Law of large numbers: Worked examples — Hinglish deep dive

1.3.14 · D3 · AI-ML › Probability & Statistics › Law of large numbers

Parent: Law of large numbers · Yeh child page drill page hai. Hum ek grid banate hain jisme Law of Large Numbers (LLN) ke har possible scenario ko cover kiya gaya hai, phir har grid cell ka ek example solve karte hain. Yahan kuch bhi assume nahi kiya ki tumne parent page yaad kiya hai — har symbol apni jagah khud earn hota hai jab woh appear karta hai.

Koi bhi example chhune se pehle, ek promise: jo bhi letter hum likhte hain uska plain-English meaning milega, aur (jahan help kare) ek picture bhi. Chalo characters ki list ek baar bana lete hain.

Definition Characters (pehle yeh padho)

$X_{i}$ — ek experiment ke $i$ -ve repeat ka outcome. " $i$ " sirf ek counter hai: $X_{1}$ pehla result hai, $X_{2}$ doosra, aur aise hi aage. Socho ek dice roll, ek coin flip, ek measured height.
$μ$ (Greek "mu") — expected value, ek single $X_{i}$ ka sach-much ka long-run average. Dekho Expected Value. Isko outcome histogram ke balance point ki tarah picture karo.
$σ^{2}$ (sigma-squared) — variance, ek $X_{i}$ kitna spread-out hai. Dekho Variance and Standard Deviation. $σ$ khud (bina square ke) standard deviation hai, $μ$ se typical distance.
$\overset{ˉ}{X}_{n} = \frac{1}{n} \sum_{i = 1}^{n} X_{i}$ — sample mean, matlab apne pehle $n$ results add karo aur $n$ se divide karo. Bar ka matlab hai "ka average", subscript $n$ ka matlab hai " $n$ samples use karke".
$ϵ$ (epsilon) — ek chhoti positive "kitna close chahiye mujhe" tolerance, jaise $0.01$ .
$δ$ (delta) — "kitni failure tolerate karoonga" probability budget, jaise $0.05$ .
i.i.d. — Independent and Identically Distributed: har $X_{i}$ same tarike se draw hota hai (identical) aur koi bhi draw doosre ko influence nahi karta (independent).

Hum sirf ek tool use karte hain: Chebyshev Inequality. Use kahin bhi lagane se pehle plain words mein samjhte hain.

Intuition Chebyshev bound kyun sach hai (words, symbols nahi)

Rule ko general form mein state karne ke liye hume "koi bhi random quantity jiske baare mein hum soch rahe hain" ka naam chahiye. Usse $Y$ bolo — sirf ek stand-in letter kisi arbitrary random variable ke liye (yeh ek coin flip ho sakta hai, ya poora sample mean; hum har baar batayenge). Iska apna mean $μ$ aur variance $σ^{2}$ hota hai.

Variance $σ^{2}$ apni definition se hi, $Y$ ki mean se squared distance ka average hai. Ab poochho: us average mein se kitna un outcomes se aa sakta hai jo door (distance $\geq ϵ$ ) land karte hain mean se? Aisa har door outcome squared-distance average mein kam se kam $ϵ^{2}$ contribute karta hai. Toh agar probability ka ek fraction $p$ itna door baitha hai, toh woh outcomes akele already kam se kam $p \cdot ϵ^{2}$ average mein daal dete hain. Lekin poora average sirf $σ^{2}$ hai. Isliye $p \cdot ϵ^{2} \leq σ^{2}$ , matlab $p = P (∣ Y - μ ∣ \geq ϵ) \leq \frac{σ ^{2}}{ϵ ^{2}} .$ Yeh ek single variable $Y$ ke liye Chebyshev hai. Sample-mean version paane ke liye, isse $Y$ ke saath apply karo jahan $Y$ sample mean $\overset{ˉ}{X}_{n}$ hai, jiska variance (independence se variances add hote hain, phir $\frac{1}{n}$ unhe $\frac{1}{n ^{2}}$ se scale karta hai) $σ^{2} / n$ hai. Substitute karo aur niche diya bound milta hai. Hum yeh "far mass must fit inside the variance" wala sentence har us example mein dobara bolenge jahan yeh use hota hai.

