Exercises — Independence of random variables — formal definition

Step 1 — support check karo. KYUN: independence ke liye product (rectangular) support chahiye. Yahan support $[0,1]\times [0,1]$ hai, jo ek sachcha rectangle hai. ✅ Step 2 — formula split karne ki koshish karo. $6x^{2}y=(6x^2)\cdot (y)$, ya cleaner $=(3x^2)(2y)$. Yeh $x$ ki function times $y$ ki function hai. ✅ Step 3 — conclude karo aur marginals padho. Factorisation shortcut se, $X\perp Y$. Har piece ko density mein normalise karo: $f_X(x)={\int }_0^{1}6x^{2}ydy=6x^2\cdot \frac{1}{2}=3x^2,f_Y(y)={\int }_0^{1}6x^{2}ydx=2y\cdot \frac{1}{2}\cdot {\int }_0^{1}6x^{2}dx\Rightarrow f_Y(y)=2y.$ Check: $f_X{f}_Y=3x^2\cdot 2y=6x^{2}y=f_{X,Y}$. ✅ Independent, $f_X(x)=3x^2$, $f_Y(y)=2y$ ke saath.

Step 1 — marginals. $p_X(0)=0.1+0.2=0.3$, $p_X(1)=0.7$; $p_Y(0)=0.4$, $p_Y(1)=0.6$. Step 2 — cell $(0,0)$ test karo. $p_X(0)p_Y(0)=0.3\cdot 0.4=0.12$, lekin table mein $0.1$ hai. $0.12\ne 0.1$. ✗ Conclusion: ek failing cell kaafi hai — NOT independent.

Step 1 — normalise karo. KYUN: ek density ko 1 tak integrate hona chahiye. ${\int }_0^{\infty }{\int }_0^{\infty }c{e}^{-2x}{e}^{-3y}dxdy=c\cdot \frac{1}{2}\cdot \frac{1}{3}=\frac{c}{6}=1\Rightarrow c=6.$ Step 2 — factor karo. $6e^{-2x}{e}^{-3y}=(2e^{-2x})(3e^{-3y})$ rectangle $(0,\infty {)}^2$ par. ✅ Independent. Step 3 — marginals. $f_X(x)=2e^{-2x}$ (Exp rate 2), $f_Y(y)=3e^{-3y}$ (Exp rate 3). Product joint ke barabar hai. ✅

Step 1 — expectations ka product. KYUN: independence se $\mathbb{E}[XY]=\mathbb{E}[X]\mathbb{E}[Y]$ milta hai. $\mathbb{E}[XY]=2\cdot (-1)=-2.$ Step 2 — sum ka variance. Generally $\mathrm{Var}(X+Y)=\mathrm{Var}(X)+\mathrm{Var}(Y)+2\mathrm{Cov}(X,Y)$. Independence ⇒ $\mathrm{Cov}=0$. $\mathrm{Var}(X+Y)=1+4+0=5.$ (Covariance term ke liye Covariance and correlation dekho.)

Step 1 — support dekho (figure). KYUN pehle: shortcut theorem ke liye rectangular support zaroori hai. Shaded region ek triangle ($x\le y$) hai, box nahi. Pehle se suspicious: agar $X=0.8$ pata hai toh $Y\ge 0.8$ force hoti hai — information leak ho rahi hai. Step 2 — marginals honestly compute karo. Fixed $x$ ke liye, $y$ ka range $[x,1]$ hai: $f_X(x)={\int }_x^{1}8xydy=8x\cdot \frac{1-x^2}{2}=4x(1-x^2),0\le x\le 1.$ Fixed $y$ ke liye, $x$ ka range $[0,y]$ hai: $f_Y(y)={\int }_0^{y}8xydx=8y\cdot \frac{y^2}{2}=4y^3,0\le y\le 1.$ Step 3 — compare karo. $f_X(x)f_Y(y)=4x(1-x^2)\cdot 4y^3=16x{y}^3(1-x^2)$, jo $8xy$ ke barabar nahi hai. ✗ Conclusion: NOT independent. Separable-looking formula ek decoy tha; triangular support ne coupling kar di.

Step 1 — enumerate karo. Chaar $(X,Y)$ points hain $(1,0),(0,1),(-1,0),(0,-1)$, har ek probability $\frac{1}{4}$ ke saath. Step 2 — means. $\mathbb{E}[X]=\frac{1}{4}(1+0-1+0)=0$; $\mathbb{E}[Y]=\frac{1}{4}(0+1+0-1)=0$. Step 3 — $\mathbb{E}[XY]$. Chaaon points mein se har ek par, ek coordinate $0$ hai, toh $XY=0$ hamesha. Is liye $\mathbb{E}[XY]=0$ aur $\mathrm{Cov}=\mathbb{E}[XY]-\mathbb{E}[X]\mathbb{E}[Y]=0$. Uncorrelated. ✅ Step 4 — dependence. $P(X=1)=\frac{1}{4}$ aur $P(Y=0)=\frac{1}{2}$, toh independent hote toh $P(X=1,Y=0)$ hota $\frac{1}{4}\cdot \frac{1}{2}=\frac{1}{8}$. Lekin actual value $\frac{1}{4}$ hai (point $(1,0)$). $\frac{1}{4}\ne \frac{1}{8}$. ✗ Conclusion: zero covariance, phir bhi dependent — covariance sirf linear structure detect karta hai.

