Question bank — Monte Carlo simulation — law of large numbers basis

True. Error $\text{SE}={\sigma }_g/\sqrt{n}$ ki tarah scale hoti hai, aur $\sqrt{4n}=2\sqrt{n}$, toh denominator double ho jaata hai aur SE aadhi ho jaati hai. Yahi "4× kaam = 2× accuracy" ka rule hai.

==True jab bhi $\theta =\mathbb{E}[g(X)]$ exist kare==. Linearity se, $\mathbb{E}[{\hat{\theta }}_n]=\frac{1}{n}\sum \mathbb{E}[g(X_i)]=\theta$ — $n$ par koi dependence nahi, aur unbiasedness ke liye independence ki zaroorat nahi (woh sirf variance ke liye chahiye).

False. ${\hat{\theta }}_n$ random hai; ek specific large-$n$ run bad luck se $\theta$ se ek small-$n$ run se bhi zyada door land kar sakta hai. LLN promise karta hai ki badi error ki probability shrink hoti hai, yeh nahi ki har draw improve hota hai.

False. ${\sigma }_g/\sqrt{n}$ rate dimension-independent hai — yahi Monte Carlo ka grid-based Numerical Integration par poora selling point hai, jis ka cost dimension ke saath exponentially badhta hai.

Practically False. Error sirf $\sqrt{2}\approx 1.41$ ke factor se girta hai, yaani ~30%. Ek poori decimal digit paane ke liye $100\times$ samples chahiye, toh doubling se koi khaas fark nahi padta.

False. Ek run mein agreement coincidence ho sakta hai. Accuracy ${\hat{\theta }}_n$ ki distribution ke baare mein ek statement hai — tum ise SE aur confidence interval se certify karte ho, kisi ek lucky number ko eyeball karke nahi.

Strong Law ke liye False. SLLN (Kolmogorov) sirf $\mathbb{E}|X|<\infty$ (finite mean) ke saath kaam karta hai. Chebyshev ka clean WLLN proof finite ${\sigma }^2$ use karta hai, lekin conclusion ke liye finite mean kaafi hai — neeche "why" item mein justification dekho.

True — lekin error bars galat hote hain. Bias sirf $\mathbb{E}[g(X_i)]=\theta$ par depend karta hai, jise correlation change nahi karta. Jo cheez toot ti hai woh hai $\mathrm{Var}({\hat{\theta }}_n)={\sigma }_g^2/n$; dekho Variance and Covariance — draws ke beech nonzero covariances true variance ko inflate karte hain.

Convergence in probability ka matlab hai ki $\theta$ se door hone ki chance khatam ho jaati hai, yeh nahi ki equality kabhi reach hoti hai. Finite $n$ ke liye hamesha ~${\sigma }_g/\sqrt{n}$ size ka residual randomness rahega.

Jo constant variance se bahar khicha jaata hai woh $\frac{1}{n^2}$ hai, $\frac{1}{n}$ nahi, kyunki $\mathrm{Var}(aY)=a^2\mathrm{Var}(Y)$ jahan $a=1/n$. Sahi result ${\sigma }^2/n$ hai, jo actually shrink karta hai.

Factor 4 missing hai. ${\bar{X}}_n\to P(\text{inside})=\pi /4$, toh scale karna zaroori hai: $\hat{\pi }=4{\bar{X}}_n$. Bhoolne par answer $\pi$ ki jagah $0.785$ ke aas paas aata hai.

$[0,3]$ par uniform density $1/3$ hai, $1$ nahi, toh $\mathbb{E}[g(U)]=\frac{1}{3}{\int }_0^{3}g$. Tumhe interval length se multiply karna hoga: ${\int }_0^{3}g=3\mathbb{E}[g(U)]$.

SE standard deviation ${\sigma }_g=\sqrt{\mathrm{Var}[g(X)]}$ use karta hai, variance ${\sigma }_g^2$ nahi. Yeh $\text{SE}={\sigma }_g/\sqrt{n}$ hai; estimate ka variance ${\sigma }_g^2/n$ hai.

$1.96$ factor Central Limit Theorem ki normal approximation se aata hai, jo bade $n$ ke liye chahiye. $n=5$ par CLT kaam nahi kar raha aur interval untrust worthy hai.

Agar $\mathbb{E}|g(X)|=\infty$ hai toh sample mean kabhi settle nahi hoga — LLN simply apply nahi hota. Samples add karna ek broken integrand ko nahi bacha sakta; tumhe reformulate karna hoga (e.g. Importance Sampling) warna estimator diverge karta rahega.

