Exercises — Continuous random variables — PDF, CDF, percentiles

PDF ke liye do cheezein chahiye: $f(x)\ge 0$ har jagah, aur total area $=1$.

(a) NO. $[0,1]$ par $-x$ negative hai (jaise $x=0.5$ par, $f=-0.5<0$). Density kabhi negative nahi ho sakti — density matlab "probability kitni tightly packed hai," aur tum negative amount pack nahi kar sakte. Condition 1 fail.

(b) NO. Yeh non-negative hai, theek hai. Lekin area ek rectangle hai jiska height $3$, width $1$, toh area $=3\ne 1$. Total probability zyada ho gayi. Condition 2 fail.

(c) YES. Non-negative ✓. Area = rectangle height $2$, width $0.5$, toh $2\times 0.5=1$ ✓. Dhyan do height $2>1$ hai — yeh allowed hai, kyunki sirf area ko $1$ hona chahiye.

$F(-\infty )=0$ (far left se abhi koi area collect nahi hua), $F(+\infty )=1$ (poora area collect ho gaya), aur $F$ hamesha non-decreasing hoti hai — kyunki $f\ge 0$ hai, number line mein aur add karne se area sirf badh sakta hai, kam nahi ho sakta.

$P(X=3)=0$. Ek single point ek zero-width interval hai, aur ${\int }_3^{3}f=0$. Value $f(3)$ ek density hai, probability nahi — dono ko confuse mat karo.

(a) $3x^2\ge 0$ on $[0,1]$ ✓. Area: ${\int }_0^{1}3x^{2}dx=[x^3{]}_0^1=1$ ✓. Kyun: yahi exactly do PDF conditions hain.

(b) Left edge $x=0$ se area accumulate karo: $0\le x\le 1$ ke liye, $F(x)={\int }_0^{x}3t^{2}dt=[t^3{]}_0^x=x^3.$ Support ke neeche kuch accumulate nahi hua; uske upar sab kuch ho gaya: $F(x)=\{\begin{array}{ll}0,& x<0\\ x^3,& 0\le x\le 1\\ 1,& x>1.\end{array}$ Teen pieces kyun: parent note ka "support tak restrict karo" rule — $F$ support se pehle $0$ aur baad mein $1$ hona chahiye.

(c) $P(a\le X\le b)=F(b)-F(a)$ use karo: $P(0.5\le X\le 1)=F(1)-F(0.5)=1-0.125=\mathrm{0.875.}$

PDF area accumulate hone ki rate hai, isliye Fundamental Theorem of Calculus se, $f=F^{\prime }$ (dekho Fundamental Theorem of Calculus): $f(x)=\frac{d}{dx}\left(1-e^{-3x}\right)=3e^{-3x},x\ge 0,$ aur $f(x)=0$ for $x<0$. Yeh rate $\lambda =3$ wali Exponential distribution hai.

Median $m$ solves $F(m)=0.5$: $1-e^{-3m}=0.5\Rightarrow e^{-3m}=0.5\Rightarrow -3m=\ln 0.5\Rightarrow m=\frac{\ln 2}{3}\approx \mathrm{0.231.}$ Kyun: median 50th percentile hai; hum CDF ko invert karte hain, PDF ko kabhi nahi.

(a) Area $1$ hona chahiye. Support ki width $6-2=4$ hai, toh uske upar height $c$ ka rectangle $4c$ area deta hai. $4c=1\Rightarrow c=\frac{1}{4}$ set karo.

(b) $2\le x\le 6$ ke liye, left edge $x=2$ se area accumulate karo: $F(x)={\int }_2^x\frac{1}{4}dt=\frac{1}{4}(x-2).$ Poori definition: $F(x)=\{\begin{array}{ll}0,& x<2\\ \frac{x-2}{4},& 2\le x\le 6\\ 1,& x>6.\end{array}$

Figure s01 dekho. Accent-red curve CDF hai: ek seedha ramp jo $x=2$ par height $0$ se $x=6$ par height $1$ tak utha hai. Dashed guide-lines ramp se neeche $x$-axis tak teeno quartile values $3,4,5$ par girti hain — har woh jagah hai jahan ramp height $0.25$, $0.5$, $0.75$ cross karta hai. Constant density constant slope deta hai, isliye seedha ramp milta hai.

