Law of Large Numbers — weak and strong: Worked examples

4.9.16 · D3 · Maths › Probability Theory & Statistics › Law of Large Numbers — weak and strong

Yeh page parent topic ka drill room hai. Hum Law of Large Numbers (LLN) ke har tarah ke questions ko step by step solve karte hain, taaki jab exam ya koi real problem aaye, toh tumne uski shape pehle se dekhi ho.

Shuru karne se pehle, plain-word reminders (taaki koi symbol bina samjhe na aaye):

Definition Woh symbols jo hum har jagah reuse karte hain

$X_{1}, X_{2}, \dots$ — ek random experiment (coin flip, dice roll, random draw) ke independent identical repeats ki list. "Independent" = ek outcome kabhi doosre ko influence nahi karta. "Identical" = har ek same rulebook se drawn hai.
$μ = E [X_{i}]$ — true mean, ek draw ka long-run average value jo milna chahiye. Isse bullseye samjho.
$\overset{ˉ}{X}_{n} = \frac{1}{n} \sum_{i = 1}^{n} X_{i}$ — sample mean: apne pehle $n$ results ko add karo, $n$ se divide karo. Yeh woh wandering dot hai jisko LLN kehta hai bullseye ki taraf jaao.
$σ^{2} = Var (X_{i}) = E [(X_{i} - μ)^{2}]$ — variance, ek draw ki $μ$ se average squared distance. Yeh measure karta hai ki ek draw kitna jumpy hai.
$ε$ (epsilon) — error tolerance jo hum choose karte hain: ek small positive number, $μ$ ke around "kitna close kaafi close hai" ka radius. Event $∣ \overset{ˉ}{X}_{n} - μ ∣ \geq ε$ matlab "sample mean bullseye se kam se kam $ε$ door raha."
$δ$ (delta) — probability ka target ceiling jo hum choose karte hain: ek small positive number, sabse badi failure chance jo hum accept karne ko taiyaar hain (e.g. $δ = 0.05$ matlab "miss karne ka zyada se zyada 5% chance").

Humara master tool woh WLLN bound hai jo parent note mein prove ki gayi hai: $P (∣ \overset{ˉ}{X}_{n} - μ ∣ \geq ε) \leq \frac{σ ^{2}}{n ε ^{2}}$ Isse aise padho: "sample mean ke bullseye ko kam se kam $ε$ se miss karne ka chance variance-over-(n times epsilon-squared) se cap hai." Yeh Chebyshev's Inequality se aata hai, jo Markov's Inequality se aata hai.

Definition Teen theorems by name (hypothesis → conclusion)

Alag-alag examples alag-alag laws pe lean karte hain; yahan har ek ek line mein hai taaki koi naam blindly use na ho.

Chebyshev's route (easy proof): hypothesis $σ^{2} < \infty$ (finite variance) → conclusion $P (∣ \overset{ˉ}{X}_{n} - μ ∣ \geq ε) \leq \frac{σ ^{2}}{n ε ^{2}} \to 0$ . Yahi is page ke har bound ko power karta hai.
Khinchin's Weak Law: hypothesis sirf $E ∣ X_{i} ∣ < \infty$ (finite mean, variance infinite ho sakta hai) → conclusion $\overset{ˉ}{X}_{n} \to μ$ in probability. Yeh Chebyshev ki assumption ko weak karta hai while Weak-Law conclusion rakhta hai.
Kolmogorov's Strong Law: hypothesis sirf $E ∣ X_{i} ∣ < \infty$ → conclusion $P (lim_{n} \overset{ˉ}{X}_{n} = μ) = 1$ , yaani convergence almost surely (zyada strong, whole-path statement). Toh Chebyshev ek rate saste mein deta hai lekin variance chahiye; Khinchin aur Kolmogorov ko sirf finite mean chahiye lekin zyada mehnat karte hain (maximal inequalities aur Borel–Cantelli ke zariye).

Scenario matrix

Har LLN question in case classes mein se ek mein aata hai. Neeche har example un cells ke saath tagged hai jo woh cover karta hai.

