Expectation

Context: FIT1058_MOC · the probability-weighted average of a Random Variable · linear over sums (even dependent ones) · a measure of location alongside Median and Mode

Quick Revision

• 🎯 Objective: $E(X)={\sum }_{k}k\mathrm{Pr}(X=k)$ ➔ probability-weighted average.
• 📦 Core Components: weighted sum ➔ linearity ➔ product needs independence.
• ⚡ Key Constraint: $E(X+Y)=E(X)+E(Y)$ always; $E(XY)=E(X)E(Y)$ only if independent.

📝 Core

1. The Mean

• Definition ➔ $E(X)={\sum }_{k}k\mathrm{Pr}(X=k)$ — average weighted by probability.
• Generalises ➔ ordinary average (equal weights $\frac{1}{n}$) to any distribution.

2. Key Properties

• Constant/scaling ➔ $E(c)=c$, $E(\alpha X)=\alpha E(X)$.
• Linearity ➔ $E(X+Y)=E(X)+E(Y)$ for any variables (no independence).
• Product ➔ $E(XY)=E(X)E(Y)$ only if independent.

3. Limitations

• Can be untypical ➔ $E$ may sit far from every value (skew/outliers).
• Use median then ➔ median better represents “typical”.

Key identities:

$E(X)={\sum }_{k}k\mathrm{Pr}(X=k),E(X+Y)=E(X)+E(Y)(\text{any }X,Y)$ $E(XY)=E(X)E(Y)(\text{independent only})$

When It Flips: linearity splits a complex variable into simple summands — a binomial $Z=\sum X_i$ of Bernoullis gives $E(Z)=np$ instantly. $E$ can lie far from every value (skewed data), where the median is more representative.

📊 Exam Execution Trace

Applied Exercise

Problem: Find $E(X)$ for a fair die, then $E(Z)$ for the two-dice total by linearity. Derivation Proof / Hand-Calculation Walkthrough:

$$\begin{array}{rl}E(X)& =\sum _{k=1}^{6}k\cdot \frac{1}{6}=3.5\\ E(Z)& =E(X)+E(Y)=3.5+3.5=7(\text{no independence needed})\end{array}$$

Final Extracted Output: $E(X)=3.5$, $E(Z)=7$ without touching $Z$‘s triangular distribution.

⚠️ Common Mistakes

• 💡 Linearity needs no independence ➔ $E(X+Y)=E(X)+E(Y)$ holds even for dependent variables; only the product rule requires independence.

🧠 Active Recall

Define $E(X)$ and state linearity — does it need independence?

• Hint: Additive always.

Answer

• Short answer: $E(X)={\sum }_{k}k\mathrm{Pr}(X=k)$; $E(X+Y)=E(X)+E(Y)$ for any variables.
• Why: Product differs ➔ $E(XY)=E(X)E(Y)$ needs independence.

When is expectation a poor "typical" value, and what is better?

• Hint: Skew/outliers.

Answer

• Short answer: With outliers $E$ can lie far from every value ($1,2,3,4,90$ → $E=20$).
• Why: Median robust ➔ barely moves with extremes (but has no linearity).