(c) $F(x_p)=p$ set karo aur linear equation $\frac{x-2}{4}=p\Rightarrow x=2+4p$ solve karo:

• $Q_1$: $p=0.25\Rightarrow x=2+1=3$.
• Median: $p=0.5\Rightarrow x=2+2=4$.
• $Q_3$: $p=0.75\Rightarrow x=2+3=5$. Yeh evenly spaced hain — uniform distribution ki pehchaan.

(a) Proposed density VALID NAHI HAI — yeh exercise deliberately baited hai. Corner $x=0$ par split karo kyunki $x<0$ ke liye $|x|=-x$ aur $x\ge 0$ ke liye $|x|=x$: ${\int }_{-1}^1\frac{1}{2}|x|dx=2{\int }_0^1\frac{1}{2}tdt=2\cdot \frac{1}{4}=\frac{1}{2}\ne 1.$ Toh $f(x)=\frac{1}{2}|x|$ ka total area $\frac{1}{2}$ hai — yeh PDF condition 2 fail karta hai. Task ise repair karna hai, pretend nahi karna ki yeh correct tha. $f(x)=k|x|$ likho aur $k$ choose karo taaki area $1$ ho: ${\int }_{-1}^{1}k|x|dx=2k{\int }_0^{1}tdt=2k\cdot \frac{1}{2}=k\stackrel{!}{=}1\Rightarrow k=1.$ Toh valid density hai $f(x)=|x|$ (ends par height $1$). Hum yahi corrected $f(x)=|x|$ parts (b) aur (c) ke liye use karte hain. (Symmetry hamare liye har step mein right half ko double karna possible banati hai.)

(b) Figure s02 dekho. Black V corrected PDF $f(x)=|x|$ hai. Accent-red dot corner at $x=0$ mark karta hai jahan formula $-t$ (left) se $+t$ (right) switch hota hai; shaded left wedge ka area $\frac{1}{2}$ hai, isliye jab tak hum $x=0$ par pohonchte hain humne $F(0)=\frac{1}{2}$ bank kar liya hai. Right wedge (lighter shading) remaining $\frac{1}{2}$ supply karta hai.

$-1\le x\le 0$ ke liye, left edge $-1$ se accumulate karo, jahan $|t|=-t$: $F(x)={\int }_{-1}^x(-t)dt={\left[-\frac{t^2}{2}\right]}_{-1}^x=-\frac{x^2}{2}+\frac{1}{2}=\frac{1-x^2}{2}.$ Sanity check: $F(-1)=0$ ✓, $F(0)=\frac{1}{2}$ ✓ (half area left half par hai). $0\le x\le 1$ ke liye, already banked $F(0)=\frac{1}{2}$ se continue karo, ab $|t|=t$ ke saath: $F(x)=\frac{1}{2}+{\int }_0^{x}tdt=\frac{1}{2}+\frac{x^2}{2}=\frac{1+x^2}{2}.$ Sanity check: $F(1)=\frac{2}{2}=1$ ✓.