#	Case class	Isme kya khaas hai	Example
A	Probability bound karo fixed $n$ par	seedha $σ^{2} / (n ε^{2})$ mein plug karo	Ex 1
B	$n$ ke liye invert karo (sample-size design)	bound ko $n$ ke liye solve karo, target $δ$	Ex 2
C	Non-binary variable (dice / general range)	pehle $μ, σ^{2}$ compute karo	Ex 3
D	Degenerate input $σ^{2} = 0$	ek constant "random" variable	Ex 4
E	Limiting behaviour $n \to \infty$ , aur $ε \to 0$	kaun sa knob jeetta hai	Ex 5
F	LLN fail hota hai — no finite mean	Cauchy / heavy tail	Ex 6
G	Real-world word problem	Monte Carlo estimate + error	Ex 7
H	Exam twist — Weak vs Strong on one path	dono modes distinguish karo	Ex 8
I	Fallacy trap — "streak due hai"	Gambler's fallacy quantified	Ex 9

Worked examples

Ex 1 — Cell A · Fixed $n$ par probability bound karo

Ex 2 — Cell B · Sample size $n$ ke liye invert karo

Ex 3 — Cell C · Ek non-binary variable (fair die)

Worked example Dice rolls ka average

$X_{i}$ = ek fair six-sided die ka result. $n = 200$ ke liye $P (∣ \overset{ˉ}{X}_{n} - μ ∣ \geq 0.5)$ bound karo.

Forecast: dice coins se zyada jumpy hote hain — kya cap badi hogi?

$μ$ compute karo: $μ = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = \frac{21}{6} = 3.5.$ Yeh step kyun? $μ$ diya nahi hai; hum chah faces ka average karte hain.
$σ^{2}$ compute karo: $E [X^{2}] = \frac{1 + 4 + 9 + 16 + 25 + 36}{6} = \frac{91}{6} .$ Phir $σ^{2} = E [X^{2}] - μ^{2} = \frac{91}{6} - \frac{49}{4} = \frac{182 - 147}{12} = \frac{35}{12} \approx 2.9167.$ Yeh step kyun? Variance $=$ mean-of-squares minus square-of-mean; yeh definition rearranged hai.
Plug in karo $n = 200$ , $ε = 0.5$ : $\frac{35/12}{200 \cdot 0.25} = \frac{2.9167}{50} \approx 0.0583.$

Answer: zyada se zyada $\approx 5.83%$ . Verify karo: units pure probability (dimensionless) hain ✓; die ka bada $σ^{2}$ (2.92 vs coin ka 0.25) yahan bade $ε = 0.5$ se offset hota hai, ek choti cap deta hai — sensible.

Ex 4 — Cell D · Degenerate input, $σ^{2} = 0$

Ex 5 — Cell E · Limiting behaviour: kaun sa knob jeetta hai?

Yeh dekha jaana best hai. "Bad" outcomes ka band $n$ badhne par patlaa hona chahiye, jabki $ε$ chota karna pushback karta hai.

Figure — Law of Large Numbers — weak and strong

Worked example

$n \to \infty$ vs $ε \to 0$ Coin ( $σ^{2} = \frac{1}{4}$ ). Bound compare karo jab (a) $n \to \infty$ with $ε$ fixed, (b) $ε \to 0$ with $n$ fixed.

Forecast: case (b) mein, kya guarantee survive karti hai?

Case (a): $\frac{1/4}{n ε ^{2}} n \to \infty 0.$ Yeh step kyun? $n$ denominator mein hai, toh ise badhana cap ko $0$ tak drive karta hai — yahi Weak Law hai.
Case (b): $n$ fixed ke saath, jaise $ε \to 0$ toh raw expression $\frac{1/4}{n ε ^{2}} \to + \infty.$ Yeh step kyun? $ε^{2}$ denominator mein shrink karta hai, expression ko blow up karta hai. Lekin kyunki probability bound $1$ exceed nahi kar sakti, usable bound hai $min (1, \frac{1/4}{n ε ^{2}})$ , jo $1$ par stuck hai — vacuous, kuch nahi batata.
Lesson: tumhe pehle $n \to \infty$ bhejnaa hoga noise shrink karne ke liye, aur tabhi tum chota $ε$ afford kar sakte ho. Figure mein red curve ( $ε = 0.05$ ) dekho: woh sabse upar wali hai; $ε$ half karna poori curve uthaa deta hai, lekin har curve phir bhi $0$ ki taraf crawl karta hai.

Answer: limit mein $n$ jeeta hai; $ε$ akele bound ko vacuousness mein tod deta hai. Verify karo: $n = 100, ε = 0.05$ par: $\frac{0.25}{100 \cdot 0.0025} = 1.0$ — exactly useless value $1$ ✓, chota $ε$ guarantee khaali kar deta hai yeh confirm karta hai. "In probability" kyun sahi lens hai iske liye Modes of Convergence dekho.

Ex 6 — Cell F · Jab LLN fail hota hai: Cauchy trap

Worked example Averages jo kabhi settle nahi hote

Standard Cauchy distribution se $X_{i}$ draw karo (bell-ish lekin tails itne fat hain ki mean undefined hai). Kya $\overset{ˉ}{X}_{n} \to$ kuch bhi?

Forecast: zyada draws help karenge... na?

