Exercises — Joint distributions — joint PMF - PDF, marginal, conditional

(a) Kya check karte hain: non-negativity aur total $=1$ dono (do defining conditions). Har entry $\ge 0$ hai. Sum: $0.05+0.05+0.10+0.10+0.20+0.10+0.10+0.15+0.15=\mathrm{1.00.}✓$ To ye ek valid joint PMF hai.

(b) Ye sirf padhna hai — row $X=1$, column $Y=1$ wala cell: $P(X=1,Y=1)=0.20$.

(c) Do cells kyun jodein: event $\{X=1,Y\le 1\}$ disjoint cells $\{X=1,Y=0\}$ aur $\{X=1,Y=1\}$ ka union hai. Disjoint $\Rightarrow$ add karo: $0.10+0.20=\mathrm{0.30.}$

Integral kyun nahi chahiye: ek region par constant density ka matlab hai "probability evenly spread." Total probability $=$ (height $k$) $\times$ (region ka area). Unit square ka area $1$ hai, isliye $k\times 1=1\Rightarrow k=1.$ Sanity picture: height $1$ ki ek flat sheet $1\times 1$ base par rakhi ho to volume $1$ hoga — aur density ke neeche ka volume hi total probability hai.

Recipe: $p_X(x)={\sum }_y{p}_{X,Y}(x,y)$ — ek row ke saath sum karo. $p_Y(y)={\sum }_x{p}_{X,Y}(x,y)$ — ek column mein neeche sum karo.

Rows (marginal of $X$): $p_X(0)=0.05+0.05+0.10=0.20,p_X(1)=0.10+0.20+0.10=0.40,p_X(2)=0.10+0.15+0.15=\mathrm{0.40.}$ Columns (marginal of $Y$): $p_Y(0)=0.05+0.10+0.10=0.25,p_Y(1)=0.05+0.20+0.15=0.40,p_Y(2)=0.10+0.10+0.15=\mathrm{0.35.}$ Check: $p_X$ ka sum $0.20+0.40+0.40=1.00$ aur $p_Y$ ka sum $0.25+0.40+0.35=1.00$ — dono consistent hain total probability $1$ ke saath. Hum margins mein summarize karte hain:

To $p_X=(0.20,0.40,0.40)$ aur $p_Y=(0.25,0.40,0.35)$. (Sums ko margins mein likhna hi literally wo jagah hai jahan se "marginal" naam aaya hai.)

Recipe: column $Y=1$ ko freeze karo, har entry ko us column ke total $p_Y(1)=0.40$ se divide karo. $p_{X\mid Y}(0\mid 1)=\frac{0.05}{0.40}=0.125,p_{X\mid Y}(1\mid 1)=\frac{0.20}{0.40}=0.50,p_{X\mid Y}(2\mid 1)=\frac{0.15}{0.40}=\mathrm{0.375.}$ Divide kyun karte hain: raw column ka sum $p_Y(1)=0.40$ hai, $1$ nahi. Divide karne se slice ek legitimate distribution mein rescale ho jaati hai. Check: $0.125+0.50+0.375=1.00✓$.

Pehle valid confirm karo: ${\int }_0^1{\int }_0^1(x+y)dxdy={\int }_0^1\left(\frac{1}{2}+y\right)dy=\frac{1}{2}+\frac{1}{2}=1✓$. Recipe: $y$ ko uske poore range $0$ se $1$ tak integrate out karo (ek fixed square hai, isliye bounds constant hain): $f_X(x)={\int }_0^1(x+y)dy=[xy+\frac{y^2}{2}{]}_0^1=x+\frac{1}{2},0<x<1,$ aur $f_X(x)=0$ for $x\le 0$ ya $x\ge 1$ (support ke bahar joint $0$ hai, to integral vanish ho jaata hai). Check karo ki 1 tak integrate hota hai: ${\int }_0^1(x+\frac{1}{2})dx=\frac{1}{2}+\frac{1}{2}=1✓$.

