Worked examples — Joint distributions — joint PMF - PDF, marginal, conditional

4.9.10 · D3 · Maths › Probability Theory & Statistics › Joint distributions — joint PMF - PDF, marginal, conditional

Scenario matrix

Har joint-distribution problem in cases mein se ek hoti hai. Right column us worked example ka naam deta hai jo use cover karta hai.

Hum saaton cells neeche cover karte hain.

Example 1 — Discrete table (Cell C1)

Step 1 — check karo ki valid PMF hai. Yeh step kyon? Table tabhi distribution hoti hai jab saari entries $\ge 0$ hon aur sum $1$ ho; warna saara downstream kaam bekaar hai. $0.05+0.10+0.05+0.15+0.30+0.15+0.05+0.05+0.10=\mathrm{1.00.}✓$

Step 2 — marginal of $X$ (har row ka sum). Yeh step kyon? $p_X(x)={\sum }_y{p}_{X,Y}(x,y)$ — "$X=x$ aur $Y$ kuch bhi" matlab Law of Total Probability $Y$-values ke upar. $p_X(0)=0.20,p_X(1)=0.60,p_X(2)=\mathrm{0.20.}$

Step 3 — marginal of $Y$ (har column ka sum). $p_Y(0)=0.25,p_Y(1)=0.45,p_Y(2)=\mathrm{0.30.}$

Step 4 — conditional. Yeh step kyon? $Y=1$ freeze karo: sirf us column ko dekho $(0.10,0.30,0.05)$, phir uske total $p_Y(1)=0.45$ se divide karo taaki slice re-normalize ho jaye (dekho Conditional Probability). $P(X=1\mid Y=1)=\frac{p_{X,Y}(1,1)}{p_Y(1)}=\frac{0.30}{0.45}=\frac{2}{3}\approx \mathrm{0.6667.}$

Step 5 — independence test. Yeh step kyon? Independence ek statement hai joint ke factoring ke baare mein, kabhi akele marginals ke baare mein nahi. Ek cell test karo. $p_{X,Y}(0,0)=0.05\text{vs}{p}_X(0)p_Y(0)=0.20\times 0.25=\mathrm{0.05.}$ Ek matching cell kaafi nahi — doosra check karo: $p_{X,Y}(2,2)=0.10\text{vs}{p}_X(2)p_Y(2)=0.20\times 0.30=\mathrm{0.06.}\ne$ Kyunki ek cell fail karta hai, $X,Y$ dependent hain.

Verify: rows ka sum $0.20,0.60,0.20$ (total $1$); columns ka sum $0.25,0.45,0.30$ (total $1$); conditional column $(0.10,0.30,0.05)/0.45$ ka sum $1$ hai. ✔

Example 2 — Independent case (Cell C2)

Step 1 — joint multiply karke banao. Yeh step kyon? Independence of Random Variables ka matlab hi hai $p_{X,Y}(x,y)=p_X(x)p_Y(y)$ — joint literally marginals ka outer product hota hai.

Step 2 — total check karo. $0.28+0.12+0.42+0.18=\mathrm{1.00.}✓$

Step 3 — independence ki tell-tale: conditional = marginal. Yeh step kyon? Independence ke under, $Y$ jaanna $X$ ke baare mein kuch nahi batata, isliye slice marginal ke barabar hoti hai. $P(X=1\mid Y=0)=\frac{0.42}{0.28+0.42}=\frac{0.42}{0.70}=0.6=p_X(1).✓$

Verify: conditional $0.6$ exactly marginal $p_X(1)=0.6$ ke barabar hai — independence ka fingerprint. ✔

Example 3 — Continuous, rectangular support (Cell C3)

Step 1 — $c$ nikalo total probability $=1$ se. Yeh step kyon? Har density ko apne support par $1$ integrate karna chahiye; isse constant pin hota hai. ${\int }_0^1{\int }_0^{1}c(x+y)dxdy=c{\int }_0^1(\frac{1}{2}+y)dy=c(\frac{1}{2}+\frac{1}{2})=c=1\Rightarrow c=1.$

Step 2 — marginal of $X$ ($y$ ko integrate out karo, constant bounds $0$ to $1$). Yeh step kyon? $f_X(x)=\int f_{X,Y}dy$; rectangle matlab $y$ hamesha pura $0\to 1$ chalता hai. $f_X(x)={\int }_0^1(x+y)dy=x+\frac{1}{2},0<x<1.$

Step 3 — marginal of $Y$ (symmetry se). $f_Y(y)={\int }_0^1(x+y)dx=y+\frac{1}{2},0<y<1.$

Step 4 — conditional. Yeh step kyon? Joint ko conditioning variable ke marginal se divide karo. $f_{X\mid Y}(x\mid y)=\frac{x+y}{y+\frac{1}{2}},0<x<1.$

Step 5 — independent? $f_X(x)f_Y(y)=(x+\frac{1}{2})(y+\frac{1}{2})\ne x+y=f_{X,Y}(x,y).$ Toh dependent hain — ek clean square par bhi, ek additive density factor nahi karta.