Hypothesis check karo: LLN require karta hai $E ∣ X_{i} ∣ < \infty$ . Cauchy ke liye, $\int ∣ x ∣ f (x) d x = \infty$ — no finite mean. Yeh step kyun? Har LLN theorem (Chebyshev, Khinchin, Kolmogorov) minimum finite mean assume karta hai; uske bina koi bullseye nahi hai converge karne ke liye.
Structural fact: Cauchy ki ek remarkable property hai ki $\overset{ˉ}{X}_{n}$ ka distribution har $n$ ke liye ek single $X_{1}$ ke same hai. Yeh step kyun? Cauchys ka average spread bilkul reduce nahi karta — $σ^{2} / n$ engine finite variance par rely karta hai, jo yahan bhi infinite hai.
Conclusion: $\overset{ˉ}{X}_{n}$ hamesha ke liye idhar-udhar koodta rehta hai; yeh kisi cheez par converge nahi karta.

Answer: LLN apply nahi hota — no convergence. Verify karo: $Var (\overset{ˉ}{X}_{n}) = σ^{2} / n$ ko finite $σ^{2}$ chahiye; Cauchy mein $σ^{2} = \infty$ hai, toh $\infty/ n = \infty$ — engine off hai ✓. Yeh crucial "degenerate output" case hai: LLN ek theorem with hypotheses hai, nature ka law nahi.

Ex 7 — Cell G · Real-world word problem (Monte Carlo)

Worked example Random sampling se

$\int_{0}^{1} x^{2} d x$ estimate karna $U_{i} \sim Uniform (0, 1)$ draw karo, $g (U_{i}) = U_{i}^{2}$ average karo. $\hat{I}_{n} = \frac{1}{n} \sum U_{i}^{2}$ kiske paas converge karta hai, aur $n = 10, 000, ε = 0.01$ par error bound karo?

Forecast: pehle true integral guess karo, phir yeh ki $10, 000$ samples ise $0.01$ tak pin karte hain ya nahi.

Target ko expectation ke roop mein identify karo: $E [U^{2}] = \int_{0}^{1} x^{2} d x = \frac{1}{3} .$ Yeh step kyun? $[0, 1]$ par ek integral $g (U)$ ka mean hai — isliye Monte Carlo Methods kaam karte hain: LLN integration ko averaging mein convert karta hai.
SLLN se, $\hat{I}_{n} \to \frac{1}{3}$ almost surely. Yeh step kyun? $g (U_{i})$ i.i.d. hain finite mean ke saath, toh Strong Law guarantee karta hai ki estimate hamesha ke liye $\frac{1}{3}$ par land karta hai.
Error bound. $σ^{2} = Var (U^{2}) = E [U^{4}] - (E [U^{2}])^{2} = \frac{1}{5} - \frac{1}{9} = \frac{4}{45} \approx 0.0889$ chahiye. Yeh step kyun? Chebyshev ko ek term ka variance chahiye; $E [U^{4}] = \int_{0}^{1} x^{4} d x = \frac{1}{5}$ .
Bound plug karo: $\frac{σ ^{2}}{n ε ^{2}} = \frac{4/45}{10000 \cdot 0.0001} = \frac{0.0889}{1} \approx 0.0889.$

Answer: estimate $\to \frac{1}{3}$ ; $0.01$ se miss karne ka zyada se zyada $\approx 8.89%$ chance. Verify karo: $\frac{1}{5} - \frac{1}{9} = \frac{9 - 5}{45} = \frac{4}{45}$ ✓; bound valid probability $\leq 1$ hai ✓. Tighter chahiye? $n$ upar push karo — cap $1/ n$ ki tarah girta hai.

Ex 8 — Cell H · Exam twist: ek trajectory par Weak vs Strong

Yeh picture ek random path dikhati hai $\overset{ˉ}{X}_{n}$ ka taaki tum literally dono claims dekh sako.

Worked example Kaun sa law funnel se re-escape forbid karta hai?

Ek student kehta hai: "WLLN se, ek baar $\overset{ˉ}{X}_{n}$ $ε$ ke andar $μ$ ke aa gaya, woh kabhi nahi jaayega." True ya false, aur kaun sa law ise address karta hai?

Forecast: kya "in probability" ek single path ko hamesha ke liye pin karta hai?

WLLN actually kya claim karta hai: har fixed bade $n$ ke liye, $P (∣ \overset{ˉ}{X}_{n} - μ ∣ > ε)$ chota hai. Yeh ek snapshot ke baare mein baat karta hai, poore path ke nahi. Yeh step kyun? Convergence in probability har $n$ ko alag se control karta hai — yeh rare later excursions allow karta hai.
SLLN kya add karta hai: $P (lim_{n} \overset{ˉ}{X}_{n} = μ) = 1$ — almost har poora path funnel mein enter karta hai aur reh jaata hai. Yeh step kyun? Almost-sure convergence, upar wali Borel–Cantelli Lemma summary ke zariye built, infinitely many escapes forbid karta hai.
Verdict: student ka "kabhi nahi jaayega" ek Strong Law statement hai, Weak nahi. Sirf WLLN ek specific path ke liye baad mein jump-out forbid nahi karta.

