Transformations of random variables — change-of-variable technique

4.9.14 · Maths › Probability Theory & Statistics

WHY kyu chahiye yeh?

Hum constantly nayi randomness ko purani se build karte hain: square karo ( $Y = X^{2}$ energy ke liye), scale karo ( $Y = σ X + μ$ standardize karne ke liye), invert karo, log lo. $f_{X}$ jaanna $Y$ ke baare mein questions answer karne ke liye bekaar hai jab tak hum density ko $g$ ke paar transport na kar sakein.

KYA conserve ho raha hai?

Probability mass, density nahi. $[x, x + d x]$ par baithi probability ki ek tiny slab ka mass $f_{X} (x) d x$ hai. Map ke baad woh $[y, y + d y]$ par rehti hai. Masses match karni chahiye: $mass after f_{Y} (y) ∣ d y ∣ = mass before f_{X} (x) ∣ d x ∣$

Yeh ek equation hi poori technique hai.

Derivation from scratch (monotonic $g$ )

Maano $Y = g (X)$ jahan $g$ strictly monotonic hai (toh uska inverse $x = g^{- 1} (y)$ exist karta hai). Sabse saaf derivation CDF ke through jaati hai, kyunki CDFs hamesha kaam karti hain.

Case: $g$ increasing. Tab $g (X) \leq y ⟺ X \leq g^{- 1} (y)$ , isliye $F_{Y} (y) = P (Y \leq y) = P (X \leq g^{- 1} (y)) = F_{X} (g^{- 1} (y)) .$ Chain rule se differentiate karo (yeh step kyun? density CDF ki derivative hoti hai): $f_{Y} (y) = f_{X} (g^{- 1} (y)) \cdot \frac{d}{d y} g^{- 1} (y) .$

Case: $g$ decreasing. Ab $g (X) \leq y ⟺ X \geq g^{- 1} (y)$ (inequality flip ho jaati hai), isliye $F_{Y} (y) = 1 - F_{X} (g^{- 1} (y)), f_{Y} (y) = - f_{X} (g^{- 1} (y)) \frac{d}{d y} g^{- 1} (y) .$ Yahan $\frac{d}{d y} g^{- 1} < 0$ hai, toh minus sign density ko positive banata hai.

Dono cases ek mein collapse ho jaate hain absolute value use karke:

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

Non-monotonic $g$ : branches par sum karo

Agar $g$ one-to-one nahi hai, toh kaafi saare $x$ -values ek hi $y$ par map hote hain. Har preimage $x_{i} = g_{i}^{- 1} (y)$ apna khud ka mass ka slab contribute karta hai, isliye hum unhe add karte hain: $f_{Y} (y) = \sum_{i} f_{X} (x_{i}) \frac{d x _{i}}{d y}, x_{i} ranges over all roots of g (x) = y .$

Worked examples

Bivariate / general case (Jacobian matrix)

$(U, V) = g (X, Y)$ ke liye jo invertible hai aur jiska inverse $(x, y) = h (u, v)$ hai:

J=\begin{pmatrix}\partial x/\partial u & \partial x/\partial v\\ \partial y/\partial u & \partial y/\partial v\end{pmatrix}.$$ Wahi conservation idea: $|\det J|$ map ka **area-scaling factor** hai (ek 2D stretch). --- ## Common mistakes > [!mistake] Jacobian ko poori tarah bhool jaana > **Galat idea:** "Bas $x=g^{-1}(y)$ ko $f_X$ mein plug karo: $f_Y(y)=f_X(g^{-1}(y))$." > **Kyun sahi lagta hai:** Yeh simple substitution lagti hai, jaise algebra mein hoti hai. > **Fix:** Densities mass-*per-unit-length* hain; variable badalne se length ki unit badal jaati hai, isliye tumhe $\left|\frac{dx}{dy}\right|$ se zaroor multiply karna hoga. Sanity check: tumhari $f_Y$ ko 1 integrate karna chahiye. Jacobian ke bina usually nahi karega. > [!mistake] Galat function differentiate karna > **Galat:** $\left|\frac{dy}{dx}\right| = |g'(x)|$ ko directly multiply karna. > **Kyun sahi lagta hai:** $g'$ woh function hai jo tumhe diya gaya tha, isliye yeh natural lagta hai. > **Fix:** formula ko $\left|\frac{dx}{dy}\right| = \frac{1}{|g'(x)|}$ chahiye, *inverse* ki derivative. Ye dono reciprocals hain. Jo bhi compute kar sako use karo, lekin end mein yeh $dx/dy$ hona chahiye. > [!mistake] Non-monotonic maps mein ek branch drop kar dena > **Galat:** $Y=X^2$ ko sirf $x=+\sqrt y$ ke saath treat karna. > **Fix:** $g(x)=y$ ke *saare* roots list karo aur unke contributions add karo. $x=-\sqrt y$ miss karna tumhari density aadhi kar deta hai. > [!mistake] Support / range ignore karna > **Galat:** $f_Y$ ko saare $y$ ke liye likhna. > **Fix:** $f_Y$ sirf wahan nonzero hoti hai jahan $X$ ke support mein $x$ ke liye $y=g(x)$ ho. Jaise $Y=X^2\ge 0$, toh $f_Y(y)=0$ for $y<0$. --- ## Forecast-then-Verify drill > [!recall] Compute karne se pehle predict karo > Maano $X\sim\text{Uniform}(0,1)$ aur $Y=X^2$. > 1. $f_Y$ $y=0$ ke paas badi hogi ya $y=1$ ke paas? (Forecast.) > 2. Ise compute karo. > > *Answer:* roots: sirf $x=+\sqrt y$ (kyunki $X>0$). $\frac{dx}{dy}=\frac{1}{2\sqrt y}$, $f_X=1$. Toh $f_Y(y)=\frac{1}{2\sqrt y}$ on $(0,1)$. Yeh **$y=0$ ke paas blow up karta hai** — sahi forecast: square karne par chhote $X$ values $0$ ke paas compress ho jaate hain, wahan density pile up ho jaati hai. --- ## #flashcards/maths Change-of-variable technique kaunsi ek quantity conserve karti hai? ::: Probability mass, $f_X(x)\,|dx| = f_Y(y)\,|dy|$. 1D monotonic change-of-variable formula state karo. ::: $f_Y(y)=f_X(g^{-1}(y))\left|\frac{d}{dy}g^{-1}(y)\right|$. Jacobian par absolute value kyun hai? ::: Densities nonnegative rehni chahiye; decreasing $g$ ke liye derivative negative hoti hai, aur $|\cdot|$ dono monotonic cases ko ek formula mein bhi pack karta hai. Non-monotonic $g$ ke liye $f_Y$ kaise banate hain? ::: Saare roots $x_i$ of $g(x)=y$ par sum karo: $\sum_i f_X(x_i)\left|dx_i/dy\right|$. $Y=aX+b$ ke liye $f_Y$ kya hai? ::: $f_Y(y)=\frac{1}{|a|}f_X\!\left(\frac{y-b}{a}\right)$. Agar $X\sim N(0,1)$, toh $Y=X^2$ ki distribution kya hai? ::: Chi-square with 1 df: $f_Y(y)=\frac{1}{\sqrt{2\pi y}}e^{-y/2}$, $y>0$. $U\sim\text{Uniform}(0,1)$ se $\text{Exp}(\lambda)$ kaise generate karte hain? ::: $Y=-\frac1\lambda\ln U$; tab $f_Y(y)=\lambda e^{-\lambda y}$. Bivariate case mein Jacobian factor ki jagah kya aata hai? ::: Inverse map ke Jacobian matrix ka absolute determinant, $|\det J|$ (area-scaling factor). Final density formula mein $|dy/dx|$ use karte hain ya $|dx/dy|$? ::: $|dx/dy|$ — inverse ki derivative; yeh $|g'(x)|$ ka reciprocal hai. $f_Y$ ko $g$ ke range tak restrict kyun karna chahiye? ::: Support ke bahar $g(\text{support})$ mein koi probability mass land nahi kar sakta, isliye wahan $f_Y=0$ hoti hai. --- > [!recall]- Feynman: ek 12-saal ke bachche ko explain karo > Socho tum ek ruler par fix amount ki jam (probability) lagaate ho. Ab tum ruler ko rubber ki tarah stretch karte ho: jahan stretch karo, jam patli ho jaati hai; jahan squeeze karo, jam zyada thick pile up ho jaati hai. Jam ki *amount* kabhi nahi badlti — sirf har jagah kitni thick hai woh badlti hai. Har point par "stretch amount" Jacobian $|dx/dy|$ hai. Change-of-variable formula bas yeh kehta hai: nayi thickness = purani thickness × (wahan kitna squeeze hua). Agar tumhara function ruler ko fold karta hai toh do purani jagah ek nayi jagah par aa jaati hain, tum bas jam ki dono layers add kar dete ho. > [!mnemonic] > **"Invert, Differentiate, Absolute, Sum, Support"** → **I D A S S** ("I Did A Smart Switch"). > $g$ ko Invert karo, inverse ko Differentiate karo, Absolute value lo, branches par Sum karo, Support tak restrict karo. ## Connections - [[Cumulative Distribution Function]] — safe fallback method (CDF technique) jo is formula ko *derive* karne ke liye use hoti hai. - [[Jacobian Determinant]] — $|dx/dy|$ ka multivariable generalization. - [[Normal Distribution]] — standardization $Z=\frac{X-\mu}{\sigma}$ ek linear transform hai. - [[Chi-square Distribution]] — standard normal ke $X^2$ se milta hai. - [[Inverse Transform Sampling]] — $X=F^{-1}(U)$ use karta hai, ek direct application. - [[Expectation and LOTUS]] — alternative jab tumhe sirf moments chahiye, poori density nahi. - [[Moment Generating Functions]] — transforms/sums ki distributions find karne ka ek aur raasta. ## 🖼️ Concept Map ```mermaid flowchart TD X[Known dist of X] -->|apply Y = g of X| Q[What is dist of Y?] CONS[Probability mass conserved] -->|core principle| EQ[fY dy = fX dx] Q -->|answered by| EQ EQ -->|via CDF, g monotonic| DER[Differentiate FY] DER -->|chain rule| FORM[Change-of-variable formula] FORM -->|contains| JAC[Jacobian abs dx/dy] JAC -->|is the| STRETCH[Stretch factor of density] FORM -->|g not one-to-one| SUM[Sum over branches] SUM -->|add contributions| PRE[Each preimage xi] FORM -->|example Y = aX+b| LIN[Affine: divide by abs a] ``` ## 🔬 Deep Dive > [!intuition] Aur gehraai mein jao — visual, zero se > Is topic ke step-by-step 3Blue1Brown-style breakdowns. - [[4.9.14 D1 Foundations|D1 · Foundations — har symbol zero se]] - [[4.9.14 D2 Visual Walkthrough|D2 · Visual walkthrough — pictures mein derivation]] - [[4.9.14 D3 Worked Examples|D3 · Worked examples — har scenario]] - [[4.9.14 D4 Exercises|D4 · Exercises — graded, full solutions]] - [[4.9.14 D5 Question Bank|D5 · Question bank — concept traps]]