Worked examples — Monte Carlo simulation — law of large numbers basis

4.9.25 · D3 · Maths › Probability Theory & Statistics › Monte Carlo simulation — law of large numbers basis

Symbols aane se pehle, parent note ke teen characters ki ek reminder:

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Scenario matrix

Har Monte Carlo problem inhi case classes mein se ek mein aata hai. Table ek map hai; neeche ke examples har row ko fill karte hain.

Cell	Case class	Kya cheez ise alag banati hai	Example
A	Bounded integrand on $[0,1]$	Textbook case, finite variance	Ex 1
B	Integral on $[a,b]\ne [0,1]$	Uniform ke liye rescale karna zaroori	Ex 2
C	Improper / infinite range $[0,\infty )$	Uniform impossible — sampling density badlo	Ex 3
D	Probability = indicator average (0/1 output)	$g$ sirf 0 ya 1 return karta hai (Bernoulli)	Ex 4
E	Degenerate: $g$ constant → zero variance	$\text{SE}=0$, ek hi sample mein converge	Ex 5
F	Limiting behaviour: error kaise scale karta hai?	Target accuracy ke liye $n$ plan karo	Ex 6
G	Heavy tail: variance infinite → LLN toot jaata hai	Woh ek case jo converge NAHI karta	Ex 7
H	Real-world word problem	Story ko expectation mein translate karo	Ex 8
I	Exam twist: variance reduction	Same $\theta$, chhota $\text{SE}$	Ex 9

Prerequisite links jo aap open rakhna chahein: Bernoulli Distribution, Variance and Covariance, Central Limit Theorem, Chebyshev's Inequality, Numerical Integration, Importance Sampling, Strong Law of Large Numbers. Yeh page Monte Carlo simulation — law of large numbers basis ka child hai.

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Ex 1 — Cell A: bounded integrand on $[0,1]$

Active symbol: random input $U\sim \text{Unif}(0,1)$ hai (uniform).

Forecast: true value guess karo aur guess karo ki kya $n=1000$ samples 2 decimals sahi denge. (Padhne se pehle likh lo.)

Step 1 — Integral ko expectation mein badlo. ${\int }_0^1{x}^{2}dx=\mathbb{E}[U^2],U\sim \text{Unif}(0,1).$ Yeh step kyun? $[0,1]$ par uniform density constant $1$ hai, toh ${\int }_0^{1}g(x)\cdot 1dx$ by definition $g(U)$ ka average hai. Yahan $g(x)=x^2$.

Step 2 — Estimator likho. ${\hat{\theta }}_n=\frac{1}{n}{\sum }_{i=1}^n{U}_i^2.$ Kyun? True average ko sample average se replace karo — LLN promise karta hai yeh converge karega.

Step 3 — True value & true variance (taaki pata ho kya expect karein). $\theta =[\frac{x^3}{3}{]}_0^1=\frac{1}{3},\mathbb{E}[U^4]=\frac{1}{5},{\sigma }_g^2=\mathbb{E}[U^4]-{\theta }^2=\frac{1}{5}-\frac{1}{9}=\frac{4}{45}.$ ${\sigma }_g^2$ kyun compute karein? Kyunki $\text{SE}={\sigma }_g/\sqrt{n}$ hume error size pehle se batata hai, kuch bhi run karne se pehle.

Step 4 — $n=1000$ par predicted error. $\text{SE}=\sqrt{\frac{4}{45}}/\sqrt{1000}=\frac{0.2981}{31.62}\approx \mathrm{0.00943.}$ $n=1000$ kyun plug in karein? Humne wahi sample size li jo forecast ne poocha tha, taaki hum pehle se decide kar sakein ki kya yeh 2 decimals dega — koi simulation nahi chahiye.

Verify: $\theta =0.3333\dots$, aur $\text{SE}\approx 0.0094<0.01$, toh ek typical run $\frac{1}{3}$ ke roughly $\pm 0.019$ (do SEs) ke andar land karta hai — roughly 2 decimals tak sahi. Forecast check out karta hai. Units: dimensionless (pure number).

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Ex 2 — Cell B: general interval $[a,b]$ par integral

Active symbol: random input $U\sim \text{Unif}(2,5)$ hai (uniform, lekin ab $[2,5]$ par).

Forecast: interval ki length 3 hai, 1 nahi. Guess karo: "$\times$ length" factor average ke andar jaata hai ya bahar?

