Central Limit Theorem — statement, proof sketch, significance: Worked examples

4.9.15 · D3 · Maths › Probability Theory & Statistics › Central Limit Theorem — statement, proof sketch, significanc

Kuch bhi shuru karne se pehle, teen symbols jinhe hum poore page use karenge — har ek plain words mein defined:

Ek move jo har example karta hai: $z = \frac{( your cutoff ) - ( mean of the thing )}{( SD of the thing )}, P (thing \leq cutoff) \approx Φ (z) .$

The scenario matrix

Har CLT problem inhi cells mein se kisi ek mein aata hai. Neeche ke examples us cell ke saath tagged hain jo woh cover karta hai.

Cell	Kya distinct banaata hai	Covered by
A — Symmetric discrete, sum	flat/symmetric start, bahut saron ka sum	Ex 1
B — Skewed continuous, mean	heavy right-skew start, average	Ex 2
C — Discrete count, continuity correction chahiye	Bernoulli sum, half-shifts	Ex 3
D — Two-sided ("between") probability	$Φ (z_{2}) - Φ (z_{1})$ chahiye	Ex 4
E — Inverse problem ( $n$ ke liye solve karo)	error target diya, sample size dhundo	Ex 5
F — Real-world word problem	English → $μ, σ, n$ mein translate karo	Ex 6
G — Degenerate: $σ = 0$	zero variance, koi bell hi nahi	Ex 7
H — Limiting / failure: infinite variance	Cauchy Distribution, CLT toot jaata hai	Ex 8
I — Exam twist: do means ka difference	do independent averages combine karo	Ex 9

Hum signs cover karte hain (positive aur negative $z$ ), dono tails, "between" case, zero-input case, aur woh limiting case jahan theorem kaam karna band kar deta hai. Kuch bhi undiscovered nahi rehta.

Example 1 — Cell A: symmetric discrete, ek bada sum

Worked example 100 fair dice ka sum

Ek fair die $1$ – $6$ mein se har ek $\frac{1}{6}$ probability se show karta hai. Ise $n = 100$ baar roll karo, $S = \sum X_{i}$ lo. $P (S > 380)$ dhundo.

Forecast: $380$ mean $350$ se upar hai — kya yeh thoda upar hai ya bahut? Guess karo ki answer $0.4$ ke kareeb hai ya $0.04$ ke.

Ek die ka $μ$ aur $σ^{2}$ dhundo. $μ = \frac{1 + 2 + \dots + 6}{6} = 3.5$ . Variance $= E [X^{2}] - μ^{2} = \frac{91}{6} - 3. 5^{2} = \frac{35}{12} \approx 2.9167$ , toh $σ \approx 1.708$ . Yeh step kyun? CLT sirf inhi do numbers use karta hai; die ki flat shape limit ke liye irrelevant hai.
Sum tak scale karo. $E [S] = 100 \cdot 3.5 = 350$ ; $Var (S) = 100 \cdot 2.9167 = 291.67$ ; $SD (S) = 291.67 = 17.078$ . Yeh step kyun? Means add hote hain aur — kyunki rolls independent hain — variances add hote hain. SDs add nahi hote; pehle variances add karo, phir square-root lo.
Cutoff standardize karo. $z = \frac{380 - 350}{17.078} = 1.756$ . Yeh step kyun? Yeh " $380$ dice scale par" ko " $1.756$ standard errors above the mean" mein convert karta hai, jo woh language hai jो table $Φ$ bolti hai.
Tail padho. $P (S > 380) \approx 1 - Φ (1.756) = 1 - 0.9605 = 0.0395$ . Yeh step kyun? $Φ$ left area deta hai; hum right tail chahte hain, toh $1$ se subtract karo.

Verify: $z = 1.756 > 0$ matlab cutoff mean se upar hai, toh ise exceed karne ki probability $0.5$ se neeche honi chahiye — aur $0.0395$ hai. Units cancel ho jaate hain (dice minus dice over dice = pure number). Forecast: woh chota wala tha, $\approx 0.04$ . ✓

Example 2 — Cell B: heavily skewed continuous, ek sample mean

Worked example 50 exponential waiting times ka average

$X_{i} \sim Exp (λ = 2)$ (ek bahut right-skewed distribution). Iske liye, $μ = 1/ λ = 0.5$ aur $σ = 1/ λ = 0.5$ . $n = 50$ measurements lo aur unhe $\overset{ˉ}{X}$ mein average karo. $P (\overset{ˉ}{X} > 0.6)$ dhundo.