The scenario matrix

Har LLN problem in cells mein se ek mein hoti hai. Niche ke examples cell se labelled hain.

Cell	Scenario class	Kya cheez tricky banati hai	Example
A	Bounded 0/1 outcome (Bernoulli)	variance $= p (1 - p)$	Ex 1
B	Required $n$ ke liye solve karo	invert the bound	Ex 2
C	Large-spread continuous variable	big $σ$ , real units	Ex 3
D	Zero-variance / degenerate input	$σ^{2} = 0$ edge case	Ex 4
E	Limiting behaviour ( $n \to \infty$ and $ϵ \to 0$ )	race mein kaun jeeta	Ex 5
F	Monte Carlo estimate of an integral	random variable is an indicator	Ex 6
G	Word problem / "kya mera coin rigged hai?"	interpret karo, panic mat karo	Ex 7
H	Exam twist: heavy-tail / infinite variance	LLN can fail	Ex 8

Ex 1 — Cell A: bounded 0/1 outcome

Upar wala figure LLN ke peeche ki picture dikhata hai: light grey lines running head-proportion ke 40 individual random runs hain, chote $n$ par wild jitter karte hue; thick cyan line unka average hai, $n$ badhne ke saath $μ = 0.5$ ko ever more tightly hug karta hua. Do dashed amber curves $μ \pm σ / n$ "error band" (standard-error funnel) hain jo $1/ n$ ki tarah narrow hoti hai — woh funnel jisme har run khich jaata hai. Har example ke liye is funnel ko yaad rakho.

Worked example Fair coin, fixed

$n$ Ek fair coin $n = 10, 000$ baar flip karo. $X_{i} = 1$ heads ke liye, $0$ tails ke liye. Bound karo woh probability ki head-proportion $\overset{ˉ}{X}_{n}$ $0.5$ se $ϵ = 0.01$ ya zyada miss kare.

Forecast: abhi guess karo — kya bound $1%$ , $25%$ , ya $90%$ ke kareeb hai?

Step 1. $μ$ nikalo. Kyunki $X_{i}$ aadha time $1$ hai, $μ = 0.5$ . Yeh step kyun? Chebyshev ko true mean chahiye; fair coin ke liye yeh heads-probability hai.

Step 2. $σ^{2}$ nikalo. 0/1 variable ke liye, $X_{i}^{2} = X_{i}$ (kyunki $0^{2} = 0, 1^{2} = 1$ ), toh $E [X_{i}^{2}] = E [X_{i}] = 0.5$ , giving $σ^{2} = 0.5 - 0. 5^{2} = 0.25$ . Yeh step kyun? Variance $= E [X^{2}] - μ^{2}$ ; 0/1 outcomes ke liye algebra nicely collapse ho jaata hai.

Step 3. Plug in ("far mass must fit inside the variance"): $P (∣ \overset{ˉ}{X}_{n} - 0.5∣ \geq 0.01) \leq \frac{0.25}{10000 \cdot ( 0.01 ) ^{2}} = \frac{0.25}{1} = 0.25.$ Yeh step kyun? Ek bound mein direct substitution.

Verify: Units pure probability hain (dimensionless), aur $0.25 \leq 1$ — ek valid probability. Sharper CLT answer $\approx 0.046$ hai, comfortably below hamara $0.25$ , toh hamara bound consistent hai (isse over-estimate karna chahiye). ✓ Number ko honestly padho: $0.25$ sirf ek worst-case ceiling hai — 1-in-4 chance tak — nahin ek claim ki miss rare hai. Yeh sirf kehta hai "25% se bura nahi"; sach-much ki chance (CLT) genuinely chhoti $\approx 4.6%$ hai. Funnel figure mein, $n = 10, 000$ par standard-error band $\pm 0.005$ width tak shrink ho gayi hai, isliye real miss chance crude bound se itni chhoti hai.

