数学期望shùxué qīwàng

Core Concept

The expected value (or mathematical expectation) of a random variable is the weighted average of all possible values, where the weights are the probabilities.

Discrete Random Variable

For a discrete random variable $X$ taking values $x_1, x_2, \ldots, x_n$ with probabilities $p_1, p_2, \ldots, p_n$:

$E(X) = \sum_{i=1}^{n} x_i \cdot p_i = x_1p_1 + x_2p_2 + \cdots + x_np_n$

Notation

• $E(X)$ - Expected value of $X$
• $\mu$ (mu) - Often used to denote expected value
• $\overline{X}$ - Sample mean (estimate of $E(X)$)

Interpretation

The expected value represents:

• The long-run average of many independent trials
• The center of mass of the probability distribution
• The fair value in gambling/finance contexts

Important: The expected value may not be an actually possible outcome.

Properties of Expected Value

1. Linearity

$E(aX + b) = aE(X) + b$

where $a$ and $b$ are constants.

2. Sum of Random Variables

$E(X + Y) = E(X) + E(Y)$

This holds even if $X$ and $Y$ are NOT independent.

3. Product of Independent Variables

If $X$ and $Y$ are independent: $E(XY) = E(X) \cdot E(Y)$

4. Expected Value of a Constant

$E(c) = c$

Common Distributions

Distribution	Expected Value
Bernoulli($p$)	$p$
Binomial($n, p$)	$np$
Uniform($\{1,2,...,n\}$)	$\dfrac{n+1}{2}$
Geometric($p$)	$\dfrac{1}{p}$

CSCA Practice Problems

💡 Note: The following practice problems are designed based on the CSCA exam syllabus.

Example 1: Basic (Difficulty ★★☆☆☆)

A random variable $X$ has the following distribution:

$X$	1	2	3
$P$	0.2	0.5	0.3

Find $E(X)$.

Solution: $E(X) = 1 \times 0.2 + 2 \times 0.5 + 3 \times 0.3$ $= 0.2 + 1.0 + 0.9 = 2.1$

Answer: $E(X) = 2.1$

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Example 2: Intermediate (Difficulty ★★★☆☆)

If $E(X) = 3$, find $E(2X + 5)$.

Solution:

Using linearity: $E(2X + 5) = 2E(X) + 5 = 2(3) + 5 = 11$

Answer: $11$

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Example 3: Intermediate (Difficulty ★★★☆☆)

A fair coin is tossed 100 times. Let $X$ be the number of heads. Find $E(X)$.

Solution:

$X$ follows Binomial distribution with $n = 100$, $p = 0.5$.

$E(X) = np = 100 \times 0.5 = 50$

Answer: $E(X) = 50$

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Example 4: Advanced (Difficulty ★★★★☆)

A box contains 3 red and 2 white balls. Two balls are drawn without replacement. Let $X$ be the number of red balls drawn. Find $E(X)$.

Solution:

Find the distribution of $X$:

$P(X = 0) = \dfrac{C_2^2}{C_5^2} = \dfrac{1}{10}$ (both white)

$P(X = 1) = \dfrac{C_3^1 \cdot C_2^1}{C_5^2} = \dfrac{6}{10}$ (one red, one white)

$P(X = 2) = \dfrac{C_3^2}{C_5^2} = \dfrac{3}{10}$ (both red)

$E(X) = 0 \times \frac{1}{10} + 1 \times \frac{6}{10} + 2 \times \frac{3}{10}$ $= 0 + 0.6 + 0.6 = 1.2$

Answer: $E(X) = 1.2$

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Example 5: Advanced (Difficulty ★★★★★)

If $E(X) = 2$ and $E(X^2) = 8$, find $E((X-1)^2)$.

Solution:

Expand: $E((X-1)^2) = E(X^2 - 2X + 1)$ $= E(X^2) - 2E(X) + 1$ $= 8 - 2(2) + 1 = 8 - 4 + 1 = 5$

Answer: $5$

Real-World Applications

1. Fair Games

A game is "fair" if $E(\text{profit}) = 0$.

Example: You pay ¥2 to flip a coin. Heads: win ¥4. Tails: win nothing. $E(\text{profit}) = 0.5 \times (4-2) + 0.5 \times (0-2) = 1 - 1 = 0$ This is a fair game.

2. Insurance

Insurance companies use expected value to set premiums.

3. Investment Analysis

Expected return helps compare investment options.

Common Mistakes

❌ Mistake 1: Confusing E(X) with Most Likely Value

Wrong: $E(X)$ is the value that occurs most often ✗

Correct: $E(X)$ is the weighted average; mode is the most frequent value ✓

❌ Mistake 2: Forgetting Probabilities Must Sum to 1

Before calculating, verify: $\sum p_i = 1$

❌ Mistake 3: Wrong Linearity Application

Wrong: $E(X^2) = (E(X))^2$ ✗

Correct: Generally $E(X^2) \neq (E(X))^2$. The difference is the variance! ✓

Relationship with Variance

$\text{Var}(X) = E(X^2) - (E(X))^2$

Or equivalently: $E(X^2) = \text{Var}(X) + (E(X))^2$

Study Tips

1. ✅ Remember the formula: $E(X) = \sum x_i p_i$
2. ✅ Master linearity: $E(aX+b) = aE(X)+b$
3. ✅ Know common distributions: Binomial expected value is $np$
4. ✅ Don't confuse with variance: $E(X^2) \neq (E(X))^2$

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💡 Exam Tip: When given a probability distribution table, first verify probabilities sum to 1, then apply the definition directly!