Exercises — Central Limit Theorem — statement, proof sketch, significance

$P(Z>z)=1-\Phi (z)$ kyun hota hai? Bell ke neeche total area 1 hai. $\Phi (z)$ $z$ ke baayein taraf ka area hai. Jo left mein nahi hai woh right mein hai, toh right-tail area $1-\Phi (z)$ hai. ::: $P(Z>z)=1-P(Z\le z)=1-\Phi (z)$.

CLT ko teen cheezein chahiye: independent, identically distributed, aur finite variance (${\sigma }^2<\infty$).

• (a) ✅ Dice rolls i.i.d. hain ${\sigma }^2=\frac{35}{12}<\infty$ ke saath. 200 ka mean beautifully normal hai.
• (b) ❌ Cauchy distribution ka koi finite mean nahi aur infinite variance hai. Recipe toot jaati hai — Cauchy draws ka average khud Cauchy hai, kabhi settle nahi karta. Yeh classic CLT counterexample hai.
• (c) ✅ Heights i.i.d. hain finite variance ke saath, toh ${\bar{X}}_{40}$ approximately normal hai (aur $n=40>30$, rule of thumb se comfortably aage).

• Sample mean ${\bar{X}}_{64}$: centre $=\mu =10$, standard error $=\frac{\sigma }{\sqrt{n}}=\frac{4}{\sqrt{64}}=\frac{4}{8}=0.5$.
• Sum $S$: centre $=n\mu =64\cdot 10=640$, spread $=\sigma \sqrt{n}=4\cdot 8=32$. Dhyaan do average ka spread ($0.5$) ek single value ke spread ($4$) se aath guna chhota hai — yahi $1/\sqrt{64}=1/8$ kaam kar raha hai.

Standard error $=\frac{\sigma }{\sqrt{n}}=\frac{10}{\sqrt{100}}=1$. Boundary $52$ ko standardize karo: $z=\frac{{\bar{X}}_n-\mu }{\sigma /\sqrt{n}}=\frac{52-50}{1}=\mathrm{2.00.}$ Ab "greater than" ko table lookup mein convert karo. Table $\Phi (z)=P(Z\le z)$ deti hai — baayein taraf ka area. Hum $z=2$ ke daayein taraf ka area chahte hain, jo poore minus left part hai (total area 1 hai): $P({\bar{X}}_{100}>52)=P(Z>2.00)=1-\Phi (2.00)=1-0.9772=\mathrm{0.0228.}$ Lagbhag $2.3\%$ chance ki 100 draws ka average 52 se zyada hoga. (Yahan koi symmetry assume nahi ki gayi — bas "left area + right area = 1.")

• $E[S]=n\mu =100\cdot 3.5=350$; $\mathrm{SD}(S)=\sigma \sqrt{n}=1.7078\sqrt{100}=17.078$.
• $z=\frac{S-n\mu }{\sigma \sqrt{n}}=\frac{330-350}{17.078}=-1.171$.
• Negative $z$ ke liye hum bell ki 0 ke baare mein mirror symmetry use karte hain: $-1.171$ par left-tail area $+1.171$ par right-tail area ke barabar hai, yaani $\Phi (-1.171)=1-\Phi (1.171)$. Isliye $P(S<330)=\Phi (-1.171)=1-\Phi (1.171)\approx 1-0.8792=\mathrm{0.1208.}$ Lagbhag $12\%$ chance ki 100-roll total 330 se neeche hoga. (Yahan symmetry use ki gayi hai, aur explicitly state ki gayi hai, kyunki table sirf positive $z$ list karta hai.)

Bell curve par ek central 95% region har taraf $z=1.96$ standard errors tak jaata hai (isliye $z_{0.025}=1.96$: yeh har tail mein $2.5\%$ chhodta hai). Toh $c=1.96\times (\text{standard error})=1.96\times 1=\mathrm{1.96.}$ Isliye ${\bar{X}}_{100}$ lagbhag 95% time $(48.04,51.96)$ mein land karta hai. Yahi Confidence Intervals ka seed hai.