Step 1 — $[2,5]$ par uniformly sample karo. Ek point $U\sim \text{Unif}(2,5)$ ki density $\frac{1}{b-a}=\frac{1}{3}$ hai (constant, lekin ab $\frac{1}{3}$ nahi $1$). Yeh step kyun? Lambe interval par, "evenly spread" ka matlab hai har unit of length ko probability $\frac{1}{b-a}$ milti hai, toh density shrink ho jaati hai.

Step 2 — Expectation se integral recover karo. $\mathbb{E}[\sqrt{U}]={\int }_2^5\sqrt{x}\cdot \frac{1}{3}dx=\frac{1}{3}\theta \Rightarrow \theta =(b-a)\mathbb{E}[\sqrt{U}]=3\mathbb{E}[\sqrt{U}].$ Kyun? Density $\frac{1}{3}$ integral ko divide karti hai; ise undo karne ke liye average ko $(b-a)=3$ se multiply karo. Length factor average ke BAHAR hai.

Step 3 — Estimator. ${\hat{\theta }}_n=(b-a)\cdot \frac{1}{n}{\sum }_{i=1}^n\sqrt{U_i}=3\cdot \frac{1}{n}\sum \sqrt{U_i}.$ Exactly yeh form kyun? Step 2 ne prove kiya $\theta =3\mathbb{E}[\sqrt{U}]$; hum simply true average $\mathbb{E}[\sqrt{U}]$ ko sample average $\frac{1}{n}\sum \sqrt{U_i}$ se replace karte hain (LLN) aur constant $3$ ko bahar rakhte hain — yahi estimator hum actually run karenge.

Step 4 — Check ke liye true value. ${\int }_2^5{x}^{1/2}dx=[\frac{2}{3}{x}^{3/2}{]}_2^5=\frac{2}{3}(5^{1.5}-2^{1.5})=\frac{2}{3}(11.1803-2.8284)=\mathrm{5.5679.}$ Exact integral yahan kyun compute karein? Is case mein ek elementary antiderivative $\frac{2}{3}{x}^{3/2}$ hai, toh hum true $\theta$ by hand nikaal sakte hain aur ise yardstick ki tarah use kar sakte hain jise Monte Carlo estimate match karna chahiye — validation, method nahi.

Verify: $\theta \approx 5.5679$. Sanity check: interval ki width 3 hai aur $\sqrt{x}$ $\sqrt{2}\approx 1.41$ se $\sqrt{5}\approx 2.24$ tak range karta hai, toh integral $3\times 1.41=4.24$ aur $3\times 2.24=6.71$ ke beech hona chahiye. Humara $5.57$ aaram se andar hai. ✔

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Ex 3 — Cell C: $[0,\infty )$ par improper integral

Active symbol: random input $X\sim \text{Exp}(1)$ hai (exponential), uniform $U$ se $X=-\ln U$ ke zariye bana.

Forecast: "$[0,\infty )$ par uniformly sample karo" possible nahi — infinite line par koi even spread nahi hoti. Guess karo hum isse kaise bachte hain.

Figure s01 — kya dekh rahe ho: horizontal axis $x$ hai jo $0$ se infinity ki taraf jaata hai; vertical axis probability density hai. Red curve sampling density $p(x)=e^{-x}$ hai. Neeche black tick marks 40 actual samples hain jo isse draw kiye gaye — notice karo woh $x=0$ ke paas crowd karte hain (jahan red curve tall hai) aur right ki taraf thin ho jaate hain (jahan yeh flat ho jaati hai). Yeh Step 1 ke peeche ki picture hai: hum wahan sample karte hain jahan density kehti hai mass hai.

Step 1 — Built-in density pehchano. Function $p(x)=e^{-x}$ for $x\ge 0$ ek genuine probability density hai: yeh positive hai aur ${\int }_0^{\infty }{e}^{-x}dx=1$. Yeh Exponential(1) distribution hai. Yeh step kyun? Kyunki hum infinite line par uniform nahi bana sakte, hum instead ek aisi density se sample karte hain jo already wahan rehti hai. Figure mein red curve samples ko 0 ke paas pack karti hai aur infinity ki taraf thin karti hai, exactly jahan $e^{-x}$ kehta hai "mass" hai.