Forecast: raw exponential ek taraf jhuka hua hai — kya iska average sirf 50 terms ke baad bhi roughly bell-shaped hoga? Guess karo haan ya nahi.

Standard error nikalo. $\frac{σ}{n} = \frac{0.5}{50} = 0.0707$ . Yeh step kyun? Hum sum nahi, average kar rahe hain, toh spread shrink hota hai $1/ n$ ki tarah. Yeh $\overset{ˉ}{X}$ ka SD hai.
Standardize karo. $z = \frac{0.6 - 0.5}{0.0707} = 1.414$ . Yeh step kyun? Same bridge pehle wala: $0.6$ ko standard-error units mein express karo.
Right tail. $P (\overset{ˉ}{X} > 0.6) \approx 1 - Φ (1.414) = 1 - 0.9214 = 0.0786$ .

Verify: raw variable skewed hai, phir bhi CLT apply hota hai kyunki iska variance finite hai. Answer $0.0786 < 0.5$ consistent hai $0.6$ ke mean se upar hone ke saath. Forecast: haan — averaging skew ko tame kar deta hai. ✓ (Yeh Cell B hai: skewness allowed hai, sirf zyada $n$ chahiye hoga agar answer tail mein super accurate hona ho.)

Example 3 — Cell C: discrete count, continuity correction

Figure — Central Limit Theorem — statement, proof sketch, significance

Worked example Normal se Binomial (de Moivre–Laplace)

$X \sim Bin (n = 200, p = 0.3)$ — 200 independent Bernoulli trials ka sum. $P (X \leq 50)$ dhundo.

Forecast: $50$ expected count $n p = 60$ se neeche hai. Left tail ya right tail? $0.5$ se upar ya neeche?

Count ka Mean aur SD. $μ = n p = 200 \cdot 0.3 = 60$ ; $σ = n p (1 - p) = 200 \cdot 0.3 \cdot 0.7 = 42 = 6.4807$ . Yeh step kyun? Ek Bernoulli $(p)$ ka variance $p (1 - p)$ hota hai; 200 unhe add karo.
Continuity correction apply karo. $P (X \leq 50)$ ke liye cutoff $50.5$ use karo, $50$ nahi. Yeh step kyun? Figure s01 mein red bar dekho: har integer count ek width 1 ka bar hai integer par centred. " $50$ " ka bar $49.5$ se $50.5$ tak stretch karta hai. Ise poora smooth curve ke neeche capture karne ke liye, boundary iske right edge, $50.5$ par baithe. Yeh miss karna CLT ka single most common error hai.
Standardize karo. $z = \frac{50.5 - 60}{6.4807} = - 1.4659$ . Yeh step kyun? Negative sign note karo — cutoff mean se neeche hai, toh $z < 0$ . Signs matter karte hain: yeh tumhe bell ke sahi side par place karta hai.
Left tail directly padho. $P (X \leq 50) \approx Φ (- 1.4659) = 0.0713$ . Yeh step kyun? $Φ$ already left area deta hai, toh yahan koi subtraction nahi. Symmetry use karo $Φ (- a) = 1 - Φ (a)$ : $1 - Φ (1.466) = 1 - 0.9287 = 0.0713$ .

Verify: below-mean cutoff ⇒ probability below $0.5$ ⇒ $0.0713$ ✓. Correction ke bina hum $50$ use karte, $z = - 1.543$ milta aur $Φ = 0.0614$ — bar width ignore karne se ek noticeable error.

Example 4 — Cell D: ek "between" (two-sided) probability

Worked example Sample mean ek window mein land karna

Example 1 ke 100 dice reuse karo lekin average $\overset{ˉ}{X} = S /100$ ke liye pucho. $P (3.4 < \overset{ˉ}{X} < 3.6)$ dhundo.

Forecast: yeh window mean $3.5$ par centred hai. Kya answer bada hona chahiye (bell ka zyada hissa) ya tiny?

