15.3 Discrete Random Variables, Probability Distributions, Expectation, and Variance

15.3 Discrete Random Variables, Probability Distributions, Expectation, and Variance

1. Discrete Random Variables and Distribution Tables

A random variable $X$ is a real-valued function defined on a sample space $\Omega$, assigning a real number $X(\omega)$ to each elementary outcome $\omega \in \Omega$. A random variable is discrete if its set of possible values {$x_1, x_2, \dots, x_n, \dots$} is finite or countably infinite.

1.1 Probability Distribution Table (Distribution Series)

The probability distribution of a discrete random variable $X$ taking values $x_1, x_2, \dots, x_n$ is represented by the table:

where $p_i = P(X = x_i)$.

1.2 Essential Axiomatic Properties

1. Non-negativity: $p_i \ge 0$ for all $i = 1, 2, \dots, n$.
2. Normalization: $\sum_{i=1}^n p_i = p_1 + p_2 + \dots + p_n = 1$.

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2. Common Discrete Probability Distributions

2.1 Two-Point (Bernoulli) Distribution $B(1, p)$

A random variable $X$ has a 0-1 (Two-point) distribution if it takes only two values, 1 ("Success") and 0 ("Failure"), with probabilities $P(X = 1) = p$ and $P(X = 0) = 1 - p = q$.

2.2 Binomial Distribution $B(n, p)$

When $X$ represents the number of successes in $n$ independent Bernoulli trials with success probability $p$: $P(X = k) = C_n^k p^k (1-p)^{n-k}, \quad k = 0, 1, 2, \dots, n$

• Sampling Mechanism: Sampling with replacement from a population containing two types of items.

2.3 Hypergeometric Distribution $H(N, M, n)$

Consider a finite population of $N$ items containing $M$ items of Type I (e.g., defective goods) and $N - M$ items of Type II. If $n$ items are drawn randomly without replacement, the number $X$ of Type I items drawn follows a Hypergeometric distribution, denoted $X \sim H(N, M, n)$: $P(X = k) = \frac{C_M^k C_{N-M}^{n-k}}{C_N^n}, \quad k = \max(0, n - (N-M)), \dots, \min(n, M)$

Key Comparison: Binomial vs. Hypergeometric

• Hypergeometric: Sampling without replacement from a finite population ($N$). The probability of success changes with each draw.
• Binomial: Sampling with replacement (or sampling without replacement when population $N \to \infty$ or $N$ is very large relative to sample size $n$, e.g., $n/N < 0.05$).

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3. Mathematical Expectation (Mean) of Discrete Random Variables

3.1 Definition

The mathematical expectation (or expected value, mean) of a discrete random variable $X$ with distribution $P(X = x_i) = p_i$ is defined as: $E(X) = \mu = \sum_{i=1}^n x_i p_i = x_1 p_1 + x_2 p_2 + \dots + x_n p_n$ Interpretation: $E(X)$ represents the weighted average outcome or center of mass of the distribution across infinite independent repetitions of the experiment.

3.2 Properties of Expectation

1. Constant: $E(c) = c$ for any constant $c$.
2. Linear Transformation: $E(aX + b) = a E(X) + b$ for any constants $a, b$.
3. Additive Property: For any two random variables $X$ and $Y$, $E(X + Y) = E(X) + E(Y)$ (regardless of independence).
4. Multiplicative Property: If $X$ and $Y$ are independent, $E(XY) = E(X) E(Y)$.

3.3 Expectations of Standard Distributions

• Two-point Distribution $B(1, p)$: $E(X) = 1 \cdot p + 0 \cdot (1-p) = p$.
• Binomial Distribution $B(n, p)$: Expressing $X = \sum_{i=1}^n X_i$ where $X_i \sim B(1, p)$ are independent: $E(X) = E\left(\sum_{i=1}^n X_i\right) = \sum_{i=1}^n E(X_i) = n p$
• Hypergeometric Distribution $H(N, M, n)$: $E(X) = n \times \frac{M}{N}$

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4. Variance and Standard Deviation

4.1 Definition

The variance of $X$, denoted by $D(X)$ or $Var(X)$, measures the dispersion or spread of the random variable around its mean $E(X)$: $D(X) = E\left[ (X - E(X))^2 \right] = \sum_{i=1}^n (x_i - E(X))^2 p_i$

Computational Formula (frequently used in calculations): $D(X) = E(X^2) - [E(X)]^2 = \left( \sum_{i=1}^n x_i^2 p_i \right) - [E(X)]^2$

The standard deviation is $\sigma(X) = \sqrt{D(X)}$.