Step 1 — convolution kyun. Independent $X,Y$ ke liye, $Z=X+Y$ ki density convolution hai (dekho Sums of independent random variables — convolution): $f_Z(z)={\int }_{-\infty }^{\infty }{f}_X(t)f_Y(z-t)dt.$ Independence hi allow karta hai ki $f_X(t)f_Y(z-t)$ ko integral ke andar multiply karein. Step 2 — plug in karo. $f_X=f_Y=1$ on $[0,1]$, baaki $0$. Integrand $1$ tabhi hoga jab $0\le t\le 1$ aur $0\le z-t\le 1$, yaani $\max (0,z-1)\le t\le \min (1,z)$. Step 3 — case split (saare cases!).

• $z<0$ ya $z>2$: koi overlap nahi, $f_Z=0$.
• $0\le z\le 1$: $t\in [0,z]$, length $z$, toh $f_Z(z)=z$.
• $1\le z\le 2$: $t\in [z-1,1]$, length $2-z$, toh $f_Z(z)=2-z$. Yeh triangular density hai jo $z=1$ par peak karti hai (figure). Step 4 — evaluate karo. $f_Z(0.5)=0.5$; $f_Z(1)=2-1=1$ (peak).

Step 1 — discrete convolution. $P(X+Y=n)={\sum }_{k=0}^{n}P(X=k)P(Y=n-k)$ independence use karke. Step 2 — classic identity. Independent Poissons add hote hain: $X+Y\sim \text{Poisson}(\lambda +\mu )=\text{Poisson}(5)$. (Proof: sum ${\sum }_k\frac{e^{-\lambda }{\lambda }^k}{k!}\frac{e^{-\mu }{\mu }^{n-k}}{(n-k)!}=e^{-(\lambda +\mu )}\frac{(\lambda +\mu {)}^n}{n!}$ binomial theorem se.) Step 3 — read off karo. $\mathbb{E}[X+Y]=\lambda +\mu =5$. $P(X+Y=0)=e^{-5}\approx 0.006738$.

Goal. Dikhaao $P(g(X)\in C,h(Y)\in D)=P(g(X)\in C)P(h(Y)\in D)$ sabhi Borel $C,D$ ke liye. Step 1 — sets ko pull back karo. KYUN: $g(X)$ ke baare mein questions ko $X$ ke baare mein questions mein convert karo. Maano $A=g^{-1}(C)=\{x:g(x)\in C\}$ aur $B=h^{-1}(D)$. Yeh Borel sets hain kyunki $g,h$ measurable hain. Tab $\{g(X)\in C\}=\{X\in A\},\{h(Y)\in D\}=\{Y\in B\}.$ Step 2 — $X\perp Y$ ki definition apply karo. Independence har pair of Borel sets ke liye hold karta hai, khaaskar $A,B$ ke liye: $P(X\in A,Y\in B)=P(X\in A)P(Y\in B).$ Step 3 — wapas translate karo. $P(g(X)\in C,h(Y)\in D)=P(X\in A,Y\in B)=P(X\in A)P(Y\in B)=P(g(X)\in C)P(h(Y)\in D).$ Kyunki $C,D$ arbitrary the, $g(X)\perp h(Y)$. $■$ Insight: independence kisi bhi per-variable transformation ke under preserve hoti hai — alag-alag har variable ko relabel karke dependence create nahi ki ja sakti.

Step 1 — idea. KYUN koi bhi function nahi: hamen chahiye ki yeh 1 tak integrate ho, non-negative rahe, uniform marginals ho, phir bhi $x$ aur $y$ ko couple kare. Ek perturbation trick kaam karta hai: $f_{X,Y}(x,y)=1+\alpha (2x-1)(2y-1),(x,y)\in [0,1{]}^2,$ $\alpha =\frac{1}{2}$ ke saath (koi bhi $|\alpha |\le 1$ use kar sakte hain jo ise $\ge 0$ rakhe, kyunki $|(2x-1)(2y-1)|\le 1$). Step 2 — non-negativity. $\alpha =\frac{1}{2}$ ke saath: minimum value $1-\frac{1}{2}=\frac{1}{2}>0$ hai. ✅ Step 3 — $X$ ka marginal. $y$ integrate out karo; dhyaan do ${\int }_0^1(2y-1)dy=0$: $f_X(x)={\int }_0^1[1+\alpha (2x-1)(2y-1)]dy=1+\alpha (2x-1)\cdot 0=1.$ Toh $X\sim U[0,1]$. Symmetry se $f_Y(y)=1$, $Y\sim U[0,1]$. ✅ Step 4 — dependence. Independent hote toh $f_{X,Y}=f_X{f}_Y=1$ har jagah hota. Lekin $(x,y)=(1,1)$ par: $f_{X,Y}=1+\frac{1}{2}(1)(1)=1.5\ne 1$. ✗ Conclusion: uniform marginals, rectangular support, phir bhi dependent — kyunki joint factor refuse karta hai chahe har margin innocent dikhti ho.

Upar har problem ek hi sawaal hai alag-alag kapdon mein: kya joint sach mein marginals ke product ke barabar hai, har point par, ek rectangular region par? Agar haan — independent, aur zindagi aasaan hai (multiply karo, variances add karo, sums convolve karo). Agar support bent hai (triangle, circle) ya formula split nahi hota, toh woh gossip karte hain.