Mean ka Standard error. $\frac{σ}{n} = \frac{1.708}{10} = 0.1708$ . Yeh step kyun? Hum sum se mean par switch ho gaye, toh single-die $σ$ ko $100 = 10$ se divide karo.
Dono ends standardize karo. $z_{1} = \frac{3.4 - 3.5}{0.1708} = - 0.5855, z_{2} = \frac{3.6 - 3.5}{0.1708} = + 0.5855.$ Yeh step kyun? Ek "between" question ke liye do boundaries chahiye; note karo yeh symmetric hain ( $\pm$ ) kyunki window mean ke baare mein symmetric hai.
Do left-areas subtract karo. $P = Φ (0.5855) - Φ (- 0.5855)$ . Yeh step kyun? $Φ (z_{2})$ right wall ke left mein sab kuch hai; $Φ (z_{1})$ (left wall ke left mein sab kuch) hata do; jo bachta hai woh beech ki strip hai.
Evaluate karo. $Φ (0.5855) = 0.7209$ , $Φ (- 0.5855) = 0.2791$ , toh $P = 0.7209 - 0.2791 = 0.4418$ .

Verify: window roughly $\pm 0.59$ SE wide hai, half a standard deviation se thoda zyada har side — bell ka kareeb $44%$ capture karna sensible hai (yaad raho $\pm 1$ SE $\approx 68%$ capture karta hai). Symmetric window ⇒ $z_{1} = - z_{2}$ ✓. Forecast: moderately bada, tiny nahi. ✓

Example 5 — Cell E: inverse problem, $n$ ke liye solve karo

Worked example Error target hit karne ke liye kitna data chahiye?

Aap $σ = 2$ ke saath ek mean estimate kar rahe ho. Aap chahte ho ki ek $95%$ confidence interval jiska half-width (margin of error) zyada se zyada $0.25$ ho. $n$ kitna bada hona chahiye?

Forecast: aap ek SE se tighter interval chahte ho — kya $n$ tens mein hoga, hundreds mein, ya thousands mein?

Margin of error likho. $95%$ ke liye hum $z_{0.025} = 1.96$ use karte hain (woh point jisme har tail mein $2.5%$ ho). Margin $= z_{0.025} \cdot \frac{σ}{n} = 1.96 \cdot \frac{2}{n}$ . Yeh step kyun? CLT kehta hai $\overset{ˉ}{X}$ normal hai spread $σ / n$ ke saath; interval har side $1.96$ SEs tak jaata hai.
Ise $\leq 0.25$ set karo aur solve karo. $1.96 \cdot \frac{2}{n} \leq 0.25 \Rightarrow n \geq \frac{1.96 \cdot 2}{0.25} = 15.68 \Rightarrow n \geq 15.6 8^{2} = 245.86.$ Yeh step kyun? $n$ isolate karo, phir square karo. Squaring exactly wahi wajah hai ki apna error half karne mein 4× data lagta hai.
Round up karo. $n = 246$ (sample-size requirement ko kabhi neeche round nahi kar sakte). Yeh step kyun? $n = 245$ margin $0.25$ se thoda zyada deta; bar clear karni hai.

Verify: $n = 246$ back plug karo: margin $= 1.96 \cdot 2/ 246 = 0.2499 \leq 0.25$ ✓. $n = 245$ plug karo: margin $= 0.2504 > 0.25$ ✗ — confirm karta hai ki 246 sabse chhota hai jo kaam karta hai.

Example 6 — Cell F: real-world word problem

Worked example Elevator weight limit

Ek elevator $1200$ kg ke liye rated hai. Adult passenger weights independent hain, mean $μ = 75$ kg aur SD $σ = 15$ kg ke saath (shape unknown). Agar $n = 15$ adults board karte hain, toh total weight rating exceed karne ki probability kya hai?

Forecast: $15 \times 75 = 1125$ kg expected load hai, limit se thoda neeche. Risky hai ya safe?

Sum mein translate karo. Total load $S = \sum_{i = 1}^{15} X_{i}$ . $E [S] = 15 \cdot 75 = 1125$ ; $Var (S) = 15 \cdot 1 5^{2} = 3375$ ; $SD (S) = 3375 = 58.095$ . Yeh step kyun? "Total weight exceeds" ek statement hai sum ke baare mein; means aur variances 15 independent logon par add hote hain.
Limit standardize karo. $z = \frac{1200 - 1125}{58.095} = 1.291$ . Yeh step kyun? $1200$ kg rating ko total ke standard-error units mein express karo.
Right tail. $P (S > 1200) \approx 1 - Φ (1.291) = 1 - 0.9017 = 0.0983$ .

