Transformations of random variables — change-of-variable technique: Worked examples

Before we start, three plain-word reminders so every symbol below is earned:

The scenario matrix

Every problem this topic throws lives in one of these cells. The right column names the worked example that hits it.

#	Case class	What is special about it	Hit by
A	$g$ strictly increasing	inequalities keep their direction; $\frac{d x}{d y} > 0$	Example 1
B	$g$ strictly decreasing	inequalities flip; $\frac{d x}{d y} < 0$ , the $	\cdot
C	$g$ non-monotonic / folding (two branches)	must sum over both roots $\pm$	Example 3
D	$g$ maps a whole ray to a point / degenerate & discrete mix	classic $Y = max (X, 0)$ : an atom appears	Example 4
E	Support / range boundary matters; density blows up at an edge (limiting behaviour)	Example 5
F	Word problem (real-world units) — voltage → power	Example 6
G	Exam twist: composite / affine-then-square, sign of $a$ both ways	Example 7

We work all seven. Each begins with a Forecast — commit to a guess before reading the steps.

Example 1 — Cell A (increasing map)

Worked example Increasing affine:

$Y = 2 X + 1$ , with $X \sim Uniform (0, 1)$ Forecast: The uniform density is a flat block of height $1$ on $[0, 1]$ . Stretch the axis by $2$ . Will the new block be taller or shorter, and on what interval?

Step 1 — Find the inverse $g^{- 1}$ . Why this step? The formula feeds on $g^{- 1} (y)$ , so solve $y = 2 x + 1$ for $x$ : $x = \frac{y - 1}{2}$ .

Step 2 — Differentiate the inverse. Why this step? The stretch factor is $\frac{d x}{d y}$ , not $g^{'}$ . Here $\frac{d x}{d y} = \frac{1}{2}$ , magnitude $\frac{1}{2}$ .

Step 3 — Find the new support. Why this step? $f_{Y} = 0$ outside the range of $g$ . As $x$ runs over $(0, 1)$ , $y = 2 x + 1$ runs over $(1, 3)$ .

Step 4 — Assemble. With $f_{X} \equiv 1$ on $(0, 1)$ : $f_{Y} (y) = 1 \cdot \frac{1}{2} = \frac{1}{2}, 1 < y < 3.$

Verify: integrate: $\int_{1}^{3} \frac{1}{2} d y = \frac{1}{2} \cdot 2 = 1$ . ✓ The block got shorter (height $\frac{1}{2}$ ) and wider (length $2$ ) — forecast confirmed: stretching by $2$ dilutes the density by $2$ .

Because the inequality never flipped, this is the textbook increasing case: $F_{Y} (y) = F_{X} (g^{- 1} (y))$ directly. See Cumulative Distribution Function for why the CDF route always works.

Example 2 — Cell B (decreasing map, the $∣ \cdot ∣$ earns its keep)

Worked example Decreasing affine:

$Y = 1 - X$ , with $X \sim Uniform (0, 1)$ Forecast: $Y = 1 - X$ just reflects the interval $(0, 1)$ about $\frac{1}{2}$ . Will the density change shape at all?

Step 1 — Inverse. Why? Solve $y = 1 - x$ : $x = 1 - y$ .

Step 2 — Derivative of inverse. Why? $\frac{d x}{d y} = - 1$ . This is negative — a decreasing map. The raw formula through the CDF gives a minus sign because the inequality flips: $g (X) \leq y ⟺ X \geq 1 - y$ , so $F_{Y} (y) = 1 - F_{X} (1 - y)$ . Differentiating: $f_{Y} (y) = - f_{X} (1 - y) \cdot (- 1) = f_{X} (1 - y)$ . The two minus signs cancel — exactly what $\frac{d x}{d y} = 1$ packages for us.

Step 3 — Support. $x \in (0, 1) \Rightarrow y = 1 - x \in (0, 1)$ .

Step 4 — Assemble. $f_{Y} (y) = 1 \cdot ∣ - 1∣ = 1, 0 < y < 1.$

Verify: $\int_{0}^{1} 1 d y = 1$ . ✓ A reflection is length-preserving (stretch factor $1$ ), so the density is unchanged — forecast confirmed.

Example 3 — Cell C (folding map, sum over branches)

This is the classic two-root case. Picture the parabola.

Figure — Transformations of random variables — change-of-variable technique

Worked example Square of a uniform:

$Y = X^{2}$ , with $X \sim Uniform (- 1, 1)$ Forecast: Squaring folds both $x = + 0.5$ and $x = - 0.5$ onto the same $y = 0.25$ . Two strips pour into one. Should $f_{Y}$ near $y = 0$ be large or small?

