Week 2.1: Functions of Random Variables

Prerequisites: Joint PMF (Week 1.1: Joint Probability Mass Function (Joint PMF)), Conditional Distributions (Week 1.3: Conditional Distributions) Cross-links: BSMA1002 (Stats 1) — transformations of RVs Core question: Given the distribution of $X$, what is the distribution of $Y = g(X)$?

1. Intuition: The "Push-Forward" Problem

• One-to-one functions: Each $X$ value maps to a unique $Y$ value (e.g., $Y = X + 5$, $Y = 2^X$). Probability transfers directly.
• Many-to-one functions: Multiple $X$ values map to the same $Y$ value (e.g., $Y = X^2$, $Y = (X-5)^2$). We must add the probabilities of all $X$ values that lead to the same $Y$.

2. General Method: The Table Method

Algorithm (PMF of $Y = g(X)$)

1. List all possible values of $X$ with their probabilities.
2. Compute $y = g(x)$ for each $x$.
3. Group by $y$: for each distinct $y$, sum the probabilities of all $x$ that map to it.
4. The result is $f_Y(y)$. Formula:

3. One-to-One Functions

Key result: If $g$ is one-to-one and $Y = g(X)$, then

The PMF shape "shifts" but probability mass is preserved exactly.

Example: Uniform $\{0,\dots,10\}$, $Y = X - 5$

Example: Binomial(10, 0.5), $Y = 2^X$

4. Many-to-One Functions

Example: $X \sim \text{Uniform}\{0,\dots,10\}$, $Y = (X-5)^2$

Example: Clipped Function

5. Functions of Multiple RVs

Example: Sum of Two Dice

Example: Max of Two Dice

6. Mermaid: Decision Flow for Finding $f_{g(X)}$

7. Practice Questions

Q1 (Easy)

Full Solution

One-to-one (linear). $Y$ takes values $1,4,7,10$, each with probability $1/4$.

$f_Y(1) = f_Y(4) = f_Y(7) = f_Y(10) = 1/4$.

Q2 (Medium)

Full Solution

Many-to-one. Mapping:

$Y=0$: $X=0$ (1 value) → $f_Y(0) = 1/7$ $Y=1$: $X=-1,1$ (2 values) → $f_Y(1) = 2/7$ $Y=2$: $X=-2,2$ (2 values) → $f_Y(2) = 2/7$ $Y=3$: $X=-3,3$ (2 values) → $f_Y(3) = 2/7$

Check: $1/7 + 2/7 + 2/7 + 2/7 = 1$. ✓

Q3 (Hard)

Full Solution

$f_X(x) = \binom{5}{x} (0.4)^x (0.6)^{5-x}$ for $x=0,1,2,3,4,5$.

Compute:

• $x=0 \to y=0$: $f_X(0) = (0.6)^5 = 0.07776$
• $x=1 \to y=1$: $f_X(1) = 5(0.4)(0.6)^4 = 0.2592$
• $x=2 \to y=4$: $f_X(2) = 10(0.4)^2(0.6)^3 = 0.3456$
• $x=3 \to y=9$: $f_X(3) = 10(0.4)^3(0.6)^2 = 0.2304$
• $x=4 \to y=16$: $f_X(4) = 5(0.4)^4(0.6) = 0.0768$
• $x=5 \to y=25$: $f_X(5) = (0.4)^5 = 0.01024$

This is one-to-one on $\{0,1,2,3,4,5\}$, so $f_Y(y) = f_X(\sqrt{y})$ for $y \in \{0,1,4,9,16,25\}$.

Q4 (Exam-style — Max)

Full Solution

$W = w$ when $(X,Y)$ is in the set $\{(x,y): \max(x,y)=w\}$. This means: either $X=w$ and $Y \leq w$, or $Y=w$ and $X \leq w$, but subtract the overlap $(w,w)$ counted twice.

Count of favourable outcomes: $w$ (for $X=w, Y=1..w$) $+ w$ (for $Y=w, X=1..w$) $- 1$ (double-count $(w,w)$) $= 2w-1$.

So $f_W(w) = \frac{2w-1}{36}$ for $w=1,\dots,6$.

Next topic: Week 2.2: Sum (Convolution), Min & Max of Random Variables — Deep dive into convolution, min, and max of independent RVs. Join Discord PreviousWeek 1.4: Multiple Random Variables (n > 2)NextWeek 2.2: Sum (Convolution), Min & Max of Random Variables