss_amount = -1
# =====================================================
# Simulation
# =====================================================
profits = np.random.choice(
[win_amount, loss_amount],
size=(n_players, n_rounds),
p=[win_prob, 1 - win_prob]
)
cumulative_profits = np.cumsum(profits, axis=1)
expected_profit_per_round = win_amount * win_prob + loss_amount * (1 - win_prob)
expected_path = expected_profit_per_round * np.arange(1, n_rounds + 1)
final_profits = cumulative_profits[:, -1]
# =====================================================
# Plot
# =====================================================
fig, axes = plt.subplots(1, 2, figsize=(16, 5))
# -----------------------------------------------------
# Left plot: cumulative profit paths
# -----------------------------------------------------
ax = axes[0]
for i in range(n_players):
ax.plot(
np.arange(1, n_rounds + 1),
cumulative_profits[i],
alpha=0.18,
linewidth=1
)
ax.plot(
np.arange(1, n_rounds + 1),
expected_path,
color="red",
linewidth=3,
linestyle="--",
label=r"Theoretical expectation"
)
ax.axhline(0, color="black", linewidth=1)
ax.set_xlabel("Number of rounds")
ax.set_ylabel("Cumulative profit")
ax.set_title("Cumulative Profit of 100 Players")
ax.legend()
ax.grid(True, alpha=0.3)
# -----------------------------------------------------
# Right plot: final profits
# -----------------------------------------------------
ax = axes[1]
ax.hist(
final_profits,
bins=20,
edgecolor="black",
alpha=0.75
)
ax.axvline(
np.mean(final_profits),
color="red",
linestyle="--",
linewidth=3,
label="Empirical mean"
)
ax.axvline(
expected_path[-1],
color="black",
linestyle=":",
linewidth=3,
label="Theoretical expectation"
)
ax.set_xlabel("Final profit after 200 rounds")
ax.set_ylabel("Number of players")
ax.set_title("Distribution of Final Profits")
ax.legend()
ax.grid(True, alpha=0.3)
plt.tight_layout()
plt.show()
# =====================================================
# Numerical summary
# =====================================================
print(f"Expected profit per round = {expected_profit_per_round:.2f}")
print(f"Expected profit after {n_rounds} rounds = {expected_path[-1]:.2f}")
print()
print(f"Empirical mean final profit = {np.mean(final_profits):.2f}")
print(f"Empirical variance = {np.var(final_profits):.2f}")
print()
print(f"Fraction of players who lost money = {np.mean(final_profits < 0):.3f}")
Expected profit per round = 0.10
Expected profit after 200 rounds = 20.00
Empirical mean final profit = 23.30
Empirical variance = 1397.55
Fraction of players who lost money = 0.3103.6 Why Do We Care About Expectation and Variance?
At first glance, expectation and variance may seem like purely mathematical quantities. However, they appear in almost every area of statistics, data science, machine learning, finance, engineering, and scientific research.
The expectation describes what we should expect on average if an experiment is repeated many times, while the variance measures how much uncertainty or variability is present around that average.
In practice, good decisions often require understanding both quantities.
4 Discrete Conditional Expectation
4.1 Conditional Distribution
Let \(X\) and \(Y\) be discrete random variables. For every value \(y\) satisfying \(\mathbb P(Y=y)>0\), the conditional probability mass function of \(X\) is
\[ \mathbb P(X=x\mid Y=y) = \frac{\mathbb P(X=x,Y=y)} {\mathbb P(Y=y)}. \]
For each fixed value of \(y\), these probabilities define a valid probability distribution. Indeed,
\[ \sum_x \mathbb P(X=x\mid Y=y) = 1. \]
Therefore, once \(Y=y\) is observed, we may compute probabilities and expectations exactly as we do for any ordinary discrete distribution.
4.2 Definition
The conditional expectation of \(X\) given \(Y=y\) is defined by
\[ \boxed{ \mathbb E[X\mid Y=y] = \sum_x x\,\mathbb P(X=x\mid Y=y). } \]
4.2.1 Conditional Expectation is a Random Variable
Define the function
\[ g(y)=\mathbb E[X\mid Y=y]. \]
Since this is a function of \(y\), we may replace \(y\) by the random variable \(Y\) itself.
This gives
\[ \boxed{ \mathbb E[X\mid Y]=g(Y). } \]
Therefore, conditional expectation is itself a random variable.
4.2.2 Interpretation
Before observing \(Y\), our best prediction of \(X\) is \(\mathbb E[X]\).
After observing \(Y\), we have additional information and our prediction becomes
\[ \mathbb E[X\mid Y]. \]
The conditional expectation updates our prediction using the information contained in \(Y\).