Question bank — Law of Large Numbers — weak and strong

True. Almost-sure convergence (whole-path settling) convergence in probability (per-$n$ closeness) se strictly stronger mode hai, isliye SLLN hamesha WLLN deta hai — dekho Modes of Convergence. Reverse fail hota hai.

False. WLLN sirf har fixed large $n$ ko pin down karta hai; running average phir bhi infinitely often wander out kar sakta hai, jo SLLN forbid karta hai. In-probability, almost-sure ko force nahi karta.

False. Finite ${\sigma }^2$ sirf easy Chebyshev proof ko power karta hai. Khinchin's weak law aur Kolmogorov's strong law i.i.d. variables ke liye sirf $\mathbb{E}|X_i|<\infty$ (ek finite mean) chahte hain.

Classical statements ke liye largely true — independence covariances ko vanish karta hai isliye $\mathrm{Var}({\bar{X}}_n)={\sigma }^2/n$. Weakly-dependent sequences ke liye bhi LLN hain, lekin standard theorem i.i.d. assume karta hai.

False — yeh Central Limit Theorem hai. LLN kehta hai ${\bar{X}}_n\to \mu$, ek single constant; CLT uske aas paas bachi hui wiggle describe karta hai.

False. Cauchy distribution ka koi mean nahi hota, aur uske sample means kabhi converge nahi karte — woh ek single draw jitne hi spread out rehte hain. Finite mean hi sach mein minimal condition hai.

False. Deviations ka sum typically grow karta hai (like $\sqrt{n}$). $n$ se divide karna average ko tame karta hai, sum shrink nahi hota.

Almost har sequence ke liye True (probability 1). Exceptional sequences exist karti hain (jaise sab heads) jahan yeh fail hota hai, lekin woh probability zero ka set banati hain.

False. Yeh deliberately crude hai — CLT kaafi sharper tail estimates deta hai. Chebyshev ek safe overkill hai, useful isliye kyunki sirf variance ${\sigma }^2$ chahiye.

False — yeh ek simplification hai. Yeh i.i.d. case ke liye exact hai. Sirf independent, non-identically distributed $X_i$ ke liye, Kolmogorov's strong law ki bajaye ${\sum }_i\mathrm{Var}(X_i)/i^2<\infty$ chahiye. Isliye minimal condition setting par depend karti hai.

Series $\sum 1/n$ diverge karta hai, isliye yeh bound summable nahi hai. Yahi exact reason hai kyun SLLN ko Kolmogorov's sharper maximal inequality chahiye, naive per-$n$ Chebyshev bound nahi — dekho Borel–Cantelli Lemma.

Markov's Inequality ko $Y\ge 0$ (non-negative) chahiye. Proof sketch ek line mein: $\mathbb{E}[Y]\ge {\int }_a^{\infty }yf(y)dy\ge a\mathbb{P}(Y\ge a)$, jisse milta hai $\mathbb{P}(Y\ge a)\le \mathbb{E}[Y]/a$. Pehla step $[0,a)$ mass ko drop karta hai, jo sirf tab harmless hai jab $Y$ negative nahi ja sakta.

Koi contradiction nahi hai; bound sirf vacuous hai (probability $\le$ kuch $>1$ trivially true hai). Bounds 1 se zyada ho sakti hain; iska matlab sirf yeh hai ki wahan koi information nahi hai.

Kisi bhi finite $n$ ke liye variance positive hai, isliye ${\bar{X}}_n$ genuinely random hai. Sirf limit mein spread vanish hoti hai; convergence ek statement hai $n\to \infty$ ke baare mein.

Gambler's Fallacy. Spins independent hain, isliye past outcomes future odds ko shift nahi karte. LLN dilution se balance karta hai — streak ek bade $n$ ka tiny fraction ban jaati hai — kisi corrective force se nahi.

Convergence in probability deviation ki shrinking probabilities ke baare mein hai, paths ke pointwise convergence ke baare mein nahi. Path-by-path convergence almost-sure (strong) statement hai.

Do effects ladte hain: sum ko $n$ se divide karna variance ko $1/n^2$ se scale karta hai, lekin sum $n$ independent variances stack karta hai (har ek ${\sigma }^2$), jisse milta hai $n{\sigma }^2$. Multiply karo: $\frac{1}{n^2}\cdot n{\sigma }^2={\sigma }^2/n$. Independence woh hai jo $n$ noises ko pile up hone ki bajaye partially cancel karne deta hai like $n^2$; neeche figure dekho.

WLLN ek time mein ek $n$ ko Chebyshev's Inequality se control karta hai, lekin SLLN ko ek saath poori path ka tail control karna hota hai, jiske liye ek maximal inequality chahiye taaki bad-event probabilities Borel–Cantelli Lemma ke liye summable ban sakein.

Ek integral ${\int }_0^{1}g(x)dx$, uniform $U$ ke liye $\mathbb{E}[g(U)]$ ke barabar hai, isliye SLLN guarantee karta hai ki $g(U_i)$ ka sample average almost surely integral mein converge karta hai — dekho Monte Carlo Methods.

Agar saari bad events ki total probability finite hai, toh sirf finitely many occur ho sakti hain; last wali ke baad tum hamesha good region mein rehte ho, jo exactly almost-sure convergence hai.

Streaks allowed hain aur appear bhi hoti hain; woh sirf ever-larger $n$ ke fraction ke roop mein dab jaati hain. LLN ratio ko constrain karta hai, raw counts ko nahi.

LLN ek limiting statement hai. Finite $n$ par fraction fluctuate karta hai; CLT kehta hai woh fluctuations order $1/\sqrt{n}$ ki hoti hain, isliye exact balance required nahi hai.

Average har $n$ ke liye $c$ hai, isliye yeh trivially converge karta hai — dono laws instantly hold karte hain. Zero variance (${\sigma }^2=0$) ka matlab hai Chebyshev bound saare $n\ge 1$ ke liye $0$ hai.

Yeh bilkul fail ho jaata hai: koi finite mean $\mu$ nahi hone se, ${\bar{X}}_n$ ka distribution same Cauchy distribution hai jaise ek single draw ka, isliye woh kabhi concentrate nahi hota. Yeh canonical counterexample hai.

Classical $\mathrm{Var}({\bar{X}}_n)={\sigma }^2/n$ toot jaata hai kyunki covariance terms ab vanish nahi karti. Weak dependence ke under convergence phir bhi hold kar sakti hai, lekin simple Chebyshev proof ab apply nahi hoti.

Ek stationary sequence jiske correlations itni tezi se decay karti hain — jaise ek $m$-dependent sequence (draws $m$ se zyada apart independent hain) ya ek mixing chain jahan $\mathrm{Cov}(X_1,X_k)\to 0$ jaldi hota hai. Tab $\mathrm{Var}({\bar{X}}_n)$ phir bhi $\to 0$ hota hai (covariance sum $n^2$ se slower badhta hai), isliye mild dependence ke bawajood ${\bar{X}}_n\to \mu$.

Yeh plain Chebyshev mein reduce ho jaata hai, $\mathbb{P}(|X_1-\mu |\ge \epsilon )\le {\sigma }^2/{\epsilon }^2$ — ek valid lekin aksar useless single-sample statement, kyunki ek draw mein koi averaging benefit nahi hota.

Bound ${\sigma }^2/(n{\epsilon }^2)$ infinity tak blow up ho jaata hai, yaani vacuous ho jaata hai. Tum ek fixed finite $n$ par arbitrarily tight closeness demand nahi kar sakte; tumhe $n\to \infty$ bhi bhejna hoga.