Transformations of random variables — change-of-variable technique: Visual walkthrough

4.9.14 · D2 · Maths › Probability Theory & Statistics › Transformations of random variables — change-of-variable tec

Step 0 — Pehle kuch words pe agree karna zaroori hai

Koi bhi formula aane se pehle, chaar plain-language ideas. Hum inhe kabhi bhi iss meaning ke bina use nahi karenge.

Density-per-length wali baat sabse crucial hai. Density koi probability nahi hoti. Probability tab milti hai jab density ko length se multiply karo — wahi hai jam over a slab.

Step 1 — Probability jam hai, aur jam conserved hoti hai

KYA. Hum $x$ -ruler ke along ek fixed total amount of jam (total probability $= 1$ ) spread karte hain. Point $x$ pe jam ki height density $f_{X} (x)$ hai.

KYO. Aage jo bhi aayega woh ek hi idea par based hai: ruler ko smear karne se jam create ya destroy nahi hoti. Agar hum baad mein ruler ko stretch karein, wahi jam bas lambi ya choti stretch pe baith jaati hai — patli ya moti — lekin amount unchanged rehti hai.

PICTURE. Figure s01 mein lavender bell-shaped density $f_{X}$ $x$ -ruler ke upar dikhti hai, ek narrow coral slab $[x, x + d x]$ pe khada hai; arrows uski height $f_{X} (x)$ aur width $d x$ label karte hain, aur ek caption remind karta hai ki total shaded area 1 hai. Us chhoti slab ke upar, jam ek thin rectangle hai: $height = thickness of jam f_{X} (x) \times width of the slab d x = mass in that slab .$ Har symbol: $f_{X} (x)$ batata hai ki $x$ pe jam kitni oonchi khadi hai; $d x$ hai chhoti slab ki width; unka product hai sliver ka area = probability of landing in $[x, x + d x]$ .

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

Step 2 — Machine $g$ har slab ko naye jagah move karti hai

KYA. Machine point $x$ ko point $y = g (x)$ pe bhejti hai. Toh slab $[x, x + d x]$ carry hokar $y$ -ruler pe $[y, y + d y]$ slab ke roop mein land karti hai.

KYO. Hum $y$ -ruler pe jam ki thickness chahte hain, isliye literally follow karna padega ki $x$ -jam ka har slab kahan jaata hai.

PICTURE. Figure s02 mein lavender curve $y = g (x)$ draw hai; $x$ -axis pe width $d x$ ki ek coral slab hai, dashed lines curve tak jaati hain aur $y$ -axis tak cross karti hain, width $d y$ ki ek mint slab cast karti hain; ek note point out karta hai ki jahan curve steep hoti hai, $d y > d x$ hota hai. Chhoti $x$ -slab se upar trace karo, curve se milo, across trace karo: woh horizontal shadow hai slab $[y, y + d y]$ . Uski width $d y$ generally alag hoti hai $d x$ se — yahi difference poori kahani hai. $d y = slope of the machine g^{'} (x) d x .$ Yahan $g^{'} (x)$ curve ki steepness hai: steep curve ( $∣ g^{'} ∣$ bada) ek chhote $d x$ ko bade $d y$ mein badal deta hai — slab stretch hokar wide ho jaati hai.

Step 3 — Same jam, naya width ⇒ naya thickness

KYA. Jo jam $[x, x + d x]$ mein thi, exactly wahi ab $[y, y + d y]$ mein hai. Dono masses equal set karo: $mass after f_{Y} (y) ∣ d y ∣ = mass before f_{X} (x) ∣ d x ∣ .$ Har symbol: $f_{Y} (y)$ naye ruler pe unknown thickness hai (Step 0 mein define ki gayi $Y$ ki density); $∣ d y ∣$ uski slab width; right side pe purani thickness times purani width. Equal, kyunki wahi jam hai.

Absolute values kyun? Width ek length hai — hamesha positive. Agar machine "backwards" chalti ho (decreasing $g$ ), toh raw $d y$ sirf bookkeeping se negative aa sakta hai. Dono ko $∣ \cdot ∣$ mein wrap karna humein honest rakhta hai: hum slabs ke sizes compare karte hain, signed nahi.

PICTURE. Figure s03 mein side by side do equal-area rectangles hain: ek tall narrow coral wala (the $x$ -slab) aur ek short wide mint wala (the $y$ -slab), with a lavender arrow marked $g$ ek ko doosre par carry karta hai aur equation $f_{Y} (y) ∣ d y ∣ = f_{X} (x) ∣ d x ∣$ upar likhi hai. Equal area = conserved jam.