Ex 2 — Cell B: required $n$ ke liye solve karo

Worked example Average height pin down karne ke liye kitne log chahiye?

Heights mein $μ = 170$ cm aur $σ = 10$ cm hai. $n$ kitna bada hona chahiye taki sample mean $\geq 0.95$ probability ke saath $170$ se $ϵ = 1$ cm ke andar ho?

Forecast: 1000 se zyada ya kam log?

Step 1. Demand translate karo. "1 cm ke andar prob $\geq 0.95$ ke saath" ka matlab hai miss probability $\leq δ = 0.05$ hai: $P (∣ \overset{ˉ}{X}_{n} - 170∣ \geq 1) \leq 0.05$ . Yeh step kyun? Chebyshev miss bound karta hai, hit nahi; pehle convert karo. Yahan $δ$ hamara failure budget hai.

Step 2. Guaranteed worst case (the "far mass" bound) ko budget ke niche force karo: $\frac{σ ^{2}}{n ϵ ^{2}} \leq δ \Rightarrow \frac{100}{n \cdot 1} \leq 0.05.$ Yeh step kyun? Agar over-estimate bhi $\leq 0.05$ hai, toh true miss chance zaroor hai.

Step 3. Solve karo: $n \geq 100/0.05 = 2000$ . Yeh step kyun? Algebra — $n$ isolate karo.

Verify: $n = 2000$ back plug karo: $100/ (2000 \cdot 1) = 0.05$ exactly. ✓ (CLT ko sirf ~385 chahiye; Chebyshev deliberately conservative hai — dekho Confidence Intervals.)

Ex 3 — Cell C: large-spread continuous variable

Worked example Noisy sensor readings

Ek sensor ek value report karta hai true mean $μ = 20$ aur standard deviation $σ = 8$ units ke saath. Tum $n = 64$ readings average karte ho. Chance bound karo ki average $\geq 2$ units off ho.

Forecast: $50%$ se niche ya upar?

Step 1. Identify karo: $σ^{2} = 8^{2} = 64$ , $n = 64$ , $ϵ = 2$ . Yeh step kyun? Chebyshev kisi bhi finite-variance distribution ke liye kaam karta hai — humhe kabhi sensor ki exact shape nahi chahiye thi.

Step 2. Standard error se scale ka feel lo: $SE = σ / n = 8/ 64 = 8/8 = 1$ . Toh average typically $1$ unit wobble karta hai, aur hamara target $ϵ = 2$ "2 standard errors out" hai. Yeh step kyun? Standard error sirf ek feel-for-the-scale quantity hai (Ex 1 se funnel half-width); yeh batata hai $ϵ = 2$ kaafi far miss hai, lekin yeh khud probability nahi hai. "Kitna door" ko "kitna likely" mein turn karne ke liye hume Step 3 mein guaranteed Chebyshev bound par wapas jaana padega — yeh switch hai, aur isliye hum yeh karte hain.

Step 3. Distance ko probability bound mein turn karo ("far mass must fit inside the variance"): $P (∣ \overset{ˉ}{X}_{n} - 20∣ \geq 2) \leq \frac{σ ^{2}}{n ϵ ^{2}} = \frac{64}{64 \cdot 2 ^{2}} = \frac{64}{256} = 0.25.$ Yeh step kyun? Sirf Chebyshev bound spread ( $SE$ ) ko guaranteed chance-of-missing mein convert karta hai; standard error akela nahi kar sakta.

Verify: $ϵ = 2$ exactly $2$ standard errors hai ( $2/1$ ). "2 SE" par Chebyshev $1/ 2^{2} = 0.25$ deta hai — matches. ✓ Units: probability ke andar value units cancel hote hain. ✓

Ex 4 — Cell D: degenerate / zero-variance input

Worked example Ek "constant" experiment

Har $X_{i}$ exactly $7$ ke barabar hai (ek broken thermometer 7 par stuck). Toh $μ = 7$ , $σ^{2} = 0$ . Bound kya kehta hai $P (∣ \overset{ˉ}{X}_{n} - 7∣ \geq ϵ)$ ke baare mein kisi bhi $ϵ > 0$ ke liye?