Answer: claim ek Strong-Law property hai, jo Weak Law ko galat credit deta hai — toh jaisa kaha gaya (WLLN credit karte hue) woh false hai. Verify karo: figure mein, yellow path ko large $n$ par briefly $\pm ε$ green band se bahar jaate aur wapas aate dekho — WLLN yeh tolerate karta hai, SLLN kehta hai aisi events eventually hona band ho jaati hain. Strong $\Rightarrow$ Weak, kabhi ulta nahi.

Ex 9 — Cell I · Fallacy trap: kya tail "due" hai?

Worked example 10 heads ke baad, average kitna "off" hai quantify karo

Ek fair coin 10 baar heads aata hai ek baar mein ( $n = 10$ , sab heads). $\overset{ˉ}{X}_{10}$ kितना $μ$ se off hai, aur kya agla flip ise correct karne ki koshish karta hai?

Forecast: kya deviation streak ki wajah se shrink hota hai, ya uske bawajood?

Current deviation: $\overset{ˉ}{X}_{10} = 1$ , toh $\overset{ˉ}{X}_{10} - μ = 1 - 0.5 = 0.5.$ Yeh step kyun? Das heads fraction $1$ deta hai; true rate $0.5$ subtract karo.
Deviations ka sum: $\sum_{i = 1}^{10} (X_{i} - μ) = 10 \cdot 0.5 = 5.$ Yeh raw surplus coin ke zariye neeche nahi dhakela jaata. Yeh step kyun? $X_{i}$ independent hain — coin ki koi memory nahi, toh future flips mein koi "correcting force" nahi hai. Gambler's Fallacy dekho.
Average kaise heal hota hai — dilution se: agar agle $990$ flips perfectly average hain (aadhe heads), toh $\overset{ˉ}{X}_{1000} = \frac{10 + 495}{1000} = \frac{505}{1000} = 0.505.$ Yeh step kyun? $5$ ka surplus ab $1000$ flips mein share ho gaya hai: deviation $0.5$ se $0.005$ tak gira bina streak "cancelled" hue.

Answer: streak reverse nahi hoti; woh outvoted hoti hai jaisa $n$ badhta hai. Deviation $0.5 \to 0.005$ gayi. Verify karo: $\frac{505}{1000} - \frac{1}{2} = 0.005$ ✓, exactly $\frac{5}{1000}$ — growing $n$ par fixed surplus. Yeh "tails due hai" ka rigorous refutation hai.

Active recall

Recall "Sabse chota

$n$ solve karo" kaun se cell mein aata hai, aur algebra kya hai? Cell B. $\frac{σ ^{2}}{n ε ^{2}} \leq δ$ set karo (jahan $δ$ tumhara chosen probability ceiling hai) aur invert karo: $n \geq \frac{σ ^{2}}{δ ε ^{2}}$ .

Recall Cauchy distribution ke liye LLN kyun fail hota hai?

Koi finite mean nahi ( $E ∣ X ∣ = \infty$ ), toh koi $μ$ nahi hai converge karne ke liye; $\overset{ˉ}{X}_{n}$ sab $n$ ke liye Cauchy-distributed rehta hai.

Recall 10 heads ki streak: kya coin pushback karta hai? Average kaise recover karta hai?

Koi push-back nahi (independence, coin ki koi memory nahi). Recovery dilution se hoti hai — fixed surplus ek bade $n$ ka tiny fraction ban jaata hai, toh deviation shrink hota hai chahe surplus khud kabhi disappear na ho.

Recall Khinchin's aur Kolmogorov's laws batao aur yeh Chebyshev proof se kaise differ karte hain.

Dono ko sirf $E ∣ X ∣ < \infty$ chahiye. Khinchin convergence in probability deta hai (Weak); Kolmogorov convergence almost surely deta hai (Strong). Chebyshev ke proof ko extra hypothesis $σ^{2} < \infty$ chahiye lekin explicit rate $σ^{2} / (n ε^{2})$ reward karta hai.

Constant

X_{i} = 7

ke liye

\overset{ˉ}{X}_{n}

ka deviation

har

n

ke liye exactly

0

;

σ^{2} = 0

bound vanish kar deta hai.

WLLN ke liye woh knob jo infinity tak jaana chahiye

n

; sirf

ε

shrink karna bound ko vacuous bana deta hai.

Probability bound jo 1 ya usse zyada evaluate ho

vacuous hai — sirf

P \leq 1

kehta hai, jo hamesha true hai.