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

4.9.14 · D3 · Maths › Probability Theory & Statistics › Transformations of random variables — change-of-variable tec

Shuru karne se pehle, teen seedhi-saadhi reminders taaki neeche har symbol samajh aaye:

Scenario matrix

Is topic ka har problem inhi cells mein se kisi ek mein aata hai. Right column us worked example ka naam batata hai jo use cover karta hai.

#	Case class	Usmein kya khaas hai	Kaun cover karta hai
A	$g$ strictly increasing	inequalities ka direction nahi badalti; $\frac{d x}{d y} > 0$	Example 1
B	$g$ strictly decreasing	inequalities flip hoti hain; $\frac{d x}{d y} < 0$ , $	\cdot
C	$g$ non-monotonic / folding (do branches)	dono roots $\pm$ par sum karna padta hai	Example 3
D	$g$ poori ray ko ek point par map karta hai / degenerate & discrete mix	classic $Y = max (X, 0)$ : ek atom aata hai	Example 4
E	Support / range boundary matter karti hai; density blow up hoti hai edge par (limiting behaviour)	Example 5
F	Word problem (real-world units) — voltage → power	Example 6
G	Exam twist: composite / affine-then-square, $a$ ke sign dono taraf	Example 7

Hum saaton work karte hain. Har ek Forecast se shuru hota hai — steps padhne se pehle ek guess karo.

Example 1 — Cell A (increasing map)

Worked example Increasing affine:

$Y = 2 X + 1$ , with $X \sim Uniform (0, 1)$ Forecast: Uniform density $[0, 1]$ par height $1$ ka flat block hai. Axis ko $2$ se stretch karo. Kya naya block taller ya shorter hoga, aur kis interval par?

Step 1 — Inverse $g^{- 1}$ nikalo. Ye step kyun? Formula ko $g^{- 1} (y)$ chahiye, toh $y = 2 x + 1$ ko $x$ ke liye solve karo: $x = \frac{y - 1}{2}$ .

Step 2 — Inverse ko differentiate karo. Ye step kyun? Stretch factor $\frac{d x}{d y}$ hai, $g^{'}$ nahi. Yahan $\frac{d x}{d y} = \frac{1}{2}$ , magnitude $\frac{1}{2}$ .

Step 3 — Naya support nikalo. Ye step kyun? $g$ ki range ke bahar $f_{Y} = 0$ hoga. Jaise $x$ , $(0, 1)$ par chalta hai, $y = 2 x + 1$ , $(1, 3)$ par chalta hai.

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

Verify karo: integrate karo: $\int_{1}^{3} \frac{1}{2} d y = \frac{1}{2} \cdot 2 = 1$ . ✓ Block shorter ho gaya (height $\frac{1}{2}$ ) aur wider ho gaya (length $2$ ) — forecast confirmed: $2$ se stretch karne par density $2$ se dilute hoti hai.

Kyunki inequality kabhi flip nahi hui, yeh textbook increasing case hai: $F_{Y} (y) = F_{X} (g^{- 1} (y))$ directly. Dekho Cumulative Distribution Function ki CDF route kyun hamesha kaam karti hai.

Example 2 — Cell B (decreasing map, $∣ \cdot ∣$ apna kaam karta hai)

Worked example Decreasing affine:

$Y = 1 - X$ , with $X \sim Uniform (0, 1)$ Forecast: $Y = 1 - X$ bas interval $(0, 1)$ ko $\frac{1}{2}$ ke baare mein reflect karta hai. Kya density ki shape bilkul bhi badlegi?

Step 1 — Inverse. Kyun? $y = 1 - x$ solve karo: $x = 1 - y$ .

Step 2 — Inverse ka derivative. Kyun? $\frac{d x}{d y} = - 1$ . Yeh negative hai — ek decreasing map. CDF ke through raw formula mein minus sign aata hai kyunki inequality flip hoti hai: $g (X) \leq y ⟺ X \geq 1 - y$ , toh $F_{Y} (y) = 1 - F_{X} (1 - y)$ . Differentiate karo: $f_{Y} (y) = - f_{X} (1 - y) \cdot (- 1) = f_{X} (1 - y)$ . Dono minus signs cancel ho jaate hain — bilkul wahi jo $\frac{d x}{d y} = 1$ hamein deta hai.

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

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

Verify karo: $\int_{0}^{1} 1 d y = 1$ . ✓ Reflection length-preserving hoti hai (stretch factor $1$ ), toh density unchanged rehti hai — forecast confirmed.

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

Yeh classic two-root case hai. Parabola imagine karo.

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

Worked example Uniform ka square:

$Y = X^{2}$ , with $X \sim Uniform (- 1, 1)$ Forecast: Square karne se $x = + 0.5$ aur $x = - 0.5$ dono ek hi $y = 0.25$ par fold ho jaate hain. Do strips ek mein girti hain. Kya $y = 0$ ke paas $f_{Y}$ badi ya chhoti hogi?

