Random Variable

Context: FIT1058_MOC · a numerical function of a random outcome · "$X=k$" is an event · has a probability distribution, summarised by Expectation

Quick Revision

• 🎯 Objective: a function $X:U\to \mathbb{R}$ of a random outcome ➔ "$X=k$" is an event.
• 📦 Core Components: distribution $\mathrm{Pr}(X=k)$ ➔ sums/independence ➔ summarised by Expectation.
• ⚡ Key Constraint: write $\mathrm{Pr}(X=k)$, not $\mathrm{Pr}(X)$ — $X$ has a distribution, not one probability.

📝 Core

1. The Variable

• Definition ➔ a function $X:U\to \mathbb{R}$; randomness is in the drawn outcome $x$, $X$ deterministic.
• Event ➔ "$X=k$" is $\{x:X(x)=k\}$, $\mathrm{Pr}(X=k)={\sum }_{x:X(x)=k}\mathrm{Pr}(x)$.

2. Distribution

• Definition ➔ the values of $X$ with their probabilities.
• Notation ➔ $\mathrm{Pr}(X=k)$, not $\mathrm{Pr}(X)$ ($X$ is not an event).
• Self-contained ➔ values + distribution are all you need.

3. Combining

• Sum ➔ $\mathrm{Pr}(X+Y=k)={\sum }_{i+j=k}\mathrm{Pr}((X=i)\wedge (Y=j))$.
• Independence ➔ $\mathrm{Pr}((X=x)\wedge (Y=y))=\mathrm{Pr}(X=x)\mathrm{Pr}(Y=y)$ for all $x,y$.

Key identities:

$\mathrm{Pr}(X=k)={\sum }_{x:X(x)=k}\mathrm{Pr}(x),\mathrm{Pr}(X+Y=k)={\sum }_{i+j=k}\mathrm{Pr}((X=i)\wedge (Y=j))$

⚖️ Core Decision Matrix

Object	Has	Note
event $\{X=k\}$	a probability	$\mathrm{Pr}(X=k)$
random variable $X$	a distribution	not one number
sum $X+Y$	a new distribution	convolution
independence	per-value product	all $(x,y)$

When It Flips: a sample space breaks outcomes into equiprobable atoms; a random variable extracts a useful number, lumping many outcomes (all pairs summing to 9). The distribution of $X+Y$ generally differs from the parts.

📊 Exam Execution Trace

Applied Exercise

Problem: $Z=$ sum of two fair dice — find $\mathrm{Pr}(Z=9)$ and $\mathrm{Pr}(Z=4)$. Derivation Proof / Hand-Calculation Walkthrough:

$$\begin{array}{rl}\mathrm{Pr}(Z=9)& =\frac{4}{36}=\frac{1}{9},\mathrm{Pr}(Z=4)=\frac{3}{36}=\frac{1}{12}\end{array}$$

Final Extracted Output: each = (pairs with that sum)/36; $Z$‘s distribution is non-uniform.

⚠️ Common Mistakes

• 💡 Never "$\mathrm{Pr}(X)$" ➔ $X$ ranges over many values; only events like $\{X=k\}$ have a probability. A variable is never independent of itself.

🧠 Active Recall

What is a random variable, and why write $\mathrm{Pr}(X=k)$ rather than $\mathrm{Pr}(X)$?

• Hint: Function, not event.

Answer

• Short answer: $X:U\to \mathbb{R}$; "$X=k$" is an event; $X$ has a whole distribution.
• Why: Many values ➔ only events have single probabilities.

How is the distribution of $X+Y$ found, and when are $X,Y$ independent?

• Hint: Convolution + per-value product.

Answer

• Short answer: $\mathrm{Pr}(X+Y=k)={\sum }_{i+j=k}\mathrm{Pr}((X=i)\wedge (Y=j))$; independent iff the product holds for every $(x,y)$.
• Why: New distribution ➔ two uniform dice give a triangular total.