Step 2 — Integrand ko density $\times$ leftover mein split karo. $\theta ={\int }_0^{\infty }\underset{\text{density }p(x)}{\underbrace{e^{-x}}}\underset{g(x)}{\underbrace{\cos x}}dx={\mathbb{E}}_{X\sim \text{Exp}(1)}[\cos X].$ Kyun? Density ko hum jo bhi multiply karte hain woh function ban jaata hai jise hum average karte hain. Yahan leftover simply $\cos x$ hai, aur active input ab $X\sim \text{Exp}(1)$ hai (uniform nahi).

Step 3 — Exponential samples generate karo. $U\sim \text{Unif}(0,1)$ draw karo aur $X=-\ln U$ set karo. Yeh standard trick (inverse-CDF) uniform ko Exponential(1) mein badal deti hai. Kyun? Humare paas sirf uniform RNGs hain; $-\ln U$ unhe density mein reshape karta hai jo chahiye. Yeh Importance Sampling ka seed hai.

Step 4 — Estimator. ${\hat{\theta }}_n=\frac{1}{n}{\sum }_{i=1}^n\cos (X_i),X_i=-\ln U_i.$ Yeh form kyun? Step 2 ne $\theta$ ko $\mathbb{E}[\cos X]$ with $X\sim \text{Exp}(1)$ likh diya, toh $\cos (X_i)$ ka sample mean natural estimator hai — koi extra density factor nahi chahiye kyunki $e^{-x}$ absorb ho gaya hai jis tarah hum $X$ sample karte hain ussi mein.

Step 5 — True value. ${\int }_0^{\infty }{e}^{-x}\cos xdx=\frac{1}{1^2+1^2}=\frac{1}{2}.$ Yeh formula kyun? Standard result ${\int }_0^{\infty }{e}^{-ax}\cos (bx)dx=\frac{a}{a^2+b^2}$ with $a=b=1$.

Verify: $\theta =0.5$. Sanity: $\cos X$ $-1$ aur $1$ ke beech oscillate karta hai lekin samples chhote $x$ ke paas cluster karte hain jahan $\cos x\approx 1$, toh 1 se neeche ek positive value expected hai. $0.5$ ✔

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Ex 4 — Cell D: indicator average ke roop mein probability

Active symbol: random input pair $(X,Y)$ hai, har coordinate independent $\text{Unif}(0,1)$.

Forecast: unit square mein $y=x^2$ ke upar ka region — uska area ek fraction ke roop mein guess karo.

Figure s02 — kya dekh rahe ho: axes dart ke coordinates $X$ (horizontal) aur $Y$ (vertical) hain, dono $[0,1]$ mein. Black curve parabola $y=x^2$ hai. Red shaded band woh region hai jahan $Y>X^2$ — "success" region. Dots 120 random darts hain: red dots curve ke upar land kiye (score $1$), black dots neeche (score $0$). Estimate sirf woh fraction hai jo red hain, jo visually red area $=2/3$ hai.

Step 1 — Indicator define karo. $g(X,Y)=1\{Y>X^2\}=\{\begin{array}{ll}1& \text{point curve ke upar hai}\\ 0& \text{warna}\end{array}$ Yeh step kyun? Probability 0/1 indicator ki expectation hai: $P(A)=\mathbb{E}[1_A]$. $g$ ka output ek coin flip hai — yeh Bernoulli Distribution hai.

Step 2 — Estimator = "yes" darts ka fraction. ${\hat{\theta }}_n=\frac{1}{n}{\sum }_{i=1}^{n}1\{Y_i>X_i^2\}.$ Kyun? 0s aur 1s ka average sirf successes ka fraction count karta hai — exactly figure mein red-dot fraction.

Step 3 — True value. Parabola ke upar ka area = total square minus uske neeche ka area: $\theta =1-{\int }_0^1{x}^{2}dx=1-\frac{1}{3}=\frac{2}{3}.$ Subtract kyun? Poore square ka area $1$ hai; parabola ke neeche ka region ${\int }_0^1{x}^{2}dx=\frac{1}{3}$ area rakhta hai; baaki sab "above" region hai, toh $1-\frac{1}{3}$.

Step 4 — Iska variance (Bernoulli). $p=\frac{2}{3}$ ke saath: ${\sigma }_g^2=p(1-p)=\frac{2}{3}\cdot \frac{1}{3}=\frac{2}{9}\approx \mathrm{0.2222.}$ Care kyun karein? Indicator outputs ka variance hamesha $p(1-p)$ hota hai, $p=\frac{1}{2}$ par maximise hota hai. Extreme probabilities (0 ya 1 ke paas) saste estimate hote hain.