Step 1 — $g (x) = y$ ke saare roots list karo. Ye step kyun? $g$ one-to-one nahi hai, toh har preimage par sum karte hain. $0 < y < 1$ ke liye, $x^{2} = y$ deta hai $x = + y$ aur $x = - y$ ; dono support $(- 1, 1)$ mein hain.

Step 2 — Har branch ka stretch factor. Kyun? Har root $x = \pm y$ ke liye $\frac{d x}{d y} = \pm \frac{1}{2 y}$ , dono ke liye magnitude $\frac{1}{2 y}$ hai.

Step 3 — Dono pourings ko sum karo. $f_{X} (x) = \frac{1}{2}$ on $(- 1, 1)$ ke saath: $f_{Y} (y) = branch + \frac{1}{2} \cdot \frac{1}{2 y} + branch - \frac{1}{2} \cdot \frac{1}{2 y} = \frac{1}{2 y}, 0 < y < 1.$

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

Verify karo: $\int_{0}^{1} \frac{1}{2 y} d y = [y]_{0}^{1} = 1$ . ✓ Yeh $y = 0$ ke paas blow up karta hai — chhote $x$ -values squaring ke baad $0$ ke paas crowd karte hain. Forecast: $y = 0$ ke paas bada, confirmed.

Parent ke Example 2 se compare karo (normal squared → chi-square): same folding mechanic, alag $f_{X}$ .

Example 4 — Cell D (ek ray ek point par crush ho jaati hai → atom aata hai)

Worked example Rectifier:

$Y = max (X, 0)$ , with $X \sim Uniform (- 1, 1)$ Forecast: $x < 0$ wali har cheez $y = 0$ par map ho jaati hai. Yeh ek poore interval ki probability ek single point par aana hai. Kya koi density yeh describe kar sakti hai?

Step 1 — Input ke behaviour ko split karo. Ye step kyun? $g$ invertible nahi hai: $x \geq 0$ par yeh identity hai (theek hai), lekin $x < 0$ par yeh $0$ par collapse karta hai (degenerate). Dono regions alag handle karo.

Step 2 — Collapsed part atom ban jaata hai. Kyun? Event ${Y = 0}$ equals ${X \leq 0}$ , jiski probability hai $P (Y = 0) = P (X \leq 0) = \int_{- 1}^{0} \frac{1}{2} d x = \frac{1}{2} .$ Ek point par baitha finite probability ek point mass (atom) hai, koi density carry nahi kar sakti. Toh $Y$ ek mix hai: $0$ par weight $\frac{1}{2}$ ka atom, plus ek continuous piece.

Step 3 — Identity part apni density rakhta hai. Kyun? $x > 0$ par, $y = x$ , toh $\frac{d x}{d y} = 1$ aur $f_{Y} (y) = f_{X} (y) \cdot 1 = \frac{1}{2}$ for $0 < y < 1$ .

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

Verify karo: total mass $= atom \frac{1}{2} + = 1/2 \int_{0}^{1} \frac{1}{2} d y = 1$ . ✓ Forecast confirmed: ek pure density ise describe nahi kar sakti — change-of-variable formula sirf smooth branch par apply hota hai, aur degenerate branch ko atom ke roop mein count karna padta hai.

Example 5 — Cell E (boundary / limiting behaviour)

Worked example Uniform ka log:

$Y = - ln X$ , with $X \sim Uniform (0, 1)$ Forecast: Jaise $x \to 0^{+}$ , $- ln x \to + \infty$ ; jaise $x \to 1^{-}$ , $- ln x \to 0^{+}$ . Toh $Y$ ka range $(0, \infty)$ hai. Kis end ke paas density pile up hogi?

Step 1 — Inverse. Kyun? $y = - ln x$ solve karo: dono sides exponentiate karo, $x = e^{- y}$ .

Step 2 — Stretch factor. $\frac{d x}{d y}$ kyun, $g^{'}$ nahi? Formula ko inverse ka derivative chahiye: $\frac{d x}{d y} = - e^{- y}$ , magnitude $e^{- y}$ .

Step 3 — Support / limits. Limits check kyun karo? Cell E sirf edges ke baare mein hai. $x \in (0, 1) \Rightarrow y \in (0, \infty)$ . Jaise $y \to 0^{+}$ , $f_{Y} \to 1$ (finite); jaise $y \to \infty$ , $f_{Y} \to 0$ (exponentially khatam) — yahan koi blow-up nahi, Example 3 ki tarah nahi.

Step 4 — Assemble karo. $f_{X} \equiv 1$ ke saath: $f_{Y} (y) = 1 \cdot e^{- y} = e^{- y}, y > 0.$ Yeh $Exp (1)$ hai — same Inverse Transform Sampling mechanic jaise parent ke Example 3 mein.

