Question bank — Law of large numbers

Galat — ye guarantee karta hai ki proportion 0.5 ke paas aayega, lekin absolute count phir bhi dozens se off ho sakta hai; randomness kabhi poori tarah khatam nahi hoti.

Galat — ${\sigma }^2$ distribution ki ek fixed property hai; jo shrink hota hai wo sample mean ki variance hai, yaani ${\sigma }^2/n$.

Galat — WLLN ko sirf i.i.d. draws chahiye jinka finite mean ho; koi bhi shape kaam karta hai.

Galat — error $1/\sqrt{n}$ ke hisaab se scale hoti hai, isliye $n$ double karne se error sirf $\sqrt{2}\approx 1.41$ ke factor se ghatti hai; use half karne ke liye $4\times$ data chahiye.

Sahi — almost-sure convergence zyada strong hai aur mathematically convergence in probability ko imply karta hai; ulta implication nahi hota.

Galat — ye sirf ek upper bound hai (aksar bahut loose); sachi probability, jaise Central Limit Theorem se, aam taur par kaafi chhoti hoti hai.

Sahi — WLLN ke liye finite mean kaafi hai (jaise truncation arguments se); finite variance sirf Chebyshev-based proof ke liye chahiye hoti hai, law ke liye nahi.

Galat — dono ultimately sahi hain, lekin ek ki variance zyada ho sakti hai aur wo bahut dheere converge kar sakta hai, isliye finite-sample behaviour alag hoga.

Yahan standard error hai $\sigma /\sqrt{n}=0.5/\sqrt{1000}\approx 1.6\%$, toh 60% roughly 6 standard errors door hai — sachchi mein suspicious — lekin "definitely" is baat ko ignore karta hai ki rare events hote hain; tum ise test karoge, declare nahi karoge.

LLN kabhi bhi future outcomes ko compensate karne ke liye push nahi karta; draws independent hain, isliye agli flip abhi bhi 50/50 hai. Balance kaafi naye trials mein dilution se aata hai, correction se nahi.

Chebyshev route sirf ek proof hai; Weak Law phir bhi finite-mean, infinite-variance cases mein doosre methods se hold karta hai. Ek failed proof technique theorem ko khatam nahi karta.

Sirf tab jab draws independent hon — cross terms (covariances) zero hone chahiye. Correlated samples ke liye variance kaafi bada ho sakta hai aur ho sakta hai $1/n$ ki tarah shrink na kare.

Sum ka spread actually badhta hai: iska standard deviation $\sigma \sqrt{n}$ hai, jo increase karta hai. Sirf mean concentrate hota hai; sum absolute terms mein $n\mu$ se aur door bhaagta hai.

$1/\sqrt{n}$ rate dimension-independent hai — yahi toh Monte Carlo ki puri appeal hai. Exponent dimension ke saath multiply nahi hota.

$1/n$ se scale karne par variance se $1/n^2$ bahar aata hai (variance ek constant ke square se scale hoti hai); $n$ equal variances jodne par $n{\sigma }^2$ milta hai, aur $n{\sigma }^2/n^2={\sigma }^2/n$.

Ek common mean ke bina, "${\bar{X}}_n\to \mu$" ke paas converge karne ke liye koi ek $\mu$ hi nahi hai; identical distribution guarantee karta hai ki har draw ek hi target share kare.

Ek probability, ek indicator variable ki expected value ke barabar hoti hai, aur LLN kehta hai ki un indicators ka sample mean (observed frequency) us expectation pe converge karta hai.

Ek mini-batch gradient, per-example gradients ka sample mean hota hai; LLN se ye true full-batch gradient pe converge karta hai, isliye noisy steps average mein sahi direction mein point karte hain.

Interval width standard error $\sigma /\sqrt{n}$ ke proportional hai, aur $\sqrt{n}$ denominator matlab hai returns diminish hote hain — precision ki aakhri sliver ke liye disproportionately zyada samples chahiye.

Ye Bias-Variance Tradeoff hai: LLN sirf $n\to \infty$ limit mein error remove karta hai, isliye finite $n$ ke liye ek chhota controlled bias, badi variance se total error zyada reduce kar sakta hai.

Resampling ek statistic ki distribution ko kaafi resamples par average karke estimate karta hai; LLN ensure karta hai ki ye empirical averages true sampling behaviour ko approximate karein jaise bootstrap replicates ki sankhya badhti hai.

Ye fail ho jaata hai — sample mean converge nahi karta; har Cauchy sample mean phir se Cauchy-distributed hota hai, isliye zyada data kabhi help nahi karta. Finite mean zaroori precondition hai.

Ye trivially instant hai — har draw $\mu$ ke barabar hota hai, isliye sab $n\ge 1$ ke liye ${\bar{X}}_n=\mu$; Chebyshev bound $0$ hai, perfect concentration se match karta hai.

Kuch useful nahi — ${\bar{X}}_1=X_1$ hai poori variance ${\sigma }^2$ ke saath; ye law ek limiting statement hai aur ek single sample par koi guarantee nahi deta.

Tab ${\bar{X}}_n$ us ek draw ke barabar hoga chahe $n$ kuch bhi ho, uski variance ${\sigma }^2$ hi rahegi, aur ye $\mu$ pe converge nahi karega — independence hi wo cheez hai jo $1/n$ shrinkage possible banati hai.

Bound infinity tak blow up ho jaata hai (vacuous ho jaata hai), ye reflect karta hai ki $\mu$ ke saath exact equality ki demand finite $n$ par kabhi guarantee nahi ho sakti — tum sirf ek fixed positive $ϵ$ se bade deviations ko hi bound kar sakte ho.

Chebyshev sample-size formula zaroori $n$ ko kam batata hai; badi sachi ${\sigma }^2$ ka matlab hai ki usi precision ke liye proportionally zyada samples chahiye, isliye optimistic ${\sigma }^2$ ek jhoothi security ka ehsaas deta hai.