Question bank — Central Limit Theorem — statement, proof sketch, significance

False. Data apni original shape hamesha ke liye rakhta hai; sirf sampling distribution of the mean or sum (ek derived quantity) normal hoti hai.

False. Theorem ki poori power yahi hai ki inputs kuch bhi ho sakti hain — dice, coins, exponentials — bas unka variance finite hona chahiye; normality ek output hai, requirement nahi.

True. Independent normals ka sum exactly normal hota hai, toh koi limit chahiye hi nahi — CLT ki approximation har finite $n$ ke liye equality ban jaati hai.

False — yeh Law of Large Numbers hai. CLT ek finer statement hai $\mu$ ke aas-paas fluctuations ki shape ke baare mein, jab tum unhe $\sqrt{n}$ se magnify karo.

False. Standard error hai $\sigma /\sqrt{n}$, toh use halve karne ke liye $n$ ko quadruple karna padta hai — error $1/\sqrt{n}$ ki tarah shrink hoti hai, $1/n$ ki tarah nahi.

False. Cauchy ka infinite variance hai (aur defined mean bhi nahi), isliye ek core hypothesis fail ho jaati hai; $n$ Cauchy variables ka average kitna bhi bada $n$ ho, fir bhi same spread ke saath Cauchy hi rehta hai.

False. Classical CLT ko independence chahiye; strongly dependent variables ek non-normal limit ki taraf converge ho sakte hain ya stabilize hi nahi hote. Weak-dependence versions hain lekin unhe extra conditions chahiye.

True. $\mathrm{Var}(\sum X_i)=n{\sigma }^2$ toh $\mathrm{SD}=\sigma \sqrt{n}$ — yahi $\sqrt{n}$ growth hai isliye hum limit stabilize karne ke liye $\sqrt{n}$ se divide karte hain.

False. Woh histogram $X$ ki original distribution estimate karta hai. CLT bahut saare averages ke histogram ke baare mein hai, jahan har ek apne fresh sample se compute hota hai.

False. Limiting shape sirf pehle do moments use karti hai; higher moments (skewness ke zariye) sirf convergence ki speed ko affect karte hain, jaisa ki neeche Berry–Esseen mein quantify kiya gaya hai.

Skew individual $X_i$ mein rehta hai; unke 50 ko average karne se zyaadatar skew cancel ho jaata hai, aur ${\bar{X}}_{50}$ ek bell se achhi tarah approximate hota hai. Skewness convergence slow karti hai lekin finite variance ke liye use forbid nahi karti.

Tum continuity correction bhool gaye. Ek discrete count ek continuous curve se cover ho rahi hai, isliye $50.5$ use karo (bar ka right edge) uski width capture karne ke liye; ise ignore karne se probability bias hoti hai.

Tumne do alag limits mix kar di. Ek distribution ke roop mein ${\bar{X}}_n$ sach mein $\mu$ par collapse hoti hai (yeh LLN hai); CLT pehle $\sqrt{n}$ se rescale karta hai taaki spread $1$ par fixed rahe, jisse non-degenerate $N(0,1)$ milti hai.

"$n\ge 30$" ka rule of thumb mild skew assume karta hai. Heavily skewed ya heavy-tailed distributions ko approximation achhi hone se pehle saikdon ya hazaron terms chahiye ho sakte hain.

Convergence in distribution ka matlab hai $P(Z_n\le z)\to \Phi (z)$ — yeh $\Phi (z)$ ke paas jaata hai lekin kisi bhi finite $n$ ke liye kabhi equal nahi hota. Error choti hai, zero nahi.

Independence characteristic functions ko multiply karati hai, jisse ek power $[{\phi }_Y(t/\sqrt{n}){]}^n$ milti hai, $n$ ke saath product nahi. $n$ ek exponent hai, coefficient nahi.

Hum ise isliye drop karte hain kyunki humne $Y=(X-\mu )/\sigma$ ko standardize kiya hai, jisse $E[Y]=0$ exactly ho jaata hai. Yeh construction se vanish hota hai, negligible hone ki wajah se nahi.

$n$ se divide karne par deviation $0$ par over-squash ho jaata hai (LLN regime); $1$ se divide karne par woh infinity tak blow up karta hai. Sirf $\sqrt{n}$, sum ka actual standard deviation, spread ko fixed rakhta hai ek stable limit ke liye.