Verify karo: $\int_{0}^{\infty} e^{- y} d y = 1$ . ✓ Density $y = 0$ ke paas sabse zyada hai aur decay karti hai; infinite tail mein vanishing mass hai. Forecast confirmed.

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

Worked example Voltage se power. Ek noisy voltage

$V$ (volts) $1 Ω$ resistor par $V \sim N (0, 1)$ hai. Dissipated power hai $P = V^{2}$ (watts). $f_{P}$ nikalo. Forecast: Physically $P \geq 0$ hamesha, aur chhote voltages ( $0$ ke paas) sabse common hain. Power density kahan concentrate hogi?

Step 1 — Transform identify karo. Kyun? $P = g (V) = V^{2}$ — folding, do branches $v = \pm p$ for $p > 0$ . Same machinery jaise Example 3 mein, alag $f_{X}$ .

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

Step 3 — Branches sum karo. $f_{V} (v) = \frac{1}{2 π} e^{- v^{2} /2}$ ke saath ( $v$ mein even hai, toh dono roots equal terms dete hain): $f_{P} (p) = 2 \cdot \frac{1}{2 π} e^{- p /2} \cdot \frac{1}{2 p} = \frac{1}{2 π p} e^{- p /2}, p > 0.$ Yeh exactly $χ_{1}^{2}$ hai — standard normal voltage ko square karne par chi-square power milti hai.

Step 4 — Units sanity. $p$ watts mein hai; $f_{P} (p)$ ke units $1/ watt$ hain (density = probability per watt), jo $\frac{1}{p}$ ki dimension se match karta hai. ✓

Verify karo: $\int_{0}^{\infty} \frac{1}{2 π p} e^{- p /2} d p = 1$ . ✓ Mean power bhi $E [P] = E [V^{2}] = Var (V) = 1$ watt hai (yeh LOTUS ka kaam hai). Forecast confirmed: density $p = 0$ ke paas sabse zyada pile up hoti hai (chhote voltages common hain) aur ek heavy tail hai.

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

Worked example Composite transform:

$Y = (a X)^{2} = a^{2} X^{2}$ with $a \neq = 0$ and $X \sim N (0, 1)$ . Dikhao ki answer $a$ ke sign par depend nahi karta, aur $f_{Y}$ nikalo. Forecast: Kya $a \to - a$ flip karna $Y$ ki distribution change karta hai? Squaring signs ko khatam kar deti hai — compute karne se pehle guess karo.

Step 1 — Ek jaane-maane case tak reduce karo. Kyun? $Y = a^{2} X^{2}$ . Let $Z = X^{2}$ ; Example 6 / parent se, $Z \sim χ_{1}^{2}$ with $f_{Z} (z) = \frac{1}{2 π z} e^{- z /2}$ . Phir $Y = a^{2} Z$ ek increasing affine map hai (kyunki $a^{2} > 0$ hamesha) — Cell A machinery.

Step 2 — Scaling ka inverse. Kyun? $y = a^{2} z \Rightarrow z = \frac{y}{a ^{2}}$ , aur $\frac{d z}{d y} = \frac{1}{a ^{2}}$ , magnitude $\frac{1}{a ^{2}}$ . Note karo ki $a^{2}$ , $+ a$ aur $- a$ dono ke liye same hai — sign pehle hi chali gayi.

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

Step 4 — Sign check. Kyun? Jahan bhi $a$ aata hai woh $a^{2} = ∣ a ∣^{2}$ ke roop mein hai. Toh $f_{Y}$ , $a$ aur $- a$ ke liye identical hai.

Verify karo: $a = 2$ ke saath (toh $a^{2} = 4$ ): $f_{Y} (y) = \frac{1}{8 π y} e^{- y /8}$ aur $\int_{0}^{\infty} f_{Y} = 1$ . ✓ Forecast confirmed: $a$ ka sign irrelevant hai; sirf $a^{2}$ (variance scale) matter karta hai — yeh exactly ek scaled chi-square hai, yaani $N (0, a^{2})$ ko square karne ki law.

Recap: kaun sa cell kya sikhata hai

Recall Jaane se pehle self-test karo

Table cover karo. Har ek ke liye, cell ka naam aur woh ek cheez batao jo galat ho sakti hai. $Y = 3 X - 2$ ::: Cell A (increasing); naya support bhoolna mat. $Y = 5 - X$ ::: Cell B (decreasing); absolute value rakho warna density negative ho jaayegi. $Y = X^{2}$ , $X$ symmetric ::: Cell C; dono roots sum karo warna density half ho jaayegi. $Y = max (X, 0)$ ::: Cell D; positive-probability collapse ek atom create karta hai. $Y = - ln X$ ::: Cell E; range $(0, \infty)$ state karo aur check karo ki tail integrate hoti hai. $Y = (a X)^{2}$ ::: Cell G; sirf $a^{2}$ bachta hai, $a$ ka sign irrelevant hai.