Forecast: kya average kabhi wander karta hai?

Step 1. $σ^{2}$ compute karo. Saari values identical $\Rightarrow$ koi spread nahi $\Rightarrow$ $σ^{2} = 0$ . Yeh step kyun? Variance $μ$ se deviation measure karta hai; koi deviation nahi toh yeh $0$ hai.

Step 2. Plug in: $P (∣ \overset{ˉ}{X}_{n} - 7∣ \geq ϵ) \leq \frac{0}{n ϵ ^{2}} = 0.$ Yeh step kyun? Zero numerator bound ko har $n$ ke liye, even $n = 1$ ke liye, $0$ force karta hai. (Funnel picture mein, band ki zero width hai — ek single flat line.)

Step 3. Interpret karo: average pehle sample se hi $7$ par glued hai. Yeh step kyun? Yeh trivial edge case hai jahan LLN ki convergence instantaneous hai.

Verify: $\overset{ˉ}{X}_{n} = \frac{1}{n} (7 + 7 + \dots + 7) = \frac{7 n}{n} = 7$ literally. Off hone ki probability genuinely $0$ hai. ✓

Ex 5 — Cell E: limiting behaviour, race mein kaun jeeta?

Plot Chebyshev bound $σ^{2} / (n ϵ^{2})$ dikhata hai jab $n$ badhta hai, $ϵ$ choose karne ke teen tarikon ke liye. Cyan curve (case a, fixed $ϵ = 0.1$ ) axis ki taraf slide karta hai: error probability khatam ho jaati hai. Amber dashed line (case c, $ϵ = 1/ n$ ) height $1$ par flat baitha hai: bound kabhi improve nahi hota. Chhota white arrow case (b) mark karta hai, jahan fixed $n$ par $ϵ$ shrink karne se bound frame ke top se upar chala jaata hai. Algebra ke saath curves padho.

Worked example

$n$ shrink karo vs $ϵ$ shrink karo $σ^{2} = 1$ use karte hue: (a) $ϵ = 0.1$ fix karo aur $n \to \infty$ jaane do; (b) $n = 100$ fix karo aur $ϵ \to 0$ jaane do; (c) $ϵ = 1/ n$ lo $n \to \infty$ ke saath. Bound $σ^{2} / (n ϵ^{2})$ har case mein kya hota hai?

Forecast: kaun sa case zero NAHI jaata?

Step 1 (a). $\frac{1}{n \cdot 0. 1 ^{2}} = \frac{100}{n} \to 0$ as $n \to \infty$ . Yeh step kyun? Yahi Weak Law hai — tolerance fixed rakho, samples badhao, error probability khatam ho jaati hai (cyan curve axis ki taraf girta hua).

Step 2 (b). $\frac{1}{100 \cdot ϵ ^{2}} \to \infty$ as $ϵ \to 0$ . Yeh step kyun? Fixed sample ke saath zero error demand karna useless (>1) bound deta hai — white arrow top se shoot karta hua. LLN finite $n$ par exactness kabhi promise nahi karta.

Step 3 (c). $ϵ = 1/ n \Rightarrow \frac{1}{n \cdot ( 1/ n ) ^{2}} = \frac{1}{n \cdot ( 1/ n )} = 1.$ Yeh step kyun? Jab tum tolerance ko exactly natural $1/ n$ rate par shrink karte ho, toh bound $1$ par stall karta hai (flat amber line) — ek tie. Yeh precisely woh scale hai jis par Central Limit Theorem rehta hai.

Verify: (a) $100/ n \to 0$ ✓; (c) plug in $n = 25$ : $1/ (25 \cdot (1/5)^{2}) = 1/ (25 \cdot 1/25) = 1$ ✓.