Independent sums ke liye, densities messy convolution se combine hoti hain, lekin characteristic functions simply multiply karte hain. CF ek unique fingerprint bhi hai, aur pointwise CF convergence, distributional convergence imply karti hai Lévy's continuity theorem ke zariye.

${\phi }_Y$ ka Taylor expansion $0$ ke paas hai $1-\frac{s^2}{2}+o(s^2)$ (yahan $s$ expansion ka dummy argument hai); constant term aur $s^2$ term (mean $0$, variance $1$ se fixed) $n$-th power limit mein $s=t/\sqrt{n}$ ke baad survive karte hain, jabki sab kuch higher $o(1/n)$ mein sweep ho jaata hai.

Yeh classic exponential limit hai $(1+a_n/n{)}^n\to e^a$ jahan $a=-t^2/2$ hai. Isliye bell curve, jiska CF $e^{-t^2/2}$ hai, emerge hoti hai chahe inputs kuch bhi rahi hon.

Rare extreme values sum ko dominate karti hain aur average out hone se mana karti hain. Agar tail itni heavy hai ki ${\sigma }^2=\infty$, toh koi stabilizing scale nahi hai aur CLT genuinely fail ho jaati hai, jaise Cauchy Distribution ke saath.

Ek $\text{Bin}(n,p)$ variable $n$ i.i.d. Bernoulli$(p)$ trials ka sum hai — exactly woh setup jo CLT ko chahiye — toh uski distribution bell ke paas jaati hai (de Moivre–Laplace theorem).

Ye assume karte hain ki standardized sample mean approximately $N(0,1)$ hai, jo precisely wahi hai jo CLT deliver karti hai chahe underlying data non-normal ho.

Measurement error typically bahut saare tiny independent effects ka sum hota hai; CLT kehti hai aise sums approximately Gaussian hote hain chahe har effect ki individual distribution kuch bhi ho.

Haan: Berry–Esseen theorem kehta hai ${\sup }_z|P(Z_n\le z)-\Phi (z)|\le \frac{C\rho }{{\sigma }^3\sqrt{n}}$, jahan $\rho =E|X-\mu {|}^3$ teesra absolute moment hai aur $C\approx 0.47$. Error $1/\sqrt{n}$ ki tarah shrink hoti hai aur skewness ke saath badhti hai — yeh quantitative reason hai ki skewed data ko bada $n$ kyun chahiye.

Tab har $X_i$ probability one ke saath $\mu$ ke barabar hoti hai; koi randomness nahi hai average karne ke liye, $Z_n$ undefined hai (division by zero), aur theorem ki hypothesis ${\sigma }^2>0$ violate hoti hai.

Classical CLT apply nahi hoti. Properly scaled sum normal ki jagah ek non-Gaussian stable law ki taraf converge ho sakta hai, aur $\sqrt{n}$ scaling generally galat hoti hai.

Haan, ek extra uniformity condition ke under — Lindeberg–Feller Condition — jo ensure karti hai ki koi single term sum ko dominate nahi kare. Iske bina, ek large-variance term normality rok sakti hai.

Yeh exactly Cauchy rehti hai same scale ke saath — yeh concentrate nahi hoti aur kabhi normal nahi banti, yeh Law of Large Numbers aur CLT dono ki ek dramatic failure hai.

Generally nahi. Itne kam terms ke saath original distribution ki shape abhi bhi dominate karti hai; bell sirf $n$ badhne ke saath emerge hoti hai, symmetric light-tailed inputs ke liye zyada jaldi aur skewed ones ke liye zyada slowly.

Koi bhi $n\ge 1$ — independent normals ka sum ya mean exactly normal hota hai, isliye is special case mein koi minimum sample size ki zaroorat nahi.

Ek lattice distribution apna saara mass evenly spaced points par rakhti hai (jaise integers), toh uski CDF ek staircase hai. Is staircase par ek smooth normal curve fit karne ka matlab hai ki har jump ko midpoint par split kiya jaaye, yaani cutoff ko $\pm 0.5$ shift karo. Ek non-lattice continuous variable ki already smooth CDF hoti hai, toh aisi koi correction ki zaroorat nahi.