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

Kya hai: ek affine map $Y=aX+b$ jisme $a=3,b=-5$. Formula kyun laagoo hota hai: $g(x)=3x-5$ strictly increasing hai, isliye yeh one-to-one hai — ek hi branch. Invert karo: $x=\frac{y+5}{3}$. Jacobian: $\frac{dx}{dy}=\frac{1}{3}$, magnitude $\frac{1}{3}$. $\boxed{f_Y(y)=\frac{1}{3}{f}_X\left(\frac{y+5}{3}\right).}$ $\frac{1}{3}$ density ko dilute karta hai kyunki axis ko factor $3$ se stretch karne par same mass teen gune length mein phail jaata hai.

Yahan $a=\frac{1}{3},b=-\frac{4}{3}$ hai. $f_Y(y)=\frac{1}{|a|}{f}_X\left(\frac{y-b}{a}\right)=3f_X(3y+4)$ use karke normal density ke saath yeh standard normal $N(0,1)$ ban jaata hai: $f_Y(y)=\frac{1}{\sqrt{2\pi }}{e}^{-y^2/2}$. Dekho Normal Distribution. Yeh exactly standardisation hai: mean ghatao, standard deviation se divide karo.

Monotonic hai? $g(x)=e^x$ strictly increasing hai → ek branch. Range: $e^x>0$, toh support hai $y>0$. Invert: $y=e^x\Rightarrow x=\ln y$. Jacobian: $\frac{dx}{dy}=\frac{1}{y}$ ($y>0$ ke liye positive). $f_Y(y)=f_X(\ln y)\cdot \frac{1}{y}=\frac{1}{y\sqrt{2\pi }}\exp \left(-\frac{(\ln y{)}^2}{2}\right),y>0.$ Yeh standard log-normal hai. Extra $\frac{1}{y}$ Jacobian hai, yeh optional nahi hai.

$g(x)=\sqrt{x}$ $(0,1)$ par increasing hai; range $y\in (0,1)$. Invert: $x=y^2$. Jacobian $\frac{dx}{dy}=2y$. Kyunki $f_X=1$ on $(0,1)$: $f_Y(y)=1\cdot 2y=2y,0<y<1.$ Check: ${\int }_0^{1}2ydy=1.$ ✓ Density $y=1$ ki taraf badhti hai — square root lene se chhote $x$ alag alag ho jaate hain aur bade $x$ $1$ ke paas pack ho jaate hain.

Neeche figure $g(x)=x^2$ ka fold support $[-1,2]$ par dikhata hai. Aise padho: vertical axis par ek height $y$ lo, horizontal line kheeencho, aur dekho kitni baar yeh blue parabola ko hit karti hai support ke andar dashed lines $x=-1$ aur $x=2$ ke beech mein. Yellow line $y=0.5$ par do baar hit karti hai ($x=-\sqrt{y}$ aur $x=+\sqrt{y}$ dono support ke andar), jabki red line $y=2.5$ par curve ko support ke andar ek baar hit karti hai (green dot $+\sqrt{y}$ par) kyunki doosra root $-\sqrt{y}\approx -1.58$ left of $x=-1$ ho jaata hai (red ✗, bahar). Yahi wajah hai ki density ke do alag pieces hain.

• $0<y<1$ ke liye: dono $+\sqrt{y}$ aur $-\sqrt{y}$ $[-1,2]$ mein hain (yellow line). Do branches.
• $1<y<4$ ke liye: $-\sqrt{y}<-1$ support ke bahar gir jaata hai (red ✗), toh sirf $+\sqrt{y}$ bachta hai. Ek branch.

Jacobian per branch: $\left|\frac{dx}{dy}\right|=\frac{1}{2\sqrt{y}}$, aur $f_X=\frac{1}{3}$ jahan bhi defined ho. $f_Y(y)=\{\begin{array}{ll}\frac{1}{3}\cdot \frac{1}{2\sqrt{y}}\cdot 2=\frac{1}{3\sqrt{y}},& 0<y<1,\\ \frac{1}{3}\cdot \frac{1}{2\sqrt{y}}=\frac{1}{6\sqrt{y}},& 1<y<4,\\ 0,& \text{otherwise.}\end{array}$ Total mass check: ${\int }_0^1\frac{1}{3\sqrt{y}}dy+{\int }_1^4\frac{1}{6\sqrt{y}}dy=\frac{2}{3}+\frac{1}{3}=1.$ ✓

Do roots $\pm \sqrt{y}$, dono full support $\mathbb{R}$ mein. $f_X$ even hai, isliye dono equally contribute karte hain: $f_Y(y)=2\cdot \frac{1}{\sqrt{2\pi }}{e}^{-y/2}\cdot \frac{1}{2\sqrt{y}}=\frac{1}{\sqrt{2\pi y}}{e}^{-y/2},y>0.$ Yeh chi-square with 1 degree of freedom hai.