(a) Kyun: total probability $1$ honi chahiye. $0<x<y<1$ par, fixed $x$ ke liye $y$ variable $x$ se $1$ tak run karta hai (figure mein orange slice): ${\int }_0^1{\int }_x^{1}8xydydx={\int }_0^{1}8x[\frac{y^2}{2}{]}_x^{1}dx={\int }_0^{1}4x(1-x^2)dx=4(\frac{1}{2}-\frac{1}{4})=1.✓$

(b) Marginal of $X$ — $y$ collapse karo; fixed $x$ ke liye, $y\in (x,1)$: $f_X(x)={\int }_x^{1}8xydy=8x\cdot \frac{1-x^2}{2}=4x(1-x^2),0<x<1,$ aur $f_X(x)=0$ outside $(0,1)$. Marginal of $Y$ — $x$ collapse karo; fixed $y$ ke liye, $x\in (0,y)$ (triangle ki ek horizontal slice): $f_Y(y)={\int }_0^{y}8xydx=8y\cdot \frac{y^2}{2}=4y^3,0<y<1,$ aur $f_Y(y)=0$ outside $(0,1)$. Checks: ${\int }_0^{1}4x(1-x^2)dx=1$ aur ${\int }_0^{1}4y^{3}dy=1✓$.

(c) Conditional of $Y$ given $X=x$ — joint ko $f_X(x)$ se divide karo, support $y>x$ rakho: $f_{Y\mid X}(y\mid x)=\frac{8xy}{4x(1-x^2)}=\frac{2y}{1-x^2},x<y<1,$ aur $f_{Y\mid X}(y\mid x)=0$ for $y\le x$ ya $y\ge 1$. Support kyun shift hota hai: jab $X=x$ fix ho jaata hai, $Y$ sirf wahan live kar sakta hai jahan joint positive hai, yaani $y\in (x,1)$.

Test: independence $⟺f_{X,Y}(x,y)=f_X(x)f_Y(y)$ sab $(x,y)$ ke liye. $f_X(x)f_Y(y)=4x(1-x^2)\cdot 4y^3=16x{y}^3(1-x^2),$ lekin joint $8xy$ hai. Ye equal nahi hain, isliye $X,Y$ dependent hain. Gehri wajah (ye asli giveaway hai): support khud triangle $x<y$ hai — do variables ko jodne wala ek constraint. Jab bhi ek joint density ka support rectangle nahi hota (intervals ka product), variables independent nahi ho sakte, chahe formula kaisa bhi dikhta ho. Jaanna ki $X=0.9$ hai $Y>0.9$ force karta hai; definition ke hisaab se yahi dependence hai. Dekho Independence of Random Variables.

(a) Marginals (support $x<y$; fixed $x$ ke liye, $y\in (x,1)$; fixed $y$ ke liye, $x\in (0,y)$): $f_X(x)={\int }_x^{1}2dy=2(1-x),0<x<1;f_Y(y)={\int }_0^{y}2dx=2y,0<y<1,$ aur dono marginals $0$ hain outside $(0,1)$.

(b) Expectations (har marginal use karo — dekho Expectation and Variance): $E[X]={\int }_0^{1}x\cdot 2(1-x)dx=2(\frac{1}{2}-\frac{1}{3})=\frac{1}{3},$ $E[Y]={\int }_0^{1}y\cdot 2ydy=2\cdot \frac{1}{3}=\frac{2}{3}.$

(c) Cross-moment (triangle par $xy$ ko joint ke saath integrate karo): $E[XY]={\int }_0^1{\int }_0^{y}xy\cdot 2dxdy={\int }_0^{1}2y\cdot \frac{y^2}{2}dy={\int }_0^1{y}^{3}dy=\frac{1}{4}.$ Phir, Covariance and Correlation ke zariye, $\mathrm{Cov}(X,Y)=E[XY]-E[X]E[Y]=\frac{1}{4}-\frac{1}{3}\cdot \frac{2}{3}=\frac{1}{4}-\frac{2}{9}=\frac{9-8}{36}=\frac{1}{36}.$

(d) $\mathrm{Cov}=+\frac{1}{36}>0$: $X$ aur $Y$ saath mein move karte hain. Ye geometry se match karta hai — kyunki $x<y$ hamesha hota hai, ek bada $x$ ek bada $y$ force karta hai, isliye positive association hai.

Tool: chain rule $f_{X,Y}(x,y)=f_{X\mid Y}(x\mid y)f_Y(y)$.

• $f_Y(y)=1$ for $0<y<1$.
• Given $Y=y$, $X$ uniform hai $(0,y)$ par, isliye $f_{X\mid Y}(x\mid y)=\frac{1}{y}$ for $0<x<y$.

