Question bank — Transformations of random variables — change-of-variable technique

False — Jacobian $y$ ka function hota hai, isliye yeh density ko reshape karta hai; jaise $Y=X^2$ ke liye $\frac{1}{2\sqrt{y}}$ factor exactly wahi hai jo $y=0$ ke paas blow-up create karta hai, jo ki shape hai, sirf ek constant nahi.

True — $\left|dx/dy\right|=1/|a|$ $y$ mein constant hai, isliye $f_Y$ bas $f_X$ hai jo $b$ se shift hui, $|a|$ se stretch hui, aur height mein $1/|a|$ se scale hui; family (jaise normal) preserve rehti hai.

False — aapko inverse ka derivative chahiye, $\left|dx/dy\right|=1/|g^{\prime }(x)|$, isliye $|g^{\prime }|$ se divide karna chahiye, multiply nahi; $|g^{\prime }|$ se multiply karna stretch factor ko ulta kar deta hai.

False — discrete masses sirf relabel hote hain: $P(Y=g(x))=P(X=x)$ mein koi stretch factor nahi, kyunki koi "length" dilute nahi hoti; Jacobian purely ek density phenomenon hai.

True — ek monotonic $g$ mein har $y$ ke liye exactly ek root $x_1=g^{-1}(y)$ hoti hai, isliye sum mein ek hi term hoti hai aur yeh $f_X(g^{-1}(y))|dx/dy|$ mein collapse ho jaata hai.

True — equal MGFs ka matlab identical distributions hai, aur identically distributed inputs par same deterministic $g$ apply karne se identically distributed outputs milte hain.

False — branch drop karne se density har jagah half ho jaati hai, isliye yeh $1/2$ tak integrate hoti hai, 1 tak nahi; dono symmetric branches same $y$ mein mass dalti hain.

True — probability mass ka conservation matlab hai $\int f_{Y}dy=\int f_{X}dx=1$; ek failed normalization check ek reliable signal hai ki koi Jacobian ya branch missing hai.

Error hai absolute value ka missing hona; yeh $\frac{1}{|a|}$ hona chahiye, warna $a<0$ ke liye density negative aa jaati hai, jo impossible hai.

Error yeh hai ki $x^3$ ko non-injective keh rahe ho — cubing saare reals par strictly increasing hoti hai, isliye exactly ek real root $x=y^{1/3}$ hai aur sirf ek branch hai.

Do errors hain: absolute value non-monotonic hai jiske roots $x=\pm y$ hain, aur Jacobian $|dx/dy|=1$ ko har branch se multiply karna hoga; sahi answer hai $f_Y(y)=2f_X(y)$ for $y>0$.

Exponential $(-\infty ,\infty )$ ko $(0,\infty )$ par map karta hai, isliye $f_Y(y)=0$ for $y\le 0$; range restrict karna bhool jaane se density wahan assign ho jaati hai jahan koi mass aa hi nahi sakta.

Yeh actually sahi hai — dono $\pm \sqrt{y}$, $(-1,1)$ mein hain, isliye dono branches sum karna sahi hai; trap yeh hai ki yeh lagta hai pehle wale "one-sided" uniform case jaisa jahan sirf $+\sqrt{y}$ valid tha.

Formula inverse map $h=g^{-1}$ ka Jacobian use karta hai; forward wala use karne se reciprocal area-factor milta hai, jo $1/|\det J|$ se off hai jab tak aap reciprocal na lo.

Restricted domain $(-\pi /2,\pi /2)$ par, tan strictly increasing aur one-to-one hai, isliye single branch hai; periodicity tabhi matter karti hai jab support multiple periods span kare.

CDF hamesha well-defined aur monotone hoti hai chahe density awkward ho, aur $F_Y(y)=F_X(g^{-1}(y))$ differentiate karne se Jacobian automatically chain rule ke through mil jaata hai — "$dx$" ke baare mein koi hand-waving nahi chahiye.

Yeh ek integrable singularity hai: ${\int }_0^1\frac{1}{2\sqrt{y}}dy=[\sqrt{y}{]}_0^1=1$, isliye ek vanishingly thin sliver par infinite height bhi finite mass hold karta hai — squaring bahut se chhote $x$ ko ek tiny $y$-neighborhood mein compress kar deta hai.

Densities nonnegative mass-per-length hoti hain; $dx/dy$ ka sign sirf map ki direction record karta hai, aur CDF derivation mein inequality flip orientation ko already handle kar leta hai, magnitude ko true stretch factor ke roop mein chodke.

Mechanism forced hai: do symmetric normal branches har $y>0$ ko feed karti hain, aur $e^{-x^2/2}$ shape combined with $\frac{1}{2\sqrt{y}}$ Jacobian exactly $\frac{1}{\sqrt{2\pi y}}{e}^{-y/2}$ produce karta hai, jo definition se ${\chi }_1^2$ density hai.

Kyunki kisi bhi continuous $Y$ ke liye $F(Y)$ uniformly distributed hota hai; usse ulta chalate hue, ek uniform ko $F^{-1}$ mein feed karna target distribution exactly reconstruct karta hai — change-of-variable ka direct application jahan $g^{-1}=F$ hai.

Formula ko computable inverse $g^{-1}$ aur uska derivative chahiye; agar $g$ ka koi elementary inverse nahi hai (jaise $g(x)=x+e^x$), toh Jacobian aur density exist karte hain lekin closed form mein nahi likhe ja sakte.

LOTUS $\int g(x)f_X(x)dx$ directly compute karta hai bina kisi inversion ya Jacobian ke; change-of-variable machinery tabhi invoke karne layak hai jab aapko genuinely $f_Y$ khud chahiye ho, sirf ek number nahi.

Yeh break down ho jaata hai — $g$ ka koi inverse nahi hai aur $Y$ degenerate hai (5 par ek point mass), isliye koi density exist nahi karti; change-of-variable require karta hai ki $g$ support par non-constant aur invertible ho.

Inverse $x=y^{1/3}$ ka $y=0$ par infinite $dx/dy$ hai, isliye $f_Y$ wahan spike kar sakta hai; formula $y\ne 0$ ke liye pointwise hold karta hai, aur single measure-zero point probabilities affect nahi karta.

Sirf $x=-\sqrt{y}$ support mein hai, isliye square non-monotonic hone ke bawajood is domain par yeh one-to-one hai aur aap single branch use karte ho — roots ko hamesha actual support ke saath intersect karo.

Exactly zero — support mein koi $x$ wahan map nahi karta, isliye koi mass aa hi nahi sakta; yeh bhool jaana sabse common tarika hai jisse ek "valid-lagta-hua" $f_Y$ 1 tak integrate karne mein fail hota hai.

Aap ek given $y$ tak map karne wale saare $x$ ke point masses sum karte ho: $P(Y=y)={\sum }_{x:g(x)=y}P(X=x)$ — "sum over branches" ka discrete echo, lekin densities ki jagah masses ke saath.

$Y$ ek mixed variable ban jaata hai: $0$ par ek point mass (saare $X\le 0$ wahan pile up hote hain) aur $y>0$ ke liye ek continuous density; pure change-of-variable sirf continuous part handle karta hai, aur atom ko alag treat karte hain.

Haan — ek measure-zero exceptional set koi probability carry nahi karta, isliye Jacobian determinant formula almost everywhere hold karta hai aur density excluded curve se unaffected rehti hai.