CDF directly use karo (yahan densities se zyada clean hai): $P(Y\le y)=P(F^{-1}(U)\le y)=P(U\le F(y)).$ Beech ka step dono sides par $F$ apply karta hai — legal hai kyunki $F$ increasing hai isliye "$\le$" preserve hota hai. Kyunki $U$ $(0,1)$ par uniform hai, $P(U\le u)=u$ for $u\in [0,1]$. $u=F(y)\in [0,1)$ ke saath: $P(Y\le y)=F(y).$ Toh $Y$ ka exactly CDF $F$ hai. Isliye ek computer sirf uniform generator se koi bhi distribution produce kar sakta hai.

CDF invert karo: $u=1-e^{-\lambda y}\Rightarrow e^{-\lambda y}=1-u\Rightarrow y=-\frac{1}{\lambda }\ln (1-u)$. Toh $Y=-\frac{1}{\lambda }\ln (1-U)$. Kyunki $1-U$ bhi Uniform$(0,1)$ hai, tidy form $Y=-\frac{1}{\lambda }\ln U$ bhi kaam karta hai. Change-of-variable se density check ($u=e^{-\lambda y}$, $|du/dy|=\lambda e^{-\lambda y}$, $f_U=1$): $f_Y(y)=\lambda e^{-\lambda y},y>0.$ Exactly $\text{Exp}(\lambda )$.

Linear map invert karo: $x=\frac{u+v}{2},y=\frac{u-v}{2}$. Inverse ka Jacobian matrix banao: $J=\left(\begin{array}{cc}\partial x/\partial u& \partial x/\partial v\\ \partial y/\partial u& \partial y/\partial v\end{array}\right)=\left(\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& -\frac{1}{2}\end{array}\right),\det J=-\frac{1}{2},|\det J|=\frac{1}{2}.$ Determinant map ka area-scaling factor hai — yeh rotate-and-scale areas ko half karta hai. $\boxed{f_{U,V}(u,v)=\frac{1}{2}{f}_{X,Y}\left(\frac{u+v}{2},\frac{u-v}{2}\right).}$ Support: map $(x,y)\mapsto (u,v)$ poore plane ka linear bijection hai, toh $(U,V)$ saare ${\mathbb{R}}^2$ par range karta hai (jahaan transported $f_{X,Y}$ nonzero ho); $X,Y$ ke $\mathbb{R}$ par supported hone par yeh full ${\mathbb{R}}^2$ hai.

Joint density $f_{X,Y}=\frac{1}{2\pi }{e}^{-(x^2+y^2)/2}$. Inverse substitute karo. Pehle exponent expand karo, cross-terms dhyan se: ${\left(\frac{u+v}{2}\right)}^2+{\left(\frac{u-v}{2}\right)}^2=\frac{u^2+2uv+v^2}{4}+\frac{u^2-2uv+v^2}{4}=\frac{u^2+v^2}{2}.$ $\pm 2uv$ cross-terms exactly cancel ho jaate hain — yehi cheez $u$ aur $v$ ko cleanly separate hone deti hai. Ab $v$ integrate out karo: $f_U(u)={\int }_{-\infty }^{\infty }\frac{1}{2}\cdot \frac{1}{2\pi }{e}^{-(u^2+v^2)/4}dv=\frac{1}{4\pi }{e}^{-u^2/4}\underset{=2\sqrt{\pi }}{\underbrace{{\int }_{-\infty }^{\infty }{e}^{-v^2/4}dv}}.$ ${\int }_{-\infty }^{\infty }{e}^{-v^2/4}dv=2\sqrt{\pi }$ kyun: standard Gaussian integral ${\int }_{-\infty }^{\infty }{e}^{-\alpha x^2}dx=\sqrt{\pi /\alpha }$ use karo $\alpha =\frac{1}{4}$ ke saath, jo deta hai $\sqrt{\pi /(1/4)}=\sqrt{4\pi }=2\sqrt{\pi }$. Isliye $f_U(u)=\frac{1}{4\pi }{e}^{-u^2/4}\cdot 2\sqrt{\pi }=\frac{1}{\sqrt{4\pi }}{e}^{-u^2/4}.$ Yeh $N(0,2)$ hai (variance $2$). MGF cross-check: har $N(0,1)$ ka $M(t)=e^{t^2/2}$ hota hai; independence MGFs ko multiply karta hai, jo deta hai $e^{t^2/2}\cdot e^{t^2/2}=e^{t^2}$, yani $N(0,2)$ ka MGF. ✓ (Dekho Expectation and LOTUS ki MGFs kyun multiply hote hain.)