Kyunki variance ${\sigma }_g^2/n$ ki tarah girta hai, aur error ek standard deviation hai — variance ka square root — jo ${\sigma }_g/\sqrt{n}$ deta hai. Square root hi $n$ ko $\sqrt{n}$ banata hai.

Unbiasedness expectation ki linearity use karta hai, jo dependence ko ignore karta hai. Sum ka variance sirf tab variances ke sum ke barabar hota hai jab covariance terms zero hon — aur independence hi woh covariances zero karta hai (dekho Variance and Covariance).

Kyunki indicator $1_A$ $A$ par $1$ aur baaki jagah $0$ hota hai, toh $\mathbb{E}[1_A]=1\cdot P(A)+0\cdot P(A^c)=P(A)$. Isliye "hits" count karna ek probability estimate karta hai — yeh ek Bernoulli Distribution mean hai.

Chebyshev's Inequality deviation probability ko sirf variance use karke bound karta hai, toh yeh kisi bhi distribution ke liye kaam karta hai jisme finite ${\sigma }^2$ ho. Humein ${\bar{X}}_n$ ki shape jaanne ki zaroorat nahi — sirf yeh ki uska variance $\to 0$ hai.

Chebyshev route ko ${\sigma }^2<\infty$ chahiye, lekin woh ek hi raasta nahi hai. Ek truncation argument har $X_i$ ko ek bounded part (jis ka average Chebyshev se control hota hai truncating ke baad) aur ek rare-tail part (jis ka contribution $\to 0$ hota hai kyunki $\mathbb{E}|X|<\infty$ tail mass ko vanish karta hai) mein split karta hai. Kolmogorov ka SLLN ise formalize karta hai, toh sirf finite mean convergence deta hai — finite variance sirf tumhe clean $\sigma /\sqrt{n}$ rate deti hai.

$d$ dimensions mein axis per $k$ points wala grid $k^d$ cost karta hai — $d$ mein exponential. Monte Carlo ka ${\sigma }_g/\sqrt{n}$ rate $d$ se blind hai, toh ek target error tak pahunchne ki cost dimensions badhne se nahi phoolti.

Indicator Bernoulli hai jisme $p=\pi /4\approx 0.785$ hai, jo variance $p(1-p)\approx 0.169$ deta hai aur ($\times 4$ ke baad) $\text{SE}=4\sqrt{0.169/n}$. Woh mota constant plus $1/\sqrt{n}$ rate ise dhire dhire converge karta hai.

Monte Carlo output ek random number hai; uski usefulness is par depend karti hai ki woh kitna tight hai. Sirf ${\hat{\theta }}_n$ report karna chhupata hai ki woh ek digit ke liye acha hai ya paanch ke liye — SE hi claim ko falsifiable banata hai.

Tab ${\sigma }_g^2=\mathrm{Var}(g(X))=0$, toh har $n$ ke liye $\text{SE}=0$ — ek single sample exact answer deta hai, kyunki average karne ke liye kuch random bacha hi nahi.

Haan — $\mathbb{E}[{\hat{\theta }}_1]=\mathbb{E}[g(X_1)]=\theta$, toh $n=1$ par bhi unbiased hai. Lekin uska variance poora ${\sigma }_g^2$ hai: single output hi estimate hai, toh woh ek poking ki poori bounce inherit karta hai aur unbiased toh hai par bilkul imprecise.

Haan — Strong Law of Large Numbers ko sirf $\mathbb{E}|g(X)|<\infty$ chahiye, toh ${\hat{\theta }}_n\to \theta$ (truncation argument se, Chebyshev se nahi). Lekin CLT confidence interval aur ${\sigma }_g/\sqrt{n}$ rate collapse ho jaate hain kyunki ${\sigma }_g=\infty$ hai, toh tum usual tarike se error quantify nahi kar sakte.

Effective sample size $1$ hai. Effective sample size $n_{\text{eff}}$ woh hai — "tumhara correlated batch actually kitne independent draws ke barabar hai" — perfectly correlated draws ke liye har copy wohi information carry karta hai, toh $n_{\text{eff}}=1$. Naive formula ${\sigma }_g/\sqrt{n}$ report karta hai, lekin sacchi uncertainty ek single sample ki hai — yeh dangerous overconfidence hai.

Nahi. Sample variance true $p(1-p)$ ko underestimate karta hai jab tumne koi miss observe nahi kiya. Tumhara reported SE $0$ ek small-sample artifact hai; real uncertainty nonzero hai kyunki underlying ${\sigma }_g^2=p(1-p)>0$ hai.