Central limit theorem: Exercises — Hinglish deep dive

1.3.15 · D4 · AI-ML › Probability & Statistics › Central limit theorem

Do symbols baar baar aate hain, toh kisi bhi problem se pehle inhe simple words mein samajh lete hain.

L1 — Recognition

Goal: kya tum spot kar sakte ho jab CLT apply hoti hai aur uske parameters read kar sakte ho?

Exercise 1.1. Ek random variable $X$ ka $μ = 10$ , $σ^{2} = 36$ hai. Tum $n = 9$ i.i.d. draws ka mean lete ho. $\overset{ˉ}{X}_{9}$ ki approximate distribution aur uska standard error likho.

Recall Solution 1.1

CLT kehta hai $\overset{ˉ}{X}_{n} \sim N (μ, σ^{2} / n)$ .

Mean same rehta hai $μ = 10$ .
Variance shrink hota hai: $σ^{2} / n = 36/9 = 4$ .
Toh $\overset{ˉ}{X}_{9} \sim N (10, 4)$ .
Standard error $SE = σ / n = 6/3 = 2$ .

Kaisa dikhta hai: individual $X$ ki bell ka spread $σ = 6$ hai; average ki bell zyada narrow hai, spread $2$ , same point par centred.

Exercise 1.2. In settings mein se classical CLT directly kiske liye normal limit guarantee nahi karta? (a) 100 i.i.d. uniform draws ka mean, (b) Cauchy distribution (infinite variance) se 100 i.i.d. draws ka mean, (c) 100 i.i.d. Bernoulli draws ka mean.

Recall Solution 1.2

CLT ko finite variance $σ^{2} < \infty$ chahiye.

(a) Uniform: finite variance ✓ CLT apply hota hai.
(b) Cauchy: infinite variance ✗ — CLT apply nahi hota. (Iske sample means Cauchy hi rehte hain, kabhi bell-shaped nahi bante.)
(c) Bernoulli: variance $p (1 - p)$ , finite ✓ CLT apply hota hai.

Answer: (b).

L2 — Application

Goal: numbers ko standardize-then- $Φ$ mein plug karo.

Exercise 2.1. $X$ ka $μ = 50$ , $σ = 10$ hai. $n = 25$ ke liye, $P (\overset{ˉ}{X}_{25} \leq 52)$ nikalo.

Recall Solution 2.1

$SE = 10/ 25 = 10/5 = 2$ . Bound $52$ ko standardize karo: $Z = \frac{52 - 50}{2} = 1.$ "Bell ka kitna hissa $z = 1$ ke left mein hai?" Woh $Φ (1)$ hai: $P (\overset{ˉ}{X}_{25} \leq 52) = Φ (1) \approx 0.8413 .$

Exercise 2.2. Ek fair die ( $μ = 3.5$ , $σ^{2} = 2.9167$ , $σ = 1.708$ ) ko $n = 49$ baar roll kiya jaata hai. $P (\overset{ˉ}{X}_{49} \geq 3.9)$ nikalo.

Recall Solution 2.2

$SE = 1.708/ 49 = 1.708/7 = 0.2440$ . $Z = \frac{3.9 - 3.5}{0.2440} = \frac{0.4}{0.2440} = 1.639.$ Hum $z$ ke right ka area chahte hain, jo $1 - Φ (z)$ hai: $P (\overset{ˉ}{X}_{49} \geq 3.9) = 1 - Φ (1.639) \approx 1 - 0.9494 = 0.0506 .$

Picture: $3.9$ centre se $\approx 1.64$ SE upar baitha hai — bilkul 95th-percentile line par, toh sirf approximately 5% averages isse aage jaate hain.

L3 — Analysis

Goal: yeh reason karo ki jab parameters move hote hain toh answer kaise change hota hai.

Exercise 3.1. Ek sensor ki readings ka $σ = 4$ hai. Tum currently $n = 100$ readings average kar rahe ho, jo kuch standard error deta hai. Standard error ko halve karne ke liye tumhe kitni readings average karni hongi?

Recall Solution 3.1

$SE = σ / n$ . $SE$ ko halve karne ke liye $n$ ko double karna hoga, isliye $n$ ko quadruple karna hoga. $n_{new} = 4 \times 100 = 400 .$ Check: purana $SE = 4/10 = 0.4$ ; naya $SE = 4/20 = 0.2$ . Exactly halved.

$n$ kyun aur $n$ kyun nahi? Precision samples ke square root ke saath improve hoti hai — sampling ke peeche yeh classic law of diminishing returns hai. Law of Large Numbers dekho.

Exercise 3.2. Same $μ$ ke do estimators: estimator A, $n_{A} = 64$ samples average karta hai per-sample $σ_{A} = 8$ ke saath; estimator B, $n_{B} = 400$ samples average karta hai $σ_{B} = 20$ ke saath. Kaun zyada precise hai (chhota SE), aur kis ratio mein?

Recall Solution 3.2

$SE_{A} = 8/ 64 = 8/8 = 1.$ $SE_{B} = 20/ 400 = 20/20 = 1.$ Dono equally precise hain — ratio $SE_{A} / SE_{B} = 1$ .

Insight: B ke paas zyada samples hain lekin noisier data hai; dono effects exactly cancel ho jaate hain. Precision combination $σ / n$ par depend karti hai, kisi ek quantity par nahi.

L4 — Synthesis

Goal: CLT ko sample-size aur confidence-interval logic ke saath ek problem mein combine karo.