Step 1 — List all roots of $g (x) = y$ . Why this step? $g$ is not one-to-one, so we sum over every preimage. For $0 < y < 1$ , $x^{2} = y$ gives $x = + y$ and $x = - y$ ; both lie in the support $(- 1, 1)$ .

Step 2 — Stretch factor per branch. Why? Each root $x = \pm y$ has $\frac{d x}{d y} = \pm \frac{1}{2 y}$ , magnitude $\frac{1}{2 y}$ for both.

Step 3 — Sum the two pourings. With $f_{X} (x) = \frac{1}{2}$ on $(- 1, 1)$ : $f_{Y} (y) = branch + \frac{1}{2} \cdot \frac{1}{2 y} + branch - \frac{1}{2} \cdot \frac{1}{2 y} = \frac{1}{2 y}, 0 < y < 1.$

Step 4 — Support. $Y = X^{2} \in [0, 1)$ , so $f_{Y} = 0$ elsewhere.

Verify: $\int_{0}^{1} \frac{1}{2 y} d y = [y]_{0}^{1} = 1$ . ✓ It blows up near $y = 0$ — small $x$ -values crowd near $0$ after squaring. Forecast: large near $y = 0$ , confirmed.

Compare with the parent's Example 2 (a normal squared → chi-square): same folding mechanic, different $f_{X}$ .

Example 4 — Cell D (a ray crushed to a point → an atom appears)

Worked example Rectifier:

$Y = max (X, 0)$ , with $X \sim Uniform (- 1, 1)$ Forecast: Everything with $x < 0$ gets mapped to $y = 0$ . That is a whole interval of probability landing on a single point. Can a density describe that?

Step 1 — Split the input's behaviour. Why this step? $g$ is not invertible: on $x \geq 0$ it is the identity (nice), but on $x < 0$ it collapses to $0$ (degenerate). Handle the two regions separately.

Step 2 — The collapsed part becomes an atom. Why? The event ${Y = 0}$ equals ${X \leq 0}$ , which has probability $P (Y = 0) = P (X \leq 0) = \int_{- 1}^{0} \frac{1}{2} d x = \frac{1}{2} .$ A finite probability sitting on one point is a point mass (atom), not something a density can carry. So $Y$ is a mix: an atom of weight $\frac{1}{2}$ at $0$ , plus a continuous piece.

Step 3 — The identity part keeps its density. Why? On $x > 0$ , $y = x$ , so $\frac{d x}{d y} = 1$ and $f_{Y} (y) = f_{X} (y) \cdot 1 = \frac{1}{2}$ for $0 < y < 1$ .

Step 4 — Write the full law. $Y = {0 density \frac{1}{2} with probability \frac{1}{2} (atom) on (0, 1) .$

Verify: total mass $= atom \frac{1}{2} + = 1/2 \int_{0}^{1} \frac{1}{2} d y = 1$ . ✓ Forecast confirmed: a pure density cannot describe it — the change-of-variable formula only applies to the smooth branch, and the degenerate branch must be counted as an atom.

Example 5 — Cell E (boundary / limiting behaviour)

Worked example Log of a uniform:

$Y = - ln X$ , with $X \sim Uniform (0, 1)$ Forecast: As $x \to 0^{+}$ , $- ln x \to + \infty$ ; as $x \to 1^{-}$ , $- ln x \to 0^{+}$ . So $Y$ ranges over $(0, \infty)$ . Near which end will density pile up?

Step 1 — Inverse. Why? Solve $y = - ln x$ : exponentiate both sides, $x = e^{- y}$ .

Step 2 — Stretch factor. Why $\frac{d x}{d y}$ and not $g^{'}$ ? The formula needs the derivative of the inverse: $\frac{d x}{d y} = - e^{- y}$ , magnitude $e^{- y}$ .

Step 3 — Support / limits. Why check limits? Cell E is all about the edges. $x \in (0, 1) \Rightarrow y \in (0, \infty)$ . As $y \to 0^{+}$ , $f_{Y} \to 1$ (finite); as $y \to \infty$ , $f_{Y} \to 0$ (dies exponentially) — no blow-up here, unlike Example 3.

Step 4 — Assemble. With $f_{X} \equiv 1$ : $f_{Y} (y) = 1 \cdot e^{- y} = e^{- y}, y > 0.$ This is $Exp (1)$ — the same Inverse Transform Sampling mechanic as the parent's Example 3.

Verify: $\int_{0}^{\infty} e^{- y} d y = 1$ . ✓ Density is heaviest near $y = 0$ and decays; the infinite tail carries vanishing mass. Forecast confirmed.