Verify: ${\int }_0^1{f}_{X}dx={\int }_0^1(x+\frac{1}{2})dx=1$ ✔; ${\int }_0^1{f}_{X\mid Y}(x\mid y)dx=\frac{\frac{1}{2}+y}{y+\frac{1}{2}}=1$ har $y$ ke liye ✔.

Example 4 — Continuous, triangular support (Cell C4)

Step 1 — picture se bounds padho. Yeh step kyon? Region hai $\{0<x<y<1\}$: fixed $x$ ke liye, $y$ $x$ se $1$ tak jaata hai; fixed $y$ ke liye, $x$ $0$ se $y$ tak jaata hai. Constant bounds galat answer denge. Figure mein amber slice dekho.

Step 2 — confirm total $=1$. ${\int }_0^1{\int }_0^{y}8xydxdy={\int }_0^{1}8y\cdot \frac{y^2}{2}dy={\int }_0^{1}4y^{3}dy=1.✓$

Step 3 — marginal of $X$ (fixed $x$ ke liye, $y$ $x$ se $1$ tak). $f_X(x)={\int }_x^{1}8xydy=8x\cdot \frac{1-x^2}{2}=4x(1-x^2),0<x<1.$

Step 4 — marginal of $Y$ (fixed $y$ ke liye, $x$ $0$ se $y$ tak). $f_Y(y)={\int }_0^{y}8xydx=8y\cdot \frac{y^2}{2}=4y^3,0<y<1.$

Step 5 — conditional of $Y$ given $X=x$. Yeh step kyon? Hum $P(Y>0.5\mid X=0.25)$ chahte hain, toh $X$ par condition karo. Joint ko $f_X$ se divide karo; support ban jaata hai $y\in (x,1)$. $f_{Y\mid X}(y\mid x)=\frac{8xy}{4x(1-x^2)}=\frac{2y}{1-x^2},x<y<1.$

Step 6 — $x=0.25$ par requested probability. $f_{Y\mid X}(y\mid 0.25)=\frac{2y}{1-0.0625}=\frac{2y}{0.9375}.$ $P(Y>0.5\mid X=0.25)={\int }_{0.5}^1\frac{2y}{0.9375}dy=\frac{1}{0.9375}[y^2{]}_{0.5}^1=\frac{1-0.25}{0.9375}=\frac{0.75}{0.9375}=\mathrm{0.8.}$

Verify: ${\int }_x^1{f}_{Y\mid X}(y\mid x)dy=\frac{1-x^2}{1-x^2}=1$ ✔; numeric answer $0.8$ neeche re-check kiya gaya hai.

Example 5 — Degenerate / zero conditioning (Cell C5)

Step 1 — conditioning value par marginal evaluate karo. Yeh step kyon? Conditional $f_{X\mid Y}(x\mid y)=f_{X,Y}(x,y)/f_Y(y)$ sirf wahan defined hai jahan $f_Y(y)>0$. $f_Y(y)=4y^3\Rightarrow f_Y(0)=0,f_Y(1.5)=0(\text{support ke bahar}).$

Step 2 — undefined declare karo. Dono mein $0/0$ ya "$0$ over $0$-support" situation aati hai: $f_{X\mid Y}(x\mid 0)=\frac{0}{0}\text{— undefined},f_{X\mid Y}(x\mid 1.5)=\frac{0}{0}\text{— undefined.}$

Verify: $f_Y(0)=4\cdot 0^3=0$ aur support require karta hai $y<1$, isliye $y=1.5$ ki zero density hai — dono conditionals undefined hain. ✔ (Koi numeric answer nahi; "answer" hai undefined.)

Example 6 — Limiting behaviour (Cell C6)

Step 1 — $X$ given $Y=y$ ka conditional likho. Yeh step kyon? Is baar $Y$ par condition karo: joint ko $f_Y(y)=4y^3$ se divide karo, support $0<x<y$. $f_{X\mid Y}(x\mid y)=\frac{8xy}{4y^3}=\frac{2x}{y^2},0<x<y.$

Step 2 — limit $y\to 1^{-}$. $f_{X\mid Y}(x\mid y)\to 2x,0<x<1.$ Puri interval par ek wide, gently increasing slope — picture ki cyan curve.

Step 3 — limit $y\to 0^{+}$. Yeh step kyon? Shape dekho: shrinking interval $0<x<y$ par density $2x/y^2$ hai. $x=y$ par iska peak $2/y\to \infty$ hai, lekin width $y\to 0$ hai. Mass $1$ rehta hai lekin $x=0$ ke paas ek spike mein concentrate ho jaata hai. ${\int }_0^y\frac{2x}{y^2}dx=\frac{1}{y^2}[x^2{]}_0^y=1\text{har }y>0\text{ ke liye}.$ Toh $y\to 0^{+}$ par slice ek ever-taller, ever-narrower spike ban jaata hai (ek "point mass forming"), even though $y=0$ par khud object undefined hai (Example 5). Figure mein amber curves yeh squeeze dikhate hain.