Verify: units check — kg minus kg over kg ek pure $z$ deta hai. Mean load cap se $\approx 1.29$ SDs neeche baitta hai, toh roughly $10%$ overload risk plausible hai aur chinta karne layak hai. Forecast: risky-ish, kareeb $1$ in $10$ . ✓ (Note: sirf $n = 15$ ke saath, yeh normal approximation decent hai lekin exact nahi — CLT asymptotic hai.)

Example 7 — Cell G: degenerate input $σ = 0$

Worked example Agar har value identical ho toh?

Maan lo $X_{i} = 7$ hamesha — ek "random" variable jisme koi randomness nahi. Toh $μ = 7$ aur $σ^{2} = 0$ . CLT $\overset{ˉ}{X}_{n}$ ke baare mein kya kehta hai?

Forecast: kya aap standard score bhi bana sakte ho? Denominator dekho.

Sample mean compute karo. Har term $7$ hai, toh $\overset{ˉ}{X}_{n} = 7$ exactly, har $n$ ke liye. Kabhi koi spread nahi. Yeh step kyun? Identical numbers average karne par wahi number wapas milta hai; fluctuate karne ke liye kuch hai hi nahi.
Standardize karne ki koshish karo. $Z_{n} = \frac{X ˉ _{n} - μ}{σ / n} = \frac{7 - 7}{0/ n} = \frac{0}{0}$ — undefined. Yeh step kyun? Yeh dikhata hai kyun parent definition $σ^{2} > 0$ demand karta hai. Zero variance ke saath koi bell converge karne ke liye hai hi nahi; aap zero spread ko zero spread se divide kar rahe hoge.
Sahi conclusion state karo. Koi Gaussian limit nahi hai. $\overset{ˉ}{X}_{n}$ ki distribution $7$ par ek single spike hai (ek "degenerate" distribution) — yahi exactly woh hai jo Law of Large Numbers deta hai, CLT layer ke collapse hone ke saath.

Verify: theorem statement mein hypothesis " $σ^{2} > 0$ " decoration nahi hai — yeh case precisely wahi hai jo ise exclude karta hai. $\overset{ˉ}{X}_{n} \equiv 7$ ka variance $0$ hai, "spike, no bell" se match karta hai. ✓

Example 8 — Cell H: limiting failure, infinite variance

Worked example Jab CLT toot jaata hai: Cauchy distribution

Maano $X_{i}$ i.i.d. standard Cauchy hain (ek bell-looking curve fat tails ke saath). Iska koi finite mean nahi aur infinite variance hai. $n \to \infty$ ke saath $\overset{ˉ}{X}_{n}$ ka kya hota hai?

Forecast: zyada data usually help karta hai — kya ek million Cauchy samples ka average kisi value ke paas settle hoga?

CLT hypothesis check karo. Theorem finite variance $σ^{2}$ maangta hai. Cauchy ke liye, $σ^{2} = \infty$ (fat tails $\int x^{2} f (x) d x$ diverge karte hain). Toh CLT apply nahi hota. Yeh step kyun? Theorem ki har hypothesis ek promise hai jo aapko honor karni hai; ek tod do aur conclusion void hai. Hum hypotheses pehle check karte hain computation se, taki blindly ek formula mein plug na karen jo yahan licensed nahi hai.
Ek term ka characteristic function likho. Standard Cauchy ke liye, characteristic function hai $φ (t) = E [e^{i tX}] = e^{- ∣ t ∣}$ . Yeh step kyun? Cauchy ka koi usable mean ya variance nahi hai, toh parent proof ka standardize-and-Taylor recipe available nahi hai. CF woh ek fingerprint hai jo abhi bhi exist karta hai aur sums ke under cleanly behave karta hai, toh hum ise track karte hain.
Average ka CF banao. Independence se, CFs multiply hote hain, aur $X$ ko $1/ n$ se scale karna CF ka argument scale karta hai: $φ_{\overset{ˉ}{X}_{n}} (t) = [φ (\frac{t}{n})]^{n} = [e^{- ∣ t / n ∣}]^{n} = e^{- n \cdot ∣ t ∣/ n} = e^{- ∣ t ∣} .$ Yeh step kyun? Yeh exactly wahi move hai jaise parent proof ka Step 2 (independence ⇒ product ⇒ $n$ -th power), lekin average $\overset{ˉ}{X}_{n} = \frac{1}{n} \sum X_{i}$ ke saath scaling $1/ n$ hai, $1/ n$ nahi. $n$ ke do powers perfectly cancel ho jaate hain.
Limit pehchano — yeh kabhi badla hi nahi. $φ_{\overset{ˉ}{X}_{n}} (t) = e^{- ∣ t ∣}$ ek single term ke CF se identical hai. Toh $\overset{ˉ}{X}_{n}$ ki exactly same standard-Cauchy distribution hai jaise ek $X_{1}$ , har $n$ ke liye. Yeh $e^{- t^{2} /2}$ (normal ka CF) ki taraf converge nahi kar raha; yeh kisi naye cheez mein converge hi nahi karta. Yeh step kyun? Kyunki CF ek unique fingerprint hai, equal CFs ⇒ equal distributions. Ek billion Cauchy samples average karna ek dekhne se zyada accurate nahi hai — exactly figure s02 mein wild red curve.