Ex 6 — Cell F: Monte Carlo integral

Dart picture unit square dikhata hai amber curve $y = e^{- x^{2}}$ ke saath; cyan dots curve ke niche gire ( $Z_{i} = 1$ ), white dots upar gire ( $Z_{i} = 0$ ). Cyan dots ka fraction shaded area estimate karta hai, jo exactly integral hai. Jaise hum zyada darts phenkate hain cyan fraction true value par lock ho jaata hai — ek picture mein LLN.

Worked example

$\int_{0}^{1} e^{- x^{2}} d x$ ko dart-throwing se estimate karo Unit square mein random darts phenko. $X_{i}, Y_{i} \sim Uniform (0, 1)$ ; $Z_{i} = 1$ set karo agar dart curve $y = e^{- x^{2}}$ ke niche gira, warna $0$ . Integral estimate karo, phir nikalo kitne darts guarantee karte hain ki estimate truth se $ϵ = 0.001$ ke andar ho failure budget $δ = 0.05$ ke saath. Dekho Monte Carlo Methods.

Forecast: hazaaron ya lakho darts?

Step 1. $Z_{i}$ ka average integral kyun estimate karta hai? Curve ke niche gire darts ka fraction curve ke niche ka area hai, jo exactly $\int_{0}^{1} e^{- x^{2}} d x \approx 0.7468$ hai. Toh $μ = E [Z_{i}] = \int_{0}^{1} e^{- x^{2}} d x$ . Yeh step kyun? LLN promise karta hai $\overset{ˉ}{Z}_{n} \to E [Z_{i}]$ ; humne $Z_{i}$ aise engineer kiya ki uska mean hi answer IS.

Step 2. Spread bound karo. $Z_{i}$ ek 0/1 variable hai, toh $σ^{2} = p (1 - p) \leq 0.25$ (max $p = 0.5$ par). Yeh step kyun? Worst-case variance ek safe sample count deta hai $p$ exactly jaane bina.

Step 3. Real Chebyshev sample-size condition use karo — sirf standard error nahi. Humhe miss-probability budget ke niche chahiye:

\;\Rightarrow\; n \ge \frac{\sigma^2}{\delta\,\epsilon^2}.$$ *Yeh step kyun?* Yeh honest guarantee hai. Standard error $\sigma/\sqrt n$ (pehle define kiya gaya — typical wobble) sirf estimator ki usual jitter describe karta hai, na ki ek *probability* ki yeh $\epsilon$ ke andar rahega; sirf Chebyshev woh probability deta hai. **Step 4.** Plug in $\sigma^2=0.25$, $\delta=0.05$, $\epsilon=0.001$: $$n\ge\frac{0.25}{0.05\cdot(0.001)^2}=\frac{0.25}{0.05\cdot 10^{-6}}=5{,}000{,}000.$$ *Yeh step kyun?* Arithmetic. Note karo yeh naive "standard error $<0.001$" count of $250{,}000$ se bahut bada hai — kyunki ek *probabilistic* guarantee standard deviation sirf shrink karne se bahut zyada strong hai. **Verify:** $0.25/(0.05\cdot 10^{-6}) = 0.25/(5\times10^{-8}) = 5\times10^6$ ✓. Sanity check ke liye $n=5\times10^6$ par standard error $0.5/\sqrt{5\times10^6}\approx 2.2\times10^{-4}$ hai, $\epsilon$ se safely under, stronger bound ke saath consistent. ✓ Aur $\int_0^1 e^{-x^2}dx=0.7468\ldots$ ✓.

Ex 7 — Cell G: "kya mera coin rigged hai?" word problem

Worked example 100 mein se 60 heads — panic?

Tum ek coin $100$ baar flip karte ho aur $60$ heads dekhte ho. Kya coin biased hai?

Forecast: damning evidence, ya ordinary luck?

