📊 Expected Value

The average value of a random variable and the powerful linearity property that makes many problems tractable.

🎯 Key Ideas

Expected Value Definition

For a discrete random variable $X$ with possible values $x_1, x_2, \ldots, x_n$: $$\mathbb{E}[X] = \sum_{i=1}^{n} x_i \cdot P(X = x_i)$$

Linearity of Expectation

For any random variables $X$ and $Y$: $$\mathbb{E}[X + Y] = \mathbb{E}[X] + \mathbb{E}[Y]$$

Indicator Variables

Binary variables that are 1 if an event occurs, 0 otherwise: $$\mathbb{E}[I_A] = P(A)$$

💡 Micro-Examples

Basic Expected Value

  • Problem: What’s the expected value of a fair die roll?
  • Solution: $\mathbb{E}[X] = 1 \cdot \frac{1}{6} + 2 \cdot \frac{1}{6} + \cdots + 6 \cdot \frac{1}{6} = \frac{21}{6} = 3.5$

Linearity of Expectation

  • Problem: What’s the expected value of the sum of two die rolls?
  • Solution: $\mathbb{E}[X + Y] = \mathbb{E}[X] + \mathbb{E}[Y] = 3.5 + 3.5 = 7$

Indicator Variables

  • Problem: What’s the expected number of heads in 10 coin flips?
  • Solution: Let $I_i$ be 1 if flip $i$ is heads, 0 otherwise. Then $\mathbb{E}[\text{heads}] = \mathbb{E}[I_1 + I_2 + \cdots + I_{10}] = 10 \cdot \frac{1}{2} = 5$

⚠️ Common Traps & Fixes

Trap: Assuming independence for linearity

  • Wrong: “Linearity only works for independent variables”
  • Right: Linearity works for ANY variables, independent or not!

Trap: Confusing expected value with most likely value

  • Wrong: “The expected value of a die roll is 3 or 4”
  • Right: The expected value is 3.5, which is not a possible outcome!

Trap: Forgetting to use indicators

  • Wrong: “What’s the expected number of aces in a 5-card hand?” = Complex calculation
  • Right: Use indicators! $\mathbb{E}[\text{aces}] = 5 \cdot \frac{4}{52} = \frac{5}{13}$

🏆 AMC-Style Worked Example

Problem: What’s the expected number of fixed points in a random permutation of ${1,2,3,4,5}$?

Solution:

  • Step 1: Let $I_i$ be 1 if position $i$ contains element $i$, 0 otherwise
  • Step 2: $\mathbb{E}[\text{fixed points}] = \mathbb{E}[I_1 + I_2 + I_3 + I_4 + I_5]$
  • Step 3: By linearity: $\mathbb{E}[I_1 + I_2 + I_3 + I_4 + I_5] = \mathbb{E}[I_1] + \mathbb{E}[I_2] + \mathbb{E}[I_3] + \mathbb{E}[I_4] + \mathbb{E}[I_5]$
  • Step 4: For each $i$: $\mathbb{E}[I_i] = P(\text{position } i \text{ contains element } i) = \frac{1}{5}$
  • Step 5: Answer: $5 \cdot \frac{1}{5} = 1$

Key insight: Use indicators and linearity to avoid complex casework!

🔗 Related Topics

Next: Distributions → Symmetry & Invariance