Question bank — Law of large numbers

False — it guarantees the proportion gets close to 0.5, but the absolute count can still be off by dozens; randomness never fully vanishes.

False — ${\sigma }^2$ is a fixed property of the distribution; it is the variance of the sample mean, ${\sigma }^2/n$, that shrinks.

False — the WLLN needs only i.i.d. draws with finite mean; any shape works.

False — error scales as $1/\sqrt{n}$, so doubling $n$ only cuts error by a factor of $\sqrt{2}\approx 1.41$; you need $4\times$ the data to halve it.

True — almost-sure convergence is stronger and mathematically implies convergence in probability; the reverse implication does not hold.

False — it is only an upper bound (often very loose); the true probability, e.g. via the Central Limit Theorem, is usually far smaller.

True — finite mean is enough for the WLLN (e.g. via truncation arguments); finite variance is only needed for the Chebyshev-based proof, not the law itself.

False — both are eventually correct, but one may have larger variance and converge far slower, so finite-sample behaviour differs.

The standard error here is $\sigma /\sqrt{n}=0.5/\sqrt{1000}\approx 1.6\%$, so 60% is about 6 standard errors out — genuinely suspicious — but "definitely" ignores that rare events do occur; you'd test it, not declare it.

The LLN never pushes future outcomes to compensate; draws are independent, so the next flip is still 50/50. Balance emerges by dilution over many new trials, not by correction.

The Chebyshev route is just one proof; the Weak Law still holds for finite-mean, infinite-variance cases via other methods. A failed proof technique doesn't kill the theorem.

Only if the draws are independent — the cross terms (covariances) must vanish. For correlated samples the variance can be much larger and may not shrink like $1/n$.

The sum's spread actually grows: its standard deviation is $\sigma \sqrt{n}$, which increases. Only the mean concentrates; the sum wanders further from $n\mu$ in absolute terms.

The $1/\sqrt{n}$ rate is dimension-independent — that's the whole appeal of Monte Carlo. The exponent does not multiply with dimension.

Scaling by $1/n$ pulls out $1/n^2$ from variance (variance scales by the square of a constant); summing $n$ equal variances gives $n{\sigma }^2$, and $n{\sigma }^2/n^2={\sigma }^2/n$.

Without a common mean, "${\bar{X}}_n\to \mu$" has no single $\mu$ to converge to; identical distribution guarantees every draw shares the same target.

A probability equals the expected value of an indicator variable, and the LLN says the sample mean of those indicators (the observed frequency) converges to that expectation.

A mini-batch gradient is a sample mean of per-example gradients; by the LLN it converges to the true full-batch gradient, so noisy steps point on average in the right direction.

Interval width is proportional to the standard error $\sigma /\sqrt{n}$, and the $\sqrt{n}$ denominator means returns diminish — the last sliver of precision costs disproportionately many samples.

This is the Bias-Variance Tradeoff: the LLN only removes error in the $n\to \infty$ limit, so for finite $n$ a small controlled bias may reduce total error more than large variance.

Resampling estimates a statistic's distribution by averaging over many resamples; the LLN ensures those empirical averages approximate the true sampling behaviour as the number of bootstrap replicates grows.

It fails — the sample mean does not converge; each Cauchy sample mean is again Cauchy-distributed, so more data never helps. Finite mean is the essential precondition.

It's trivially instant — every draw equals $\mu$, so ${\bar{X}}_n=\mu$ for all $n\ge 1$; the Chebyshev bound is $0$, matching perfect concentration.

Nothing useful — ${\bar{X}}_1=X_1$ with full variance ${\sigma }^2$; the law is a limiting statement and gives no guarantee at a single sample.

Then ${\bar{X}}_n$ equals that one draw regardless of $n$, its variance stays ${\sigma }^2$, and it does not converge to $\mu$ — independence is what makes the $1/n$ shrinkage possible.

The bound blows up to infinity (becomes vacuous), reflecting that demanding exact equality with $\mu$ can never be guaranteed at finite $n$ — you can only bound deviations larger than a fixed positive $ϵ$.

The Chebyshev sample-size formula understates the required $n$; a larger true ${\sigma }^2$ means you need proportionally more samples for the same precision, so an optimistic ${\sigma }^2$ gives a false sense of security.