Step 1. Fairness assume karo aur head-proportion ka standard error nikalo. $σ^{2} = 0.25$ per flip ke saath, $SE = σ / n = 0.25 / 100 = 0.5/10 = 0.05$ . Yeh step kyun? Hum test karte hain ki $0.60$ natural fluctuation ke relative kitna surprising hai — $n = 100$ par funnel half-width.

Step 2. Distance measure karo: $(0.60 - 0.50) /0.05 = 0.10/0.05 = 2$ standard errors. Yeh step kyun? "Kitne standard errors out" batata hai ki hum funnel ke edge mein kitni door hain.

Step 3. 2-standard-error deviation extreme nahi hai — yeh kaafi often hota hai. LLN kehta hai fluctuations $n$ badhne ke saath shrink karte hain lekin kabhi khatam nahi hote, toh ek single 60/100 weak evidence hai. Yeh step kyun? Gambler's-fallacy panic se bachao; dekho Confidence Intervals.

Verify: 2 standard errors par Chebyshev: $P (∣ \overset{ˉ}{X} - 0.5∣ \geq 0.10) \leq 1/ 2^{2} = 0.25$ — 1-in-4 chance tak, "impossible" ke kareeb bhi nahi. ✓ Toh 60/100 fair coin ke liye unremarkable hai.

Ex 8 — Cell H: exam twist, LLN FAIL kar sakta hai

Worked example Jab koi finite mean nahi hota

Suppose har $X_{i}$ Cauchy distribution follow karta hai — ek bell-ish curve itne heavy tails ke saath ki $μ$ aur $σ^{2}$ undefined hain (defining integrals diverge karte hain). LLN $\overset{ˉ}{X}_{n}$ ke baare mein kya kehta hai?

Forecast: kya yahan averaging help karta hai?

Step 1. Hypotheses check karo. Parent mein har LLN theorem "finite expected value $μ$ ke saath" se shuru hota hai. Cauchy ka koi finite mean nahi hai. Yeh step kyun? Ek theorem ki guarantee sirf tab apply hoti hai jab uski assumptions hold hon — hamesha check karo.

Step 2. Consequence: $\overset{ˉ}{X}_{n}$ settle nahi hota. Remarkably, Cauchy ke liye $n$ values ka sample mean ek single value jaisi same distribution rakhta hai — averaging kuch nahi khareedta. Yeh step kyun? Dikhata hai ki $1/ n$ shrinkage kabhi free nahi thi; iske liye finite $σ^{2}$ chahiye tha. (Koi funnel nahi banta — band kabhi narrow nahi hota.)

Step 3. ML ke liye lesson: Monte Carlo average ya Stochastic Gradient Descent estimate par trust karne se pehle, confirm karo ki quantity ka finite mean/variance hai. Yeh step kyun? Heavy-tailed losses ya rewards silently convergence tod sakte hain.

Verify: Koi numeric claim nahi — yeh ek hypothesis-checking counterexample hai. Takeaway: LLN ka promise conditional hai. ✓

Recall Quick self-test

Kaun se cell mein $σ^{2} = 0$ hai? ::: Cell D — constant/degenerate experiment (bound saare $n$ ke liye $0$ hai). Error $10 \times$ chhoti karne ke liye, samples kitne factor se badhne chahiye? ::: $100 \times$ , kyunki error $1/ n$ ki tarah scale karta hai. Chebyshev kisi bhi distribution shape ko kyun handle kar sakta hai? ::: Yeh sirf $μ$ aur $σ^{2}$ use karta hai, full shape nahi. Standard error kya hai aur yeh Chebyshev bound se kaise alag hai? ::: SE $= σ / n$ typical wobble hai (funnel half-width); Chebyshev bound $ϵ$ ke andar rehne ki guaranteed probability hai — zyada strong claim. LLN kab fail karta hai? ::: Jab mean/variance finite nahi hote (jaise Cauchy) — Cell H.

Connections worth chasing: Bias-Variance Tradeoff, Bootstrap Sampling, यही टॉपिक हिंदी में.