Verify: $\theta =0.6667$. Sanity: parabola neeche dip karti hai, toh square ka zyada hissa uske upar hai — $\frac{1}{2}$ se upar fraction expected hai. $\frac{2}{3}$ ✔

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Ex 5 — Cell E: degenerate, zero-variance case

Active symbol: random input $U\sim \text{Unif}(0,1)$ — lekin jaise hum dekhenge, koi fark nahi padta hum kya sample karte hain.

Forecast: is "estimate" ko exactly sahi hone ke liye kitne samples chahiye?

Step 1 — Expectation compute karo. $\mathbb{E}[7]=7.$ Kyun? Kisi cheez ka average jo hamesha 7 hai woh 7 hai — koi randomness nahi.

Step 2 — Variance compute karo. ${\sigma }_g^2=\mathbb{E}[(7-7{)}^2]=0.$ Yeh kyun matter karta hai? $\text{SE}={\sigma }_g/\sqrt{n}=0/\sqrt{n}=0$ har $n$ ke liye, $n=1$ par bhi.

Step 3 — Degenerate limit interpret karo. Ek single sample ${\hat{\theta }}_1=7$ exactly deta hai. Zyada samples add karne se kuch nahi badalta. Kyun? Zero spread ke saath koi noise nahi hai average karne ke liye — estimator sample one par hi target hit kar chuka hai.

Verify: $\theta =7$, $\text{SE}=0$. Kisi bhi length ka koi bhi run exactly 7 return karta hai. ✔

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Ex 6 — Cell F: limiting behaviour, $n$ plan karna

Active symbol: random input $U\sim \text{Unif}(0,1)$, Ex 1 jaisa hi.

Forecast: Ex 1 ne $n=1000$ par $\text{SE}\approx 0.0094$ hit kiya. Half-width ~$20\times$ shrink karne ke liye guess karo: thode zyada samples, ya BAHUT zyada?

Figure s03 — kya dekh rahe ho: dono axes logarithmic hain. Horizontal axis sample size $n$ hai ($100$ se ek million tak); vertical axis $95\%$ half-width $w=1.96{\sigma }_g/\sqrt{n}$ hai. Red line woh half-width hai — log-log plot par seedha neeche jaata line kyunki $w\propto n^{-1/2}$. Dashed black line target $w=0.001$ mark karti hai; arrow dikhata hai red line $n\approx 3.4\times 10^5$ par cross karti hai. Yeh picture hi $1/\sqrt{n}$ law hai: $w$ ko $10\times$ drop karne ke liye aap $100\times$ right slide karte ho.

Step 1 — Target likho. $95\%$ half-width (upar define ki gayi) $w=1.96{\sigma }_g/\sqrt{n}$ hai. Demand karo $1.96\cdot \frac{{\sigma }_g}{\sqrt{n}}\le \mathrm{0.001.}$ Yeh step kyun? Yeh half-width confidence interval ki "$\pm$" hai; ise $0.001$ se neeche force karna exactly woh accuracy hai jo problem maangti hai, toh hum $n$ ke liye solve karte hain jo ise achieve karta hai.

Step 2 — $n$ ke liye solve karo. $\sqrt{n}\ge \frac{1.96{\sigma }_g}{0.001}\Rightarrow n\ge {\left(\frac{1.96{\sigma }_g}{0.001}\right)}^2=\frac{1.96^2\cdot \frac{4}{45}}{0.001^2}.$ Square kyun? Kyunki error square root ke neeche rehti hai; ise undo karne se sab kuch square ho jaata hai — brutal $1/\sqrt{n}$ law ka source.

Step 3 — Numbers plug karo. ${\sigma }_g^2=\frac{4}{45}=0.08889$: $n\ge \frac{3.8416\times 0.08889}{10^{-6}}=\frac{0.34148}{10^{-6}}\approx 341,476.$ Toh $n\approx 341,477$ samples chahiye. Exactly yeh particular numbers kyun substitute karein? $1.96$ Central Limit Theorem se $95\%$ factor hai, $\frac{4}{45}$ woh variance hai jo hum Ex 1 mein already compute kar chuke, aur $10^{-6}=(0.001{)}^2$ squared target hai; inhe plug karne se abstract inequality ek concrete sample budget ban jaati hai jo hum actually afford kar sakte hain.