Unknown thickness ke liye divide karke solve karo: $f_{Y} (y) = f_{X} (x) \frac{d x}{d y} .$ Ratio $\frac{d x}{d y}$ hai squeeze factor: old width divided by new width. Wide naya slab ( $∣ d y ∣$ bada) ⇒ chhota factor ⇒ patli jam.

Step 4 — Lekin $y$ point ko name karta hai, $x$ ko nahi — machine ko invert karo

KYA. Formula abhi bhi $x$ mention karta hai. $y$ -ruler pe hum sirf $y$ jaante hain. Toh $x$ ko replace karo us $x$ se jo yeh $y$ produce karta hai: $x = g^{- 1} (y)$ .

Yeh tool — inverse — kyun? Hume chahiye "kaunsa input yeh output deta tha?". Woh sawal exactly wahi hai jo ek inverse function answer karta hai: $g^{- 1}$ $g$ ko undo karta hai. Koi aur tool "machine ko backwards chalao" ka jawab nahi deta. Yeh wahi logical move hai jo Inverse Transform Sampling aur parent note mein Cumulative Distribution Function-based derivation mein hai.

PICTURE. Figure s04 mein same lavender curve hai, lekin backwards padhi ja rahi hai: ek dashed mint line vertical axis pe ek chosen $y$ se start hoti hai, curve se milti hai, aur ek coral line $x$ -axis par girkar landing point $x = g^{- 1} (y)$ mark karti hai.

Squeeze factor $1/∣ g^{'} (x) ∣$ kyun hota hai. Hume $\frac{d}{d y} g^{- 1} (y)$ chahiye, lekin hum easily sirf $g$ ka slope, $g^{'} (x)$ , jaante hain. Woh reciprocals hain, aur yahan ek-line reason hai. By definition inverse machine ko undo karta hai, toh har $y$ ke liye: $g (g^{- 1} (y)) = y .$ Dono sides ko $y$ ke respect se differentiate karo. Right side deta hai $1$ . Left side, chain rule se (outer $g$ ko inner point par differentiate karo, times inner ki derivative), deta hai $g^{'} (g^{- 1} (y)) \cdot \frac{d}{d y} g^{- 1} (y)$ . Equal set karne par:

\quad\Longrightarrow\quad \frac{d}{dy}g^{-1}(y)=\frac{1}{g'(x)},\ \ x=g^{-1}(y).$$ Toh dono slopes multiply karke $1$ dete hain — har ek doosre ka reciprocal hai. Isliye ek steep machine ($|g'|$ bada) chhota squeeze factor deti hai: slab ko wide stretch karna jam ko dilute karta hai. Is step ke liye $g'(x)\ne0$ chahiye — exactly woh regularity jo Step 0 mein assume ki thi. Step 3 ke result mein $x=g^{-1}(y)$ substitute karo aur squeeze factor ko inverse ki derivative ke roop mein likho: > [!formula] Result, assembled (one-to-one $g$) > $$\boxed{\,f_Y(y)=f_X\!\big(g^{-1}(y)\big)\,\left|\dfrac{d}{dy}\,g^{-1}(y)\right|\,} > \;=\;f_X\!\big(g^{-1}(y)\big)\,\dfrac{1}{\big|g'(g^{-1}(y))\big|}$$ > - $g^{-1}(y)$ — "kaunsa $x$ iss $y$ ko banata tha", purane ruler ka point. > - $f_X(\cdots)$ — us point par padhi gayi purani jam thickness. > - $\left|\frac{d}{dy}g^{-1}(y)\right|=\left|\frac{dx}{dy}\right|=\frac{1}{|g'(x)|}$ — squeeze factor, slab kitni shrink hui. > > Yeh form valid hai **jab $g$ support par one-to-one ho**. Many-to-one form Step 7 ka sum hai. --- ## Step 5 — Machine ki dono directions (dono cases, drawn) **KYA.** Ek monotonic machine do tarahon mein se ek chalti hai: hamesha upar (increasing $g$) ya hamesha neeche (decreasing $g$). Hume check karna hai ki formula dono mein survive karta hai. **KYO.** Agar humne sirf increasing $g$ check kiya toh har decreasing map (jaise $Y=-X$ ya $Y=e^{-X}$) ke liye ek hidden bug reh jaata. Contract: *har* case cover karo. **PICTURE.** *Figure s05 ek side-by-side pair hai. Left: ek mint straight line upar slope karti hui — increasing $g$, "order kept", $dx/dy>0$. Right: ek coral straight line neeche slope karti hui — decreasing $g$, "ruler reversed", $dx/dy<0$, absolute value ki zaroorat hai.* - **Increasing (left).** Jaise $x$ badhta hai, $y$ badhta hai. Slab order preserve hoti hai; $\frac{dx}{dy}>0$; $|\cdot|$ kuch nahi karta. - **Decreasing (right).** Jaise $x$ badhta hai, $y$ *ghatta* hai. Slab end-for-end flip ho jaati hai; $\frac{dx}{dy}<0$. Raw, yeh **negative** thickness deta — nonsense. $|\cdot|$ sign flip karta hai, aur thickness positive rehti hai. ![[deepdives/dd-maths-4.9.14-d2-s05.png]] > [!mistake] Yeh sochna ki sign matter karta hai > $\frac{dx}{dy}$ ka sign sirf record karta hai "kya ruler flip hua?". Jam thickness kabhi negative nahi ho sakti, isliye hum size lete hain. Isliye boxed formula absolute-value bars pehanta hai — yeh silently dono upar wali pictures ko ek saath handle karta hai. --- ## Step 6 — Degenerate case: ek flat spot ($g'=0$) **KYA.** Maano kisi point par machine momentarily flat ho: $g'(x)=0$, toh $\frac{dx}{dy}=\frac{1}{g'(x)}\to\infty$. **Yeh dikhane ki zaroorat kyun.** Ek reader eventually aisa $g$ encounter karega jisme flat spot ho aur "infinity" pe panic karega. Unhe nahi karna chahiye. **PICTURE.** *Figure s06 mein lavender parabola $y=g(x)$ dikhti hai jo $x=0$ par flat hai; $0$ ke paas $x$-values ka ek wide coral band shaded hai, arrows note karte hain ki sab $y\approx0$ par crush ho jaate hain toh density wahan spike karti hai.* $x$-values ka poora band essentially ek hi $y$-value par squash ho jaata hai. Unki saari jam **wahan pile up** hoti hai — us $y$ par density genuinely spike karti hai infinity ki taraf. Yeh bug nahi; yeh ek real pile-up hai (ek spike jo phir bhi finite mass tak integrate hoti hai). ![[deepdives/dd-maths-4.9.14-d2-s06.png]] Humne parent mein ek live example dekha: $Y=X^2$ with $X\sim\text{Uniform}(0,1)$ deta hai $f_Y(y)=\frac{1}{2\sqrt y}$, jo $y\to 0$ par blow up karta hai exactly kyunki $g'(0)=0$ bahut saare chhote $x$'s ko $y\approx 0$ par squash karta hai. --- ## Step 7 — Folding case: machine do $x$'s ko ek $y$ par bhejti hai **KYA.** Agar $g$ one-to-one nahi hai (jaise $g(x)=x^2$), toh do alag slabs ek hi $y$-slab par land karti hain. Unki jam layers **stack** ho jaati hain. Step 4 ka one-to-one boxed result as-is apply nahi hota; woh ek **sum** ban jaata hai, ek term per source: $$\boxed{\,f_Y(y)=\sum_i f_X(x_i)\,\left|\frac{dx_i}{dy}\right|\,},\qquad x_i:\ g(x_i)=y.$$ Symbol by symbol: sum har us input $x_i$ ke upar run karta hai jo yeh $y$ produce karta hai; har ek apni purani thickness $f_X(x_i)$ times apna squeeze factor $\left|dx_i/dy\right|=1/|g'(x_i)|$ laata hai. Jab sirf ek hi root ho, sum mein ek term hota hai aur woh Step 4 ke box mein collapse ho jaata hai — toh yeh *general* form hai aur Step 4 uska special case hai. **Add kyun?