4.2 Properties of Variance

1. Non-negativity: $D(X) \ge 0$. $D(X) = 0$ if and only if $X$ is constant with probability 1.
2. Linear Transformation: $D(aX + b) = a^2 D(X)$ (note that constant shift $b$ does not affect variance).
3. Additive Property for Independent Variables: If $X$ and $Y$ are independent: $D(X + Y) = D(X) + D(Y), \quad D(X - Y) = D(X) + D(Y)$

4.3 Variances of Standard Distributions

• Two-point Distribution $B(1, p)$: $D(X) = p(1-p) = pq$.
• Binomial Distribution $B(n, p)$: $D(X) = n p (1-p) = npq$.
• Hypergeometric Distribution $H(N, M, n)$: $D(X) = n \frac{M}{N} \left(1 - \frac{M}{N}\right) \frac{N-n}{N-1}$ (Notice the finite population correction factor $\frac{N-n}{N-1}$ compared to the Binomial variance).

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5. Worked Gaokao Exam Examples

Example 1: Hypergeometric vs Binomial Comparison in Batch Inspection

Problem: A batch of 10 electronic components contains 3 defective items and 7 non-defective items.

1. If 3 components are selected at random without replacement, let $X$ be the number of defective items drawn. Write the probability distribution table of $X$, and calculate $E(X)$ and $D(X)$.
2. If 3 components are selected at random with replacement, let $Y$ be the number of defective items drawn. Calculate $E(Y)$ and $D(Y)$.

Detailed Solution:

1. Without Replacement ($X \sim H(10, 3, 3)$): Possible values of $X \in {0, 1, 2, 3}$. Total ways to choose 3 from 10: $C_{10}^3 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$.

   • $P(X = 0) = \frac{C_3^0 C_7^3}{120} = \frac{1 \times 35}{120} = \frac{35}{120} = \frac{7}{24}$
   • $P(X = 1) = \frac{C_3^1 C_7^2}{120} = \frac{3 \times 21}{120} = \frac{63}{120} = \frac{21}{40}$
   • $P(X = 2) = \frac{C_3^2 C_7^1}{120} = \frac{3 \times 7}{120} = \frac{21}{120} = \frac{7}{40}$
   • $P(X = 3) = \frac{C_3^3 C_7^0}{120} = \frac{1 \times 1}{120} = \frac{1}{120}$

   Check sum: $\frac{35 + 63 + 21 + 1}{120} = \frac{120}{120} = 1$.

   Expectation $E(X)$: $E(X) = 0 \left(\frac{35}{120}\right) + 1 \left(\frac{63}{120}\right) + 2 \left(\frac{21}{120}\right) + 3 \left(\frac{1}{120}\right) = \frac{0 + 63 + 42 + 3}{120} = \frac{108}{120} = \frac{9}{10} = 0.9$ Formula verification: $n \frac{M}{N} = 3 \times \frac{3}{10} = 0.9$.

   Variance $D(X)$: $E(X^2) = 0^2 \left(\frac{35}{120}\right) + 1^2 \left(\frac{63}{120}\right) + 2^2 \left(\frac{21}{120}\right) + 3^2 \left(\frac{1}{120}\right) = \frac{63 + 84 + 9}{120} = \frac{156}{120} = 1.3$. $D(X) = E(X^2) - [E(X)]^2 = 1.3 - (0.9)^2 = 1.3 - 0.81 = 0.49$

2. With Replacement ($Y \sim B(3, 0.3)$): $p = \frac{3}{10} = 0.3$, $n = 3$.

   • $E(Y) = np = 3 \times 0.3 = 0.9$.
   • $D(Y) = np(1-p) = 3 \times 0.3 \times 0.7 = 0.63$.

   Insight: Expectation is identical ($0.9$), but sampling without replacement reduces variance ($0.49 < 0.63$) due to the finite population factor $\frac{10-3}{10-1} = \frac{7}{9} \approx 0.7778$, since $0.63 \times \frac{7}{9} = 0.49$.