Verify: $f_{X\mid Y}(\cdot \mid y)$ ke neeche area $1$ hai har $y>0$ ke liye; $y=1$ par limit ${\int }_0^{1}2xdx=1$ deta hai. ✔

Example 7 — Real-world word problem (Cell C7)

Step 1 — validity check karo aur $p_Y(1)$ nikalo. Yeh step kyon? Hum $Y=1$ par condition kar rahe hain, isliye us column ka weight chahiye. Total $=0.20+0.05+0.25+0.20+0.10+0.20=1.00$ ✔. $p_Y(1)=0.05+0.20+0.20=\mathrm{0.45.}$

Step 2 — $X$ given $Y=1$ ki conditional distribution. Yeh step kyon? $Y=1$ column ko slice karo aur $0.45$ se re-normalize karo. $P(X=0\mid Y=1)=\frac{0.05}{0.45}=\frac{1}{9},P(X=1\mid Y=1)=\frac{0.20}{0.45}=\frac{4}{9},P(X=2\mid Y=1)=\frac{0.20}{0.45}=\frac{4}{9}.$

Step 3 — headline question ka answer do. $P(X=2\mid Y=1)=\frac{0.20}{0.45}=\frac{4}{9}\approx \mathrm{0.4444.}$ Baseline $p_X(2)=0.10+0.20=0.30$ se compare karo. Pastry kharidna 2 coffees ka chance badhata hai ($0.44>0.30$) — yeh saath saath chalta hai.

Step 4 — conditional expectation (dekho Expectation and Variance). Yeh step kyon? Conditional distribution ke under expected value har coffee-count ko uski conditional probability se weight karta hai. $E[X\mid Y=1]=0\cdot \frac{1}{9}+1\cdot \frac{4}{9}+2\cdot \frac{4}{9}=\frac{0+4+8}{9}=\frac{12}{9}=\frac{4}{3}\approx \mathrm{1.3333.}$

Verify: conditional probs $\frac{1}{9}+\frac{4}{9}+\frac{4}{9}=1$ ✔; $P(X=2\mid Y=1)=4/9$ aur $E[X\mid Y=1]=4/3$ neeche re-check kiye gaye hain.

Example 8 — Exam twist: joint rebuild karo (Cell C8)

Step 1 — chain rule se joint nikalo. Yeh step kyon? $f_{X,Y}(x,y)=f_{Y\mid X}(y\mid x)f_X(x)$ — slice ko jis cheez par condition kiya us uske weight se multiply karo. $f_{X,Y}(x,y)=\frac{1}{x}\cdot 2x=2,0<y<x<1.$ Triangle $0<y<x<1$ par height $2$ ki flat density.

Step 2 — sanity check: kya $1$ integrate karta hai? ${\int }_0^1{\int }_0^{x}2dydx={\int }_0^{1}2xdx=1.✓$

Step 3 — marginal of $Y$ ($x$ ko integrate out karo; fixed $y$ ke liye, $x$ $y\to 1$ chalta hai). $f_Y(y)={\int }_y^{1}2dx=2(1-y),0<y<1.$

Step 4 — reverse conditional (Bayes direction). Yeh step kyon? Ab hum $X$ given $Y=y$ chahte hain: joint ko $f_Y(y)$ se divide karo. $f_{X\mid Y}(x\mid y)=\frac{2}{2(1-y)}=\frac{1}{1-y},y<x<1.$ Yeh $(y,1)$ par uniform hai — forward "uniform on $(0,x)$" ka mirror.

Verify: joint $1$ integrate karta hai ✔; ${\int }_0^1{f}_Y(y)dy={\int }_0^{1}2(1-y)dy=1$ ✔; ${\int }_y^1{f}_{X\mid Y}(x\mid y)dx=\frac{1-y}{1-y}=1$ ✔.

Recall

Connections

• Joint distributions — joint PMF - PDF, marginal, conditional — parent jiska machinery har example yahan exercise karta hai.
• Conditional Probability — har upar wale conditional ke peechhe ka slice-and-renormalize rule.
• Independence of Random Variables — Examples 1–3 mein test kiya gaya.
• Law of Total Probability — marginals nikalne ke liye sum/integrate karne ka justification.
• Bayes' Theorem — Example 8 mein reversal direction.
• Expectation and Variance — Example 7 mein conditional expectation.
• Covariance and Correlation — natural next step jab aap joint moments compute kar sako.
• Bivariate Normal Distribution — continuous joint jo aap aage miloge.