Multiply karo: $f_{X,Y}(x,y)=\frac{1}{y}\cdot 1=\frac{1}{y},0<x<y<1,$ aur $f_{X,Y}=0$ elsewhere. Support: triangle $0<x<y<1$ (wahi triangle jaise pehle). $f_X$ recover karo — $y$ collapse karo; fixed $x$ ke liye, $y$, $x$ se $1$ tak range karta hai: $f_X(x)={\int }_x^1\frac{1}{y}dy=[\ln y{]}_x^1=-\ln x=\ln \frac{1}{x},0<x<1,$ aur $f_X(x)=0$ outside $(0,1)$. Check karo ki 1 tak integrate hota hai: ${\int }_0^1(-\ln x)dx=[x-x\ln x{]}_0^1=1✓$. Ye exactly Law of Total Probability hai jo continuous world mein carry out ki gayi hai.

(a) Symmetry se $E[X]=E[Y]=0$ (density dono axes ke baare mein symmetric hai). $E[XY]$ ke liye: $E[XY]=\frac{1}{\pi }{\iint }_{x^2+y^2<1}xydxdy.$ Har fixed $x$ ke liye, $y$ symmetrically $(-\sqrt{1-x^2},\sqrt{1-x^2})$ par run karta hai, aur ${\int }_{-a}^{a}ydy=0$. To poora integral $0$ hai: $E[XY]=0\Rightarrow \mathrm{Cov}(X,Y)=0-0\cdot 0=0.$

(b) Independence test via support: support ek disk hai, rectangle nahi. Concretely, $f_X(x)=\frac{2\sqrt{1-x^2}}{\pi }$ on $(-1,1)$ (vertical chord ki length over $\pi$) aur $0$ elsewhere, aur similarly $f_Y(y)=\frac{2\sqrt{1-y^2}}{\pi }$ on $(-1,1)$. Unka product $f_X(x)f_Y(y)=\frac{4\sqrt{1-x^2}\sqrt{1-y^2}}{{\pi }^2}$ disk par joint $\frac{1}{\pi }$ ke equal nahi hai (aur e.g. $(0.9,0.9)$ par nonzero hai jo disk ke bahar hai jahan joint $0$ hai). Isliye dependent.

(c) Resolution: covariance sirf linear association detect karta hai. Yahan dependence purely geometric/nonlinear hai: jaanna ki $X=0.9$ hai $Y$ ko $|y|<\sqrt{1-0.81}=\sqrt{0.19}$ mein squeeze karta hai, to $X$, $Y$ ki range constrain karta hai — lekin symmetrically, isliye koi linear tilt nahi. Zero covariance ka matlab hai "koi linear trend nahi," kabhi nahi "independent." Sirf Bivariate Normal Distribution ke liye zero covariance actually independence force karta hai. Dekho bhi Independence of Random Variables.

Ye Bayes' Theorem hai continuous–discrete mixed form mein. Data ka conditional PMF given $\theta$ hai $P(D=1\mid \Theta =\theta )=\theta$.

Step 1 — joint (chain rule): $f_{\Theta ,D}(\theta ,1)=P(D=1\mid \theta )f_{\Theta }(\theta )=\theta \cdot 1=\theta$ for $0<\theta <1$ (aur $0$ elsewhere).

Step 2 — data ka marginal (Law of Total Probability, $\theta$ integrate out karo): $P(D=1)={\int }_0^1\theta d\theta =\frac{1}{2}.$

Step 3 — posterior (Bayes = joint ÷ marginal): $f_{\Theta \mid D}(\theta \mid 1)=\frac{\theta }{1/2}=2\theta ,0<\theta <1,$ aur $0$ outside $(0,1)$. Check: ${\int }_0^{1}2\theta d\theta =1✓$ — ek valid density.

Step 4 — posterior mean: $E[\Theta \mid D=1]={\int }_0^1\theta \cdot 2\theta d\theta =2\cdot \frac{1}{3}=\frac{2}{3}.$ Interpretation: ek defect dekhna $\Theta$ ke baare mein hamari belief ko prior mean $\frac{1}{2}$ se $\frac{2}{3}$ tak push karta hai — exactly wahi jo ek defect ka evidence karna chahiye.