Median and Mode

Context: FIT1058_MOC · two further measures of location for a Random Variable · median = “middle” of the distribution, mode = most likely value · contrast with Expectation

Quick Revision

• 🎯 Objective: median = middle, mode = most likely value ➔ location measures beside mean.
• 📦 Core Components: median (half mass each side) ➔ mode ($\arg \max \mathrm{Pr}$).
• ⚡ Key Constraint: median is outlier-robust but has no linearity; all three locate, none spread.

📝 Core

1. The Measures

• Median ➔ $m$ with $\mathrm{Pr}(X\le m)\ge \frac{1}{2}$ and $\mathrm{Pr}(X\ge m)\ge \frac{1}{2}$.
• Mode ➔ value $g$ maximising $\mathrm{Pr}(X=g)$ (may be non-unique).
• Location ➔ both are location measures, like Expectation.

2. Median Subtleties

• ≠ mid-range ➔ mid-range $\frac{\min +\max }{2}$ ignores probabilities; median uses the whole distribution.
• Gap ➔ may fall between values (fair die → 3.5 by symmetry).

3. Mode Bounds, Doesn’t Locate

• Upper bound ➔ $\mathrm{Pr}(X=\text{mode})$ bounds every value’s probability.
• Can be far ➔ from mean/median (though clustered for well-behaved distributions).

Key identities:

$\text{median }m:\mathrm{Pr}(X\le m)\ge \frac{1}{2}\text{ and }\mathrm{Pr}(X\ge m)\ge \frac{1}{2}$ $\text{mode }g=\arg {\max }_k\mathrm{Pr}(X=k)$

⚖️ Core Decision Matrix

Measure	Definition	Property
mean (Expectation)	weighted average	linear, outlier-sensitive
median	half mass each side	robust, no linearity
mode	most likely value	bounds probabilities
all three	location	none measures spread

When It Flips: median is robust to outliers ($1,2,3,4,90$: median 3, mean 20) but has no linearity, so expectation is preferred for arithmetic. Spread needs Variance and Standard Deviation, not location.

📊 Exam Execution Trace

Manual Execution Trace

Scrabble points CDF:

Step / State	$k$	$\mathrm{Pr}(X=k)$	$\mathrm{Pr}(X\le k)$
0 (Init)	—	—	—
1	0	0.02	0.02
2	1	0.68	0.70 ≥½
3	≥2	0.30	1.00

⚠️ Common Mistakes

• 💡 Median ≠ mid-range ➔ mid-range ignores all probabilities; the median’s balance point need not be halfway between extremes.

🧠 Active Recall

Define the median and contrast it with the mid-range and the mean.

• Hint: Balance point, robust.

Answer

• Short answer: $\mathrm{Pr}(X\le m)\ge \frac{1}{2}$ and $\mathrm{Pr}(X\ge m)\ge \frac{1}{2}$; mid-range ignores probabilities; median is outlier-robust.
• Why: No linearity ➔ median of $X+Y$ ≠ sum of medians, so mean preferred for arithmetic.

What is the mode, and what does its probability tell you?

• Hint: Argmax + upper bound.

Answer

• Short answer: The most likely value; $\mathrm{Pr}(X=\text{mode})$ upper-bounds all others.
• Why: Shape-blind ➔ says nothing else; the mode can be far from mean/median.