Verify: figure mein red line half-width $\propto 1/\sqrt{n}$ dikhati hai. Half-width $0.0184$ (at $n=1000$, yaani $1.96\times 0.0094$) se $0.001$ tak jaana ek $18.4\times$ reduction hai, $18.4^2\approx 339\times$ zyada samples chahiye: $1000\times 339\approx 3.4\times 10^5$. ✔ Match karta hai. Bees-guna accuracy ~340× kaam ki cost maangti hai.

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Ex 7 — Cell G: heavy tail, LLN toot jaata hai

Active symbol: random input $X\sim \text{Cauchy}$ hai (heavy-tailed, uniform nahi).

Forecast: density 0 ke baare mein symmetric hai, toh surely mean 0 hai aur sample mean wahan converge karta hai. Guess karo kya woh reasoning safe hai.

Figure s04 — kya dekh rahe ho: horizontal axis ab tak collect kiye gaye samples $n$ ki sankhya hai; vertical axis running sample mean hai (pehle $n$ draws ka average). Black line ek Normal sample hai: woh jaldi $0$ ko hug karti hai aur wahan rehti hai — LLN kaam karta hua. Red line ek Cauchy sample hai: woh violent jumps mein lurches karta hai aur kabhi nahi settle karta, kyunki fat tails se rare gigantic draws baar baar average ko hijack karte rehte hain. Yeh "mean exist nahi karta" ka visual hai.

Step 1 — Integrability condition check karo. Weak Law ko $\mathbb{E}|X|<\infty$ chahiye. Test karo: $\mathbb{E}|X|={\int }_{-\infty }^{\infty }\frac{|x|}{\pi (1+x^2)}dx=\infty .$ Yeh step kyun? Bade $x$ ke liye integrand $\frac{|x|}{\pi x^2}=\frac{1}{\pi |x|}$ jaisa behave karta hai, aur $\int \frac{dx}{x}$ diverge karta hai (logarithmically). Mean exist nahi karta.

Step 2 — Consequence. Kyunki $\mathbb{E}|X|=\infty$, Strong Law of Large Numbers apply nahi karta. Sample mean ${\bar{X}}_n$ settle nahi karta — remarkably, $n$ Cauchy samples ka ${\bar{X}}_n$ khud standard Cauchy hai, chahe $n$ kitna bhi bada ho. Yeh kyun matter karta hai? Figure mein red trace hamesha ke liye wild jumps leta rehta hai; fat tails se occasional gigantic samples baar baar running average ko hijack karte hain.

Step 3 — Moral. Symmetry ne suggest kiya "mean $=0$", lekin mean undefined hai, toh LLN converge karne ke liye kuch hai hi nahi.

Verify: ${\int }_{-\infty }^{\infty }\frac{|x|}{\pi (1+x^2)}dx$ diverge karta hai — hum tail $\frac{1}{\pi x}$ ke divergent integral confirm karte hain partial integrals compare karke jo bina bound ke grow karte hain. ✔ (VERIFY dekho.)

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Ex 8 — Cell H: real-world word problem

Active symbol: har box ka random input teen uniforms ka triplet $U_1,U_2,U_3\sim \text{Unif}(0,1)$ hai; failures ki count $F$ ek derived random variable hai.

Forecast: sirf $10\%$ per-bulb failure ke saath, guess karo kya returns rare ($3\%$) ya common ($20\%$) hain.

Step 1 — Har box model karo. Failures ki sankhya $F\sim \text{Binomial}(3,0.1)$ (3 Bernoulli Distribution flips ka sum). Define karo $g=1\{F\ge 2\}.$ Yeh step kyun? "Return probability" $P(F\ge 2)=\mathbb{E}[1\{F\ge 2\}]$ hai — phir se ek probability indicator average ke roop mein likhi gayi.

Step 2 — Monte Carlo recipe. Har box ke liye 3 uniforms simulate karo, count karo kitne $<0.1$ hain (failures), record karo kya $\ge 2$ hain, kai boxes par average karo. Kyun? Har simulated box $g$ ka ek draw hai; LLN returned fraction ko true probability par converge karta hai. $0.1$ se neeche ka ek uniform exactly $10\%$ time hota hai, toh yeh faithfully ek bulb fail hone ko mimic karta hai.

