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

True. Error scales as $\text{SE}={\sigma }_g/\sqrt{n}$, and $\sqrt{4n}=2\sqrt{n}$, so the denominator doubles and the SE halves. This is the "4× work = 2× accuracy" rule.

==True whenever $\theta =\mathbb{E}[g(X)]$ exists==. By linearity, $\mathbb{E}[{\hat{\theta }}_n]=\frac{1}{n}\sum \mathbb{E}[g(X_i)]=\theta$ — no dependence on $n$, and no independence assumption is needed for unbiasedness (that's only needed for the variance).

False. The estimate ${\hat{\theta }}_n$ is random; a specific large-$n$ run can, by bad luck, land farther from $\theta$ than a small-$n$ run. LLN promises the probability of large error shrinks, not that every draw improves.

False. The ${\sigma }_g/\sqrt{n}$ rate is dimension-independent — that is Monte Carlo's whole selling point over grid-based Numerical Integration, whose cost explodes exponentially with dimension.

False in practice. Error only drops by factor $\sqrt{2}\approx 1.41$, i.e. ~30%. To gain a full decimal digit you need $100\times$ the samples, so doubling barely moves the needle.

False. Agreement on one run can be coincidence. Accuracy is a statement about the distribution of ${\hat{\theta }}_n$ — you certify it via the SE and a confidence interval, not by eyeballing one lucky number.

False for the Strong Law. The SLLN (Kolmogorov) holds under only $\mathbb{E}|X|<\infty$ (finite mean). Chebyshev's clean WLLN proof uses finite ${\sigma }^2$, but a finite mean already suffices for the conclusion — see the "why" item below for the justification.

True — but with wrong error bars. Bias depends only on $\mathbb{E}[g(X_i)]=\theta$, which correlation doesn't change. What breaks is $\mathrm{Var}({\hat{\theta }}_n)={\sigma }_g^2/n$; see Variance and Covariance — nonzero covariances between draws inflate the true variance.

Convergence in probability means the chance of being far from $\theta$ vanishes, not that equality is ever reached. For finite $n$ there is always residual randomness of size ~${\sigma }_g/\sqrt{n}$.

The constant pulled out of variance is $\frac{1}{n^2}$, not $\frac{1}{n}$, because $\mathrm{Var}(aY)=a^2\mathrm{Var}(Y)$ with $a=1/n$. Correct result is ${\sigma }^2/n$, which actually shrinks.

Missing the factor 4. ${\bar{X}}_n\to P(\text{inside})=\pi /4$, so you must scale: $\hat{\pi }=4{\bar{X}}_n$. Forgetting it gives an answer near $0.785$, not $\pi$.

On $[0,3]$ the uniform density is $1/3$, not $1$, so $\mathbb{E}[g(U)]=\frac{1}{3}{\int }_0^{3}g$. You must multiply by the interval length: ${\int }_0^{3}g=3\mathbb{E}[g(U)]$.

SE uses the standard deviation ${\sigma }_g=\sqrt{\mathrm{Var}[g(X)]}$, not the variance ${\sigma }_g^2$. It is $\text{SE}={\sigma }_g/\sqrt{n}$; the variance of the estimate is ${\sigma }_g^2/n$.

The $1.96$ factor comes from the normal approximation of the Central Limit Theorem, which needs large $n$. At $n=5$ the CLT hasn't kicked in and the interval is untrustworthy.

If $\mathbb{E}|g(X)|=\infty$ the sample mean never settles — LLN simply does not apply. Adding samples cannot rescue a broken integrand; you must reformulate (e.g. Importance Sampling) or the estimator diverges.

Because variance falls as ${\sigma }_g^2/n$, and the error is a standard deviation — the square root of variance — giving ${\sigma }_g/\sqrt{n}$. The square root is what turns $n$ into $\sqrt{n}$.

Unbiasedness uses linearity of expectation, which ignores dependence. Variance of a sum only equals the sum of variances when covariance terms vanish — and independence is what makes those covariances zero (see Variance and Covariance).

Because the indicator $1_A$ equals $1$ on $A$ and $0$ elsewhere, so $\mathbb{E}[1_A]=1\cdot P(A)+0\cdot P(A^c)=P(A)$. This is why counting "hits" estimates a probability — it's a Bernoulli Distribution mean.

Chebyshev's Inequality bounds deviation probability using only the variance, so it works for any distribution with finite ${\sigma }^2$. We never need to know the shape of ${\bar{X}}_n$ — just that its variance $\to 0$.

The Chebyshev route needs ${\sigma }^2<\infty$, but it's not the only route. A truncation argument splits each $X_i$ into a bounded part (whose average is controlled by Chebyshev after truncating) plus a rare-tail part (whose contribution $\to 0$ because $\mathbb{E}|X|<\infty$ makes the tail mass vanish). Kolmogorov's SLLN formalizes this, so finite mean alone gives convergence — finite variance only buys you the clean $\sigma /\sqrt{n}$ rate.

A grid with $k$ points per axis in $d$ dimensions costs $k^d$ — exponential in $d$. Monte Carlo's ${\sigma }_g/\sqrt{n}$ rate is blind to $d$, so its cost to reach a target error doesn't blow up as dimensions grow.

The indicator is Bernoulli with $p=\pi /4\approx 0.785$, giving variance $p(1-p)\approx 0.169$ and (after the $\times 4$) $\text{SE}=4\sqrt{0.169/n}$. That fat constant plus the $1/\sqrt{n}$ rate makes it converge slowly.

Monte Carlo output is a random number; its usefulness depends on how tight it is. Reporting ${\hat{\theta }}_n$ alone hides whether it's good to one digit or five — the SE is what makes the claim falsifiable.

Then ${\sigma }_g^2=\mathrm{Var}(g(X))=0$, so $\text{SE}=0$ for every $n$ — a single sample gives the exact answer, since there's nothing random left to average out.

Yes — $\mathbb{E}[{\hat{\theta }}_1]=\mathbb{E}[g(X_1)]=\theta$, so it's unbiased even at $n=1$. But its variance is the full ${\sigma }_g^2$: the single output is the estimate, so it inherits the whole bounce of one poke and is unbiased yet wildly imprecise.

Yes — the Strong Law of Large Numbers only needs $\mathbb{E}|g(X)|<\infty$, so ${\hat{\theta }}_n\to \theta$ (via the truncation argument, not Chebyshev). But the CLT confidence interval and the ${\sigma }_g/\sqrt{n}$ rate collapse because ${\sigma }_g=\infty$, so you can't quantify the error the usual way.

Effective sample size is $1$. The effective sample size $n_{\text{eff}}$ is "how many independent draws your correlated batch is really worth" — for perfectly correlated draws every copy carries the same information, so $n_{\text{eff}}=1$. The naive formula reports ${\sigma }_g/\sqrt{n}$, but the true uncertainty is that of a single sample — dangerous overconfidence.

No. The sample variance underestimates the true $p(1-p)$ when you've observed no misses. Your reported SE of $0$ is a small-sample artifact; the real uncertainty is nonzero because the underlying ${\sigma }_g^2=p(1-p)>0$.