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Probability feels intuitive until you try to make a single sentence precise. The fastest way to learn what the definitions really mean is to watch one plausible inference fail in a clean, controlled setting. Counterexamples do not just refute; they expose the boundary of a concept, and they teach you how to reason without importing hidden assumptions.

A perfect example is the widespread belief that “uncorrelated” is basically the same as “independent.” In applications, people often check a covariance, see a zero, and conclude that there is no relationship. The counterexample below shows exactly what is true, what is not, and how to replace the false inference with a correct toolbox.

The setup: a very simple probability space

Let $X$ be uniformly distributed on $[-1,1]$. Concretely, take the probability space $([-1,1],\mathcal{B},\mathbb{P})$ where $\mathcal{B}$ is the Borel $\sigma$-algebra and $\mathbb{P}$ is Lebesgue measure normalized so $\mathbb{P}([-1,1])=1$. Define the random variable

Now define a second random variable by

This is almost too simple, which is why it is such a good teacher.

A few basic facts are immediate:

• $Y$ is measurable because it is a continuous function of $X$.
• $Y$ takes values in $[0,1]$.
• Knowing $X$ determines $Y$, since $Y$ is a function of $X$.

Already you should feel tension: if one variable is a function of the other, how could they be “unrelated”? The answer is that the usual “unrelatedness test” based on covariance measures only linear dependence. The counterexample makes that statement exact.

Compute the covariance and see it vanish

First compute the means. By symmetry,

Next,

Now compute $\mathbb{E}[XY]$:

But the integrand $x^3$ is odd and the density is symmetric, so the integral is $0$. Therefore

So $X$ and $Y$ are uncorrelated.

At this point many people would say: “Great, the variables do not influence each other.” That inference is false.

Show they are not independent

Independence means that for all Borel sets $A,B$,

Because $Y=X^2$, the event $\{Y\le 1/4\}$ is exactly the same as $\{|X|\le 1/2\}$. Choose

Then

So

On the other hand,

If $X$ and $Y$ were independent, we would have

but the true value is $1/2$. Therefore $X$ and $Y$ are not independent.

This is the whole counterexample: uncorrelated does not imply independent.

What the counterexample is really saying

The key lesson is not a slogan; it is a structural fact about what covariance can see.

The covariance

measures whether $Y$ has a linear trend in $X$ after centering. In this example, the dependence of $Y$ on $X$ is quadratic, and symmetry cancels the linear term. Covariance is blind to that.

A more informative way to state the boundary is:

• Independence is a statement about products of events and factors of $\sigma$-algebras.
• Uncorrelatedness is a statement about one specific bilinear functional, the covariance.

Uncorrelatedness is weaker because it tests one moment identity, while independence tests an entire algebra of identities.

The $\sigma$-algebra view: dependence is about information

A good replacement for the false inference is to interpret dependence as shared information. Define

the $\sigma$-algebras generated by $X$ and $Y$. Independence of $X$ and $Y$ is equivalent to independence of $\mathcal{F}_X$ and $\mathcal{F}_Y$: every event determined by $X$ is independent of every event determined by $Y$.

In the present example, $\mathcal{F}_Y\subseteq \mathcal{F}_X$ because $Y$ is a function of $X$. That inclusion means: whatever you can learn from $Y$, you can learn from $X$. In particular, the events $\{Y\le t\}$ are events about $|X|$, and they are certainly not independent of events that also involve $|X|$.

That is the sharp contrast:

• Covariance $=0$ tells you a specific centered linear correlation vanishes.
• $\mathcal{F}_Y\subseteq \mathcal{F}_X$ tells you the variables are tied together by a deterministic constraint.

The counterexample is successful because it makes both facts simultaneously visible.

Conditional expectation makes the dependence quantitative

If you want a numeric test of “how much $Y$ depends on $X$,” conditional expectation is the right object.

Because $Y=X^2$ is a function of $X$,

So the conditional mean of $Y$ given $X$ is not constant; it varies with $X$. That is dependence in a very strong sense.

By contrast, if $X$ and $Y$ were independent, we would have

So a precise, checkable substitute for the false inference is:

• If $\mathbb{E}[Y\mid X]$ is almost surely constant, then $Y$ has no mean dependence on $X$.
• If $\mathbb{E}[Y\mid X]$ varies, dependence is present, even if covariance is zero.

In the counterexample, $\mathbb{E}[Y\mid X]$ varies maximally because it equals $Y$.

A “moment upgrade” that really does imply independence

You might ask: if covariance is too weak, what can you check instead?

For Gaussian random vectors, uncorrelated does imply independent. The reason is not magic; it is structure. The joint law of a multivariate Gaussian is determined completely by its mean vector and covariance matrix. So if the covariance matrix is block-diagonal, the joint density factors, and independence follows.

The correct way to carry that lesson is:

• In special families of distributions, low-order moments can determine the entire law.
• Outside those families, moments can miss nonlinear dependence.

This distinction explains why “uncorrelated implies independent” appears true in some data workflows: those workflows assume a near-Gaussian model, often implicitly.

How to build your own counterexamples

This example is the simplest member of a larger pattern.

If you want uncorrelated but dependent variables, a reliable method is:

• Start with a symmetric $X$ with $\mathbb{E}[X]=0$.
• Let $Y=g(X)$ where $g$ is an even function, so $Xg(X)$ is odd and $\mathbb{E}[Xg(X)]=0$.
• Choose $g$ nonconstant so that $Y$ genuinely depends on $X$.

For instance, $Y=X^2$, $Y=|X|$, or $Y=\mathbf{1}_{\{|X|>1/2\}}$ all work with symmetric $X$. The symmetry forces zero covariance, while the functional relation forces dependence.

This method teaches a deeper habit: when you see a cancellation, ask whether it is structural (true independence) or geometric (symmetry).

What this changes in practice

The counterexample has practical consequences in modeling, statistics, and even pure probability.

Correlation tests are not dependence tests

If you test dependence by correlation alone, you can miss:

• Quadratic