Verify: self-reproducing CF $e^{- ∣ t ∣} \to e^{- ∣ t ∣}$ (na ki $e^{- t^{2} /2}$ ) prove karta hai ki limit Cauchy hai, normal nahi. Theorem ka finite-variance clause exactly wahi hai jo fail karta hai, aur Lindeberg–Feller Condition woh general test hai jo formally aisi cases ko rule out karta hai. ✓

Example 9 — Cell I: exam twist, do means ka difference

Worked example Do independent samples compare karna

Machine A parts produce karta hai mean length $μ_{A} = 50$ mm, SD $σ_{A} = 4$ mm ke saath; aap $n_{A} = 64$ sample karte ho. Machine B ka $μ_{B} = 48$ mm, SD $σ_{B} = 3$ mm hai; aap $n_{B} = 36$ sample karte ho. Dono samples independent hain. $P (\overset{ˉ}{X}_{A} - \overset{ˉ}{X}_{B} > 3)$ dhundo.

Forecast: means ka true difference $50 - 48 = 2$ mm hai. Kya $3$ mm exceed karna common hai ya rare?

Difference ka Mean. $E [\overset{ˉ}{X}_{A} - \overset{ˉ}{X}_{B}] = μ_{A} - μ_{B} = 50 - 48 = 2$ . Yeh step kyun? Expectation linear hai: difference ka mean, means ka difference hota hai.
Difference ka Variance. Har sample mean CLT se $\approx$ normal hai variance $σ^{2} / n$ ke saath: $Var (\overset{ˉ}{X}_{A}) = \frac{16}{64} = 0.25, Var (\overset{ˉ}{X}_{B}) = \frac{9}{36} = 0.25.$ Kyunki samples independent hain, difference ke variances add hote hain: $Var (\overset{ˉ}{X}_{A} - \overset{ˉ}{X}_{B}) = 0.25 + 0.25 = 0.5$ . SD $= 0.5 = 0.7071$ . Yeh step kyun? Independent $U, V$ ke liye $Var (U - V) = Var (U) + Var (V)$ — minus sign variances subtract nahi karta (variance independent noise add karne se kabhi shrink nahi ho sakta).
Standardize karo. $z = \frac{3 - 2}{0.7071} = 1.4142$ . Yeh step kyun? Do normals ka difference khud normal hai, toh ek aakhri standardization ise finish karta hai.
Right tail. $P > 1 - Φ (1.4142) = 1 - 0.9214 = 0.0786$ .

Verify: cutoff $3$ mean $2$ se ek true-difference-plus-a-bit upar baitha hai, $\approx 1.41$ SDs out ⇒ answer $0.5$ se kaafi neeche. Notice karo $0.0786$ Example 2 ke tail se match karta hai kyunki dono mein $z = 1.414$ tha — ek achha internal consistency check. ✓

Recall Kaun si cell kaun si thi? (quick self-test)

Symmetric discrete sum ::: Cell A (Ex 1 — dice) Skewed continuous mean ::: Cell B (Ex 2 — exponential) Discrete count jisme $+ 0.5$ shift chahiye ::: Cell C (Ex 3 — binomial) Do bounds ke "between" mein ::: Cell D (Ex 4 — $Φ (z_{2}) - Φ (z_{1})$ ) Sample size $n$ ke liye solve karo ::: Cell E (Ex 5 — $n$ paane ke liye square karo) English word problem ::: Cell F (Ex 6 — elevator sum) Zero variance, koi bell nahi ::: Cell G (Ex 7 — $0/0$ , degenerate) Infinite variance, CLT fail ::: Cell H (Ex 8 — Cauchy) Do means ka difference ::: Cell I (Ex 9 — variances phir bhi add hote hain)