Exercise 4.1. Tum ek mini-batch gradient estimate kar rahe ho. Per-sample gradient variance $σ^{2} = 0.25$ (toh $σ = 0.5$ ). Tum chahte ho ki estimate true gradient ke $\pm 0.02$ ke andar ho 95% confidence ke saath. Kitna batch size $n$ chahiye?

Recall Solution 4.1

95% two-sided confidence critical value $z^{⋆} = 1.96$ use karta hai (kyunki $Φ (1.96) = 0.975$ , har tail mein $2.5%$ chodta hai). Hum chahte hain ki half-width $z^{⋆} \cdot SE$ $\leq 0.02$ ho: $1.96 \cdot \frac{0.5}{n} \leq 0.02.$ $n$ ke liye solve karo: $n \geq \frac{1.96 \times 0.5}{0.02} = 49 \Rightarrow n \geq 4 9^{2} = 2401 .$

Training ke liye kya matlab hai: tolerance ko quarter karne ke liye (jaise $\pm 0.04$ -type windows se), batch cost multiply hoti hai — yeh Stochastic Gradient Descent aur Sample Size Calculation mein compute-vs-precision tradeoff ka ek concrete example hai.

Exercise 4.2. Ek naye recommender ki true conversion rate $p = 0.05$ hai (Bernoulli, toh $σ^{2} = p (1 - p)$ ). 95% confidence par rate ko $\pm 0.005$ ke andar estimate karne ke liye kitne users $n$ chahiye?

Recall Solution 4.2

$σ^{2} = 0.05 \times 0.95 = 0.0475$ . Required: $1.96 \cdot \frac{0.0475}{n} \leq 0.005$ . $n \geq \frac{1.96 0.0475}{0.005} .$ Dono sides square karo (cleaner): $n \geq \frac{1.9 6 ^{2} \times 0.0475}{0.00 5 ^{2}} = \frac{3.8416 \times 0.0475}{0.000025} = \frac{0.18248}{0.000025} = 7299.3.$ Upar round karo (tum ek user ko fractionally sample nahi kar sakte): $n = 7300$ .

Yeh A/B-test planning ke peeche ka standard formula hai — Confidence Intervals aur Hypothesis Testing dekho.

L5 — Mastery

Goal: judge karo jab tools break ho jaate hain, aur reasoning repair karo.

Exercise 5.1. Ek colleague ek heavily skewed distribution se $n = 4$ samples ka mean ke liye 95% confidence interval compute karta hai, CLT normal approximation plug in karke. Kya yeh valid hai? Tum uski jagah kya karte? Conceptual answer do, koi numbers nahi.

Recall Solution 5.1

Reliably valid nahi. CLT ek limit hai jab $n \to \infty$ ; chhote $n$ ke liye ek skewed source ke sample mean ka khud skewed hona possible hai — bell abhi bana nahi hai, toh $Φ$ -based intervals miscalibrated hain (true coverage $\neq =$ 95%). Repairs:

$n$ badhaao taaki approximation kick in kare (skew convergence slow karta hai; tumhe hundreds ki zaroorat ho sakti hai).
Ek resampling approach use karo jo koi normality assumption nahi karta — Bootstrap Methods — interval empirically pane ke liye.
Ya exact/simulation methods use karo jaise Monte Carlo Methods.

Key mastery point: CLT ki accuracy dono $n$ aur source ki shape (skew/heavy tails) par depend karti hai.

Exercise 5.2 (capstone). Per-sample loss values ka unknown mean $μ$ hai lekin known $σ = 3$ hai. Tum $n = 144$ samples average karte ho aur $\overset{ˉ}{X} = 7.4$ observe karte ho. (a) $μ$ ke liye 95% confidence interval do. (b) Isse use karke, 5% level par $H_{0} : μ = 8$ vs $H_{1} : μ \neq = 8$ test karo.

Recall Solution 5.2

(a) $SE = σ / n = 3/ 144 = 3/12 = 0.25$ . 95% CI: $\overset{ˉ}{X} \pm z^{⋆} SE = 7.4 \pm 1.96 (0.25) = 7.4 \pm 0.49$ . $[6.91, 7.89] .$

(b) $H_{0}$ ke under test statistic (using $μ_{0} = 8$ ): $Z = \frac{X ˉ - μ _{0}}{SE} = \frac{7.4 - 8}{0.25} = \frac{- 0.6}{0.25} = - 2.4.$ $∣ Z ∣ = 2.4 > 1.96 = z^{⋆}$ , toh hum 5% level par $H_{0}$ reject karte hain. Consistency check: $μ_{0} = 8$ CI $[6.91, 7.89]$ ke bahar hai — level $α$ par two-sided test exactly tab reject karta hai jab $μ_{0}$ , $(1 - α)$ CI se bahar girta hai. Dono methods agree karte hain, jaise hona chahiye. Hypothesis Testing dekho.

Two-sided p-value: $2 (1 - Φ (2.4)) = 2 (1 - 0.9918) = 0.0164 < 0.05$ . ✓

Recall Self-test cloze

Sample mean ka standard error == $σ / n$ == hai. Precision quadruple karne ke liye (SE ko quarter karne ke liye) $n$ ko 16 se multiply karo. Ek two-sided 95% test $H_{0}$ ko tab reject karta hai jab $μ_{0}$ , 95% confidence interval ke outside ho. CLT fail hoti hai jab source variance infinite ho (jaise Cauchy).

Related tools jinhe tumne ab connect kar liya hai: Normal Distribution, Law of Large Numbers, Confidence Intervals, Hypothesis Testing, Sample Size Calculation, Bootstrap Methods, Monte Carlo Methods, Stochastic Gradient Descent, Bias-Variance Tradeoff, Maximum Likelihood Estimation.