Example 6 — Cell F (word problem, real units)

Worked example Voltage to power. A noisy voltage

$V$ (volts) across a $1 Ω$ resistor is $V \sim N (0, 1)$ . The power dissipated is $P = V^{2}$ (watts). Find $f_{P}$ . Forecast: Physically $P \geq 0$ always, and tiny voltages (near $0$ ) are the most common. Where does power density concentrate?

Step 1 — Identify the transform. Why? $P = g (V) = V^{2}$ — folding, two branches $v = \pm p$ for $p > 0$ . Same machinery as Example 3, different $f_{X}$ .

Step 2 — Stretch factor per branch. $v = \pm p \Rightarrow \frac{d v}{d p} = \pm \frac{1}{2 p}$ , magnitude $\frac{1}{2 p}$ .

Step 3 — Sum the branches. With $f_{V} (v) = \frac{1}{2 π} e^{- v^{2} /2}$ (even in $v$ , so both roots give equal terms): $f_{P} (p) = 2 \cdot \frac{1}{2 π} e^{- p /2} \cdot \frac{1}{2 p} = \frac{1}{2 π p} e^{- p /2}, p > 0.$ That is exactly $χ_{1}^{2}$ — squaring a standard normal voltage yields chi-square power.

Step 4 — Units sanity. $p$ is in watts; $f_{P} (p)$ has units of $1/ watt$ (density = probability per watt), matching $\frac{1}{p}$ 's dimension. ✓

Verify: $\int_{0}^{\infty} \frac{1}{2 π p} e^{- p /2} d p = 1$ . ✓ Also the mean power is $E [P] = E [V^{2}] = Var (V) = 1$ watt (this is LOTUS in action). Forecast confirmed: density piles up near $p = 0$ (small voltages are common) and has a heavy tail.

Example 7 — Cell G (exam twist: affine-then-square, both signs of $a$ )

Worked example Composite transform:

$Y = (a X)^{2} = a^{2} X^{2}$ with $a \neq = 0$ and $X \sim N (0, 1)$ . Show the answer does not depend on the sign of $a$ , and find $f_{Y}$ . Forecast: Does flipping $a \to - a$ change the distribution of $Y$ ? Squaring kills signs — guess before computing.

Step 1 — Reduce to a known case. Why? $Y = a^{2} X^{2}$ . Let $Z = X^{2}$ ; from Example 6 / parent, $Z \sim χ_{1}^{2}$ with $f_{Z} (z) = \frac{1}{2 π z} e^{- z /2}$ . Then $Y = a^{2} Z$ is an increasing affine map (since $a^{2} > 0$ always) — Cell A machinery.

Step 2 — Inverse of the scaling. Why? $y = a^{2} z \Rightarrow z = \frac{y}{a ^{2}}$ , and $\frac{d z}{d y} = \frac{1}{a ^{2}}$ , magnitude $\frac{1}{a ^{2}}$ . Note $a^{2}$ is the same for $+ a$ and $- a$ — the sign is already gone.

Step 3 — Assemble. $f_{Y} (y) = f_{Z} (\frac{y}{a ^{2}}) \cdot \frac{1}{a ^{2}} = \frac{1}{2 π \cdot y / a ^{2}} e^{- y / (2 a^{2})} \cdot \frac{1}{a ^{2}} = \frac{1}{2 π a ^{2} y} e^{- y / (2 a^{2})}, y > 0.$

Step 4 — Sign check. Why? Everywhere $a$ appears it is as $a^{2} = ∣ a ∣^{2}$ . So $f_{Y}$ for $a$ and $- a$ are identical.

Verify: with $a = 2$ (so $a^{2} = 4$ ): $f_{Y} (y) = \frac{1}{8 π y} e^{- y /8}$ and $\int_{0}^{\infty} f_{Y} = 1$ . ✓ Forecast confirmed: the sign of $a$ is irrelevant; only $a^{2}$ (the variance scale) matters — this is exactly a scaled chi-square, i.e. the law of $N (0, a^{2})$ squared.

Recap: which cell teaches what

Recall Self-test before you leave

Cover the table. For each, name the cell and the one thing that could go wrong. $Y = 3 X - 2$ ::: Cell A (increasing); don't forget new support. $Y = 5 - X$ ::: Cell B (decreasing); keep the absolute value or density goes negative. $Y = X^{2}$ , $X$ symmetric ::: Cell C; sum both roots or you halve the density. $Y = max (X, 0)$ ::: Cell D; a positive-probability collapse creates an atom. $Y = - ln X$ ::: Cell E; state the range $(0, \infty)$ and check the tail integrates. $Y = (a X)^{2}$ ::: Cell G; only $a^{2}$ survives, sign of $a$ irrelevant.