** Mass conservation per-*source* hai: $x$-jam ki do independent strips ek hi $y$-bin mein jaati hain, isliye total = pour 1 + pour 2. **PICTURE.** *Figure s07 mein lavender parabola $y=x^2$ negative aur positive $x$-axes ko same $y$ par fold karti hui dikhti hai; ek chosen $y$ se dashed lines dono roots $-\sqrt y$ (coral) aur $+\sqrt y$ (mint) se milti hain, caption mein likha hai ki do layers add hoti hain.* Yeh machinery [[Chi-square Distribution]] ($X\sim$ [[Normal Distribution]], $Y=X^2$) ke peechhe hai. ![[deepdives/dd-maths-4.9.14-d2-s07.png]] $X\sim N(0,1)$ ke liye, dono roots $\pm\sqrt y$ equal thickness dete hain (bell symmetric hai), toh milta hai $2\times\frac{1}{\sqrt{2\pi}}e^{-y/2}\cdot\frac{1}{2\sqrt y}=\frac{1}{\sqrt{2\pi y}}e^{-y/2}$ — exactly $\chi^2_1$. --- ## The one-picture summary *Figure s08 poori derivation ko ek canvas par compress karta hai: ek tall narrow coral $x$-slab, ek lavender arrow marked $g$ (stretch) use carry karta hua, ek short wide mint $y$-slab same area ke saath, boxed formula $f_Y(y)=f_X(g^{-1}(y))\,|dx/dy|$ top par, aur ek footer teen special cases list karta hai: fold ⇒ add layers, flat spot ⇒ spike, $|\cdot|$ ⇒ positive rakhta hai.* ![[deepdives/dd-maths-4.9.14-d2-s08.png]] > [!recall]- Feynman retelling — poora walkthrough plain words mein > Ek rubber ruler ke along jam ka fixed blob smear karo; har mark par woh kitna ooncha khada hai woh density $f_X$ hai. Ek machine $g$ ruler ko naya shape mein stretch karti hai, aur naye ruler par jam ki height $f_Y$ hai. Jahan machine steep hai woh ruler ka ek chhota piece ek *wide* stretch mein kheench leti hai, toh wahi jam ab patli spread ho jaati hai; jahan gentle hai jam moti bunch ho jaati hai. Jam ki amount kabhi nahi badlti — sirf thickness — toh naya thickness equal hai purani thickness times kitna woh spot squeeze hua, $|dx/dy|$; aur woh squeeze factor hai $1/|g'|$, machine ke slope ka reciprocal (kyunki pehle forward phir backward chalana tumhe wahi wapas lata hai, unke slopes ka product ek hota hai). Kyunki ek naya mark $y$ yaad nahi rakhta ki kaunse purane mark $x$ ne use banaya, hum machine ko $g^{-1}$ se backwards chalate hain source dhundne ke liye. Agar machine downhill chali toh ruler flip ho jaata hai, jo sirf thickness ko negative banane ki dhamki deta hai, isliye hum absolute-value bars se size lete hain. Agar machine kisi jagah flat ho toh woh poori band ko ek mark par crush karti hai aur jam wahan spike karti hai. Aur agar woh do marks ko ek par fold kare, toh hum bas do jam layers add kar dete hain. Yahi poora change-of-variable technique hai — ek law: jam conserved hai. > [!recall] Quick self-check > Factor $\left|\tfrac{dx}{dy}\right|$ kyun hai aur $\left|\tfrac{dy}{dx}\right|$ kyun nahi? ::: Hume *naya* thickness paane ke liye old-width per new-width chahiye; woh ratio $dx/dy$ hai, slope $g'(x)$ ka reciprocal. > $\tfrac{d}{dy}g^{-1}(y)=1/g'(x)$ kyun? ::: $g(g^{-1}(y))=y$ ko chain rule se differentiate karo; dono slopes multiply karke 1 dete hain, toh har ek doosre ka reciprocal hai. > Fold mein average ki jagah add kyun? ::: Do independent strips apna poora mass ek hi bin mein dalti hain; total mass sum hai, mean nahi. > Flat spot par density spike kyun karti hai? ::: $g'\!=\!0$ bahut saare $x$'s ko ek $y$ par squash karta hai, unki saari jam wahan pile karke thickness $\to\infty$ kar deta hai. --- ## #flashcards/maths Squeeze factor ke ratio mein upar kaunsi slab-width jaati hai? ::: Purani width $dx$; factor hai $|dx/dy|$ = old-width per new-width. $x$-side aur $y$-side par kaunsi geometric quantity equal hoti hai? ::: Jam rectangle ka area (thickness × width) — conserved probability mass. Jab $g$ decreasing ho, absolute value kya fix karta hai? ::: $dx/dy$ mein ek sign flip jo otherwise negative (impossible) density deta. End mein $x=g^{-1}(y)$ kyun substitute karte hain? ::: $y$-ruler par hum sirf $y$ jaante hain; inverse batata hai kaunsa $x$ use banata tha taaki hum purani thickness padh sakein. Density kya hoti hai jahan $g'(x)=0$ ho? ::: Woh infinity ki taraf spike karti hai — $x$-values ka ek band ek $y$ par crush hota hai, jam pile ho jaati hai. 1D formula hold karne ke liye $g$ mein kya regularity chahiye? ::: Continuous, differentiable, piecewise strictly monotonic, with $g'\ne0$ (nonzero slope) har invertible stretch par.