Step 3 — Check karne ke liye exact value. $P(F\ge 2)=P(F=2)+P(F=3).$ $P(F=2)=\left(\genfrac{}{}{0ex}{}{3}{2}\right)(0.1{)}^2(0.9{)}^1=3\cdot 0.01\cdot 0.9=0.027,$ $P(F=3)=\left(\genfrac{}{}{0ex}{}{3}{3}\right)(0.1{)}^3=\mathrm{0.001.}$ $\theta =0.027+0.001=\mathrm{0.028.}$ Exactly kyun compute karein? Ek chhote discrete model wale word problem ko binomial formula $\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k(1-p{)}^{n-k}$ se exactly check kiya ja sakta hai, toh hum simulator ko ground truth ke saath validate karte hain blindly trust karne ki jagah.

Verify: $\theta =0.028$ (roughly $2.8\%$) — returns rare hain, "rare" forecast jeeta. Units: ek probability, dimensionless, $[0,1]$ mein. ✔

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Ex 9 — Cell I: exam twist, variance reduction

Active symbol: random input $U\sim \text{Unif}(0,1)$ hai; antithetic partner $1-U$ bhi $\text{Unif}(0,1)$ hai lekin same draw se bana.

Forecast: dono same $\theta$ dete hain. Guess karo ki pairing help karta hai, hurt karta hai, ya kuch nahi karta.

Step 1 — True value (check ke liye). $\theta ={\int }_0^1{e}^{x}dx=e-1\approx \mathrm{1.71828.}$ Yahan se shuru kyun karein? $e^x$ ka elementary antiderivative $e^x$ hai, toh hum exact $\theta$ pin kar sakte hain aur baad mein confirm kar sakte hain ki dono estimators same target par aim karte hain.

Step 2 — Plain estimator variance. $g(x)=e^x$, aur ${\sigma }_g^2=\mathbb{E}[e^{2U}]-{\theta }^2=\frac{e^2-1}{2}-(e-1{)}^2.$ Numerically $\frac{e^2-1}{2}=3.19453$ aur $(e-1{)}^2=2.95249$, toh ${\sigma }_g^2\approx 0.24204$. Kyun? $\mathbb{E}[e^{2U}]={\int }_0^1{e}^{2x}dx=\frac{e^2-1}{2}$; variance hai second moment minus mean squared, standard $\mathrm{Var}=\mathbb{E}[g^2]-(\mathbb{E}[g]{)}^2$.

Step 3 — Antithetic estimator. Har draw ke liye $h(U)=\frac{1}{2}(e^U+e^{1-U})$ use karo, phir $h$'s ka average lo. Yeh step kyun? $e^U$ aur $e^{1-U}$ negatively correlated hain (jab ek bada hota hai doosra chhota), toh unka average chhoti variance rakhta hai aur same mean $\theta$ maintain karta hai. Yeh Importance Sampling-style variance reduction ka workhorse hai.

Step 4 — Antithetic variance (per pair). $\mathrm{Var}(h)=\frac{1}{4}(\mathrm{Var}(e^U)+\mathrm{Var}(e^{1-U})+2\mathrm{Cov}(e^U,e^{1-U})).$ Symmetry se dono variances ${\sigma }_g^2$ ke barabar hain; covariance negative hai: $\mathrm{Cov}(e^U,e^{1-U})=\mathbb{E}[e^U{e}^{1-U}]-{\theta }^2=e-(e-1{)}^2=2.71828-2.95249=-\mathrm{0.23421.}$ Toh $\mathrm{Var}(h)=\frac{1}{4}(2\times 0.24204+2\times (-0.23421))=\frac{1}{4}(0.01566)=\mathrm{0.003914.}$ $\mathbb{E}[e^U{e}^{1-U}]=e$ kyun? Kyunki $e^U{e}^{1-U}=e^{U+(1-U)}=e^1$ constant hai — exponents hamesha 1 mein add ho jaate hain! Isliye correlation itna strongly negative hai aur variance almost cancel ho jaati hai.

Verify: plain per-sample variance $\approx 0.2420$; antithetic per-pair variance $\approx 0.00391$. Ratio $\approx 62\times$ chhoti variance — ek massive win. Same target $\theta =e-1\approx 1.71828$. ✔

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