(c) Symmetry se middle band $F(0.5)-F(-0.5)$ hai: $F(0.5)=\frac{1+0.25}{2}=\frac{1.25}{2}=0.625,F(-0.5)=\frac{1-0.25}{2}=\frac{0.75}{2}=0.375,$ $P(-0.5\le X\le 0.5)=0.625-0.375=\mathrm{0.25.}$

(a) Left edge $x=1$ se accumulate karo: $F(x)={\int }_1^x\frac{2}{t^3}dt={\left[-\frac{1}{t^2}\right]}_1^x=1-\frac{1}{x^2},x\ge 1,$ aur $0$ for $x<1$. PDF integrate karne se CDF kyun milta hai: definition se $F(x)=P(X\le x)={\int }_{-\infty }^{x}f(t)dt$ — $x$ par CDF exactly $f$ ke neeche $x$ tak ka total area hai. Kyunki support $1$ par start hota hai, $1$ se neeche koi area nahi hai, isliye accumulation left edge $t=1$ par start hota hai; $1$ se $x$ tak $2/t^3$ integrate karna literally abhi tak collect hua saari density sum karta hai, jo "at most $x$ hone ki probability" ka exactly matlab hai.

(b) Median: $F(m)=0.5\Rightarrow 1-\frac{1}{m^2}=0.5\Rightarrow \frac{1}{m^2}=0.5\Rightarrow m^2=2\Rightarrow m=\sqrt{2}\approx 1.414$.

(c) 90th percentile: $1-\frac{1}{x^2}=0.9\Rightarrow \frac{1}{x^2}=0.1\Rightarrow x^2=10\Rightarrow x=\sqrt{10}\approx 3.162$.

(d) Dekho Expectation and Variance of continuous RVs: $E[X]={\int }_1^{\infty }x\cdot \frac{2}{x^3}dx={\int }_1^{\infty }\frac{2}{x^2}dx={\left[-\frac{2}{x}\right]}_1^{\infty }=0-(-2)=2.$ Yeh exist karta hai aur $2$ ke barabar hai. (Heavy tails mean ko diverge kara sakti hain, lekin yahan $1/x^2$ integrand itna fast decay karta hai ki mean finite hai.)

(a) Definition se 90th percentile $\Phi (x_{0.9})=0.90$ solve karta hai, toh $x_{0.9}=1.2816$.

(b) 10th percentile $\Phi (x_{0.1})=0.10$ solve karta hai. Symmetry use karke, $\Phi (-1.2816)=1-\Phi (1.2816)=1-0.90=0.10$, toh $x_{0.1}=-1.2816$. Kyun: standard normal $0$ ke baare mein symmetric hai, isliye 10th aur 90th percentiles mirror images hain.

(c) $Q_3=x_{0.75}=0.6745$ aur $Q_1=x_{0.25}$. Symmetry se $x_{0.25}=-x_{0.75}=-0.6745$. Toh $Q_3-Q_1=0.6745-(-0.6745)=\mathrm{1.349.}$

(a) $F$ kyun invert karte hain: recipe kahti hai $X=F^{-1}(U)$, toh humein woh rule chahiye jo, ek target cumulative probability $u$ diya ho, woh $x$ return kare jiska CDF $u$ ke barabar ho. Yahi exactly "$F(x)=u$ ko $x$ ke liye solve karo" hai. $u=1-e^{-2x}$ set karo aur step by step unwind karo: $e^{-2x}=1-u\Rightarrow -2x=\ln (1-u)\Rightarrow x=-\frac{1}{2}\ln (1-u).$ Toh $F^{-1}(u)=-\frac{1}{2}\ln (1-u)$. Har manipulation $F$ ka ek operation peel back karti hai ($1$ se subtract karo, log lo, $-2$ se divide karo) taaki $x$ ko desired probability $u$ ke terms mein express kiya ja sake — yahi "CDF ko invert karna" ka matlab hai.

(b) $u=0.5$ plug in kyun karte hain: generator ka output $U$ wahi cumulative probability hai jo hum $F^{-1}$ ko dete hain. $u=0.5$ feed karna poochta hai "woh $x$ kaun sa hai jiske neeche half probability hai?" — yaani median: $X=-\frac{1}{2}\ln (1-0.5)=-\frac{1}{2}\ln (0.5)=\frac{\ln 2}{2}\approx \mathrm{0.3466.}$ Jaise expected tha, $[0,1]$ ka middle point $X$ ke median par map hota hai.