• $\mu =np=200(0.3)=60$; $\sigma =\sqrt{np(1-p)}=\sqrt{200(0.3)(0.7)}=\sqrt{42}=6.4807$.
• Continuity correction — yeh kahan se aata hai. Count $X$ sirf whole number ho sakta hai, isliye iska probability isolated spikes $0,1,2,\dots$ par rehta hai. Un spikes ko area ke roop mein draw karne ke liye (taaki ek smooth curve unhe approximate kar sake), hum har integer $k$ ko $[k-0.5,k+0.5]$ span karte ek bar mein widen karte hain — width 1 ka rectangle jiska area $P(X=k)$ ke barabar hai. Smooth normal curve phir in bars ke tops se match ki jaati hai. Kyunki event $\{X\le 50\}$ $k=50$ par poori bar include karta hai, iska area us bar ke right edge, $50.5$ tak jaata hai — $50$ tak nahi. Figure dekho: 50 par butter-coloured bar poori tarah shaded region ke andar hai, aur lavender dashed line uske right edge par baith raha hai. $z=\frac{50.5-60}{6.4807}=-\mathrm{1.466.}$ $P(X\le 50)\approx \Phi (-1.466)=1-\Phi (1.466)\approx 1-0.9287=\mathrm{0.0713.}$

Kyunki $\mathrm{Var}(\sum (X_i-\mu ))=n{\sigma }^2$ aur ek random variable ko $n$ se divide karne se uska variance $n^2$ se divide ho jaata hai: $\mathrm{Var}(W_n)=\frac{n{\sigma }^2}{n^2}=\frac{{\sigma }^2}{n}\stackrel{n\to \infty }{\to }0.$ Spread zero ho jaata hai — $W_n\to 0$, ek single point. Yahi Law of Large Numbers hai, bell curve nahi. $n$ se divide karna "bahut zyada squash" karta hai. Non-degenerate spread rakhne ke liye tumhe $\sqrt{n}$ se divide karna hoga, kyunki yahi sum ke standard deviation ki $\sqrt{n}$ growth ko exactly cancel karta hai. Yeh Goldilocks scaling hai.

Pehle, "skewness" ka matlab kya hai. Skewness ek single number hai jo lopsidedness measure karta hai — ek tail doosre se kitni lambi hai. Iska formula standardized third moment hai: $\gamma =E\left[{\left(\frac{X-\mu }{\sigma }\right)}^3\right].$ Ek symmetric distribution ka $\gamma =0$ hota hai (left aur right cube mein cancel ho jaate hain); ek right-skewed ka $\gamma >0$ hota hai (lamba right tail cube ko average par positive banata hai). Fair coin ka Bernoulli$(0.5)$ ka $\gamma =0$ hai; exponential ka $\gamma =2$ hai. Ab jawab: skewed wala (exponential) ko bada $n$ chahiye. CLT ka limit pehle do moments se aage sab kuch throw away karta hai, lekin finite $n$ ke liye bachi hui skewness bell ko distort karti hai. Approximation error $\frac{\gamma }{\sqrt{n}}$ ki tarah shrink karta hai (Berry–Esseen flavour), toh bada $\gamma$ same closeness tak pahunchne ke liye bada $n$ demand karta hai. Symmetric input ($\gamma =0$) fast converge karta hai; exponential ka $\gamma =2$ ise slow banata hai. Exactly kab kai small independent (necessarily identical nahi) pieces ka sum normal jaata hai, yeh Lindeberg–Feller Condition pin down karta hai.

95% margin of error hai $z_{0.025}\cdot \frac{\sigma }{\sqrt{n}}=1.96\cdot \frac{12}{\sqrt{n}}$. Require karo ki yeh $\le 2$ ho: $1.96\cdot \frac{12}{\sqrt{n}}\le 2\Rightarrow \sqrt{n}\ge \frac{1.96\cdot 12}{2}=11.76\Rightarrow n\ge 11.76^2=\mathrm{138.3.}$ Upar round karo (data bahut kam nahi ho sakta): $n=139$. Dhyaan do: margin 2 se 1 karne ke liye tumhe $\sqrt{n}$ do guna bada chahiye, yaani $4\times$ data — "$1/\sqrt{n}$ diminishing returns."

Har mean approximately normal hai: $\mathrm{Var}({\bar{X}}_A)=\frac{{\sigma }_A^2}{n_A}=\frac{36}{50}=0.72,\mathrm{Var}({\bar{X}}_B)=\frac{{\sigma }_B^2}{n_B}=\frac{81}{80}=\mathrm{1.0125.}$ Independent quantities ke variances difference ke liye bhi add hote hain (subtract karne se wobble mein help nahi milti): $\mathrm{Var}(D)=0.72+1.0125=1.7325,\mathrm{SD}(D)=\sqrt{1.7325}=\mathrm{1.3162.}$ Yeh standard deviation ek two-sample z-test ka engine hai.