(c) Is step ke liye $F$ ka apne support par strictly increasing hona zaroori hai taaki $F^{-1}$ ek genuine (single-valued, invertible) function ke roop mein exist kare. Yahan $f(x)=2e^{-2x}>0$ for all $x\ge 0$, toh $F$ strictly increasing hai $[0,\infty )$ par aur genuinely invertible hai — yahi equivalence $F^{-1}(U)\le x⟺U\le F(x)$ ko license karta hai. Yeh guaranteed hone par: $P(X\le x)=P(F^{-1}(U)\le x)=P(U\le F(x))=F(x),$ jahan last step yeh use karti hai ki $U$ uniform hai, toh $P(U\le p)=p$. Output ka CDF exactly $F$ hai, jaisa chahiye tha.

Do unknowns $k,b$ ke liye do equations chahiye.

Equation 1 (valid PDF — total area $=1$): ${\int }_0^{b}kxdx=\frac{k{b}^2}{2}=1\Rightarrow k=\frac{2}{b^2}.$

CDF: $0\le x\le b$ ke liye, left edge se accumulate karo aur $k=2/b^2$ substitute karo: $F(x)={\int }_0^{x}ktdt=\frac{k{x}^2}{2}=\frac{2}{b^2}\cdot \frac{x^2}{2}=\frac{x^2}{b^2}.$ (Dhyan do: $F(x)=x^2/b^2$ hai, $x^2$ nahi — $b^2$ drop nahi hona chahiye.)

Equation 2 (median at $2$): $F(2)=0.5$ set karo: $\frac{2^2}{b^2}=\frac{4}{b^2}=0.5\Rightarrow b^2=8\Rightarrow b=2\sqrt{2}\approx \mathrm{2.828.}$

Phir Equation 1 se, $k=\frac{2}{b^2}=\frac{2}{8}=\frac{1}{4}.$

Answer: $k=\frac{1}{4}$, $b=2\sqrt{2}$, yaani $f(x)=\frac{1}{4}x$ on $[0,2\sqrt{2}]$. Check: median jahan $\frac{x^2}{8}=0.5\Rightarrow x^2=4\Rightarrow x=2$ ✓; total area $\frac{1}{4}\cdot \frac{(2\sqrt{2}{)}^2}{2}=\frac{1}{4}\cdot \frac{8}{2}=\frac{1}{4}\cdot 4=1$ ✓.

(a) Do rectangles: $0.5\times 1+0.25\times 2=0.5+0.5=1$ ✓.

(b) Figure s03 dekho. Left panel PDF hai — do flat levels (height $0.5$ phir $0.25$). Right panel accent-red CDF hai: yeh $[0,1]$ par ek steep ramp (slope $0.5$) ki tarah chadhta hai, phir $[1,3]$ par ek gentler ramp (slope $0.25$) mein bend hota hai; ramp ka slope exactly local density hai. Red dot $(1,0.5)$ par woh seam mark karta hai jahan $F$ abhi $0.5$ tak pohonchi hai — yahi median hai.

• $x<0$: $F=0$.
• $0\le x\le 1$: $F(x)={\int }_0^{x}0.5dt=0.5x$. $x=1$ par, $F=0.5$.
• $1<x\le 3$: banked $0.5$ carry karo, phir second rectangle add karo: $F(x)=0.5+{\int }_1^{x}0.25dt=0.5+0.25(x-1)$. $x=3$ par, $F=0.5+0.5=1$ ✓.
• $x>3$: $F=1$.

(c) Hum $F(m)=0.5$ chahte hain. Lekin $F$ exactly $x=1$ par $0.5$ pohonchti hai (pehle ramp ka end). Toh median $m=1$ hai. (Half area $[0,1]$ mein hai, half $(1,3]$ mein — corner exactly balance point hai.)