Equal true means ke under, $D$ 0 par centred hai $\mathrm{SD}(D)=1.3162$ ke saath. Standardize karo: $z=\frac{3-0}{1.3162}=\mathrm{2.279.}$ Two-sided 5% cutoff hai $|z|>1.96$. Kyunki $2.279>1.96$, observed gap surprising hai — hum 5% level par "means equal" reject karenge. (Two-sided $p$-value $\approx 2(1-\Phi (2.279))\approx 0.0226$.)

Tail integrate karke mean aur variance check karo.

• Mean: $E[X]={\int }_1^{\infty }x\cdot \frac{2}{x^3}dx={\int }_1^{\infty }2x^{-2}dx=[-2x^{-1}{]}_1^{\infty }=2$. Finite. ✅
• Variance ko $E[X^2]={\int }_1^{\infty }{x}^2\cdot \frac{2}{x^3}dx={\int }_1^{\infty }2x^{-1}dx=[2\ln x{]}_1^{\infty }=\infty$ chahiye. Diverge karta hai. ❌ Kyunki ${\sigma }^2=\infty$ hai, classical (finite-variance) CLT apply nahi hota. Sample mean 2 par converge karta hai (mean exist karta hai, toh LLN hold karta hai), lekin iske fluctuations Gaussian nahi hain aur $1/\sqrt{n}$ ki tarah shrink nahi karte — yeh ek heavy-tailed stable law follow karte hain. Moral: "mean hai" kaafi nahi hai; tumhe finite variance chahiye.

Pehle, notation $o(\cdot )$ ("little-o") par ek baat. $g(s)=o(s^2)$ likhna matlab hai $g$ $s\to 0$ par $s^2$ se strictly faster shrink karta hai, yaani $g(s)/s^2\to 0$. Yeh "terms jo limit mein matter karne ke liye bahut chhote hain" ke liye ek bookkeeping tag hai — yahan, second moment ke baad sab kuch. Scaling $n^{-a}$ ke saath, characteristic function ka Taylor expansion 0 ke paas deta hai ${\phi }_Y(t/n^a)=1-\frac{t^2}{2n^{2a}}+o\left(n^{-2a}\right),$ toh ${\phi }_{Z_n}(t)={\left(1-\frac{t^2}{2n^{2a}}+\cdots \right)}^n.$ Limit $(1+\frac{c_n}{n}{)}^n\to e^c$ ke finite aur non-zero hone ke liye hamen bracket ke correction ko $(\text{constant})/n$ ki tarah behave karna hoga, yaani $n^{2a}=n$, isliye $2a=1$, toh $a=\frac{1}{2}$.

• Agar $a>\frac{1}{2}$: exponent $\to 0$, $\phi \to 1$, limit ek point mass hai (over-squashed → LLN).
• Agar $a<\frac{1}{2}$: correction blow up karta hai, $\phi \to 0$, koi proper limit nahi (under-squashed → diverges). Sirf $a=\frac{1}{2}$ $e^{-t^2/2}$ par land karta hai, standard-normal fingerprint. Yahi $\sqrt{n}$ ka CF-level reason hai.

• Sum ka centre: $n\mu =2500\cdot (-0.05)=-125$.
• Sum ka spread: $\sigma \sqrt{n}=1.10\cdot \sqrt{2500}=1.10\cdot 50=55$.
• Boundary $0$ ko $z=\frac{S-n\mu }{\sigma \sqrt{n}}$ use karke standardize karo: $z=\frac{0-(-125)}{55}=\frac{125}{55}=\mathrm{2.2727.}$
• Hum right-tail area chahte hain (ahead hona matlab $S>0$), toh total minus left: $P(S>0)=P(Z>2.2727)=1-\Phi (2.2727)\approx 1-0.9885=\mathrm{0.0115.}$ 2500 plays ke baad ahead hone ka lagbhag 1.2% chance. Tiny per-play edge, kai plays se multiply hokar aur mean ke relative $\sqrt{n}$ scaling se shrunk, losing almost certain bana deta hai. Isliye casinos profit karte hain: CLT aggregate ko negative mean ke around concentrate karta hai.

Ek average standardize karne ke liye, deviation ko kis se divide karo? ::: standard error $\sigma /\sqrt{n}$ se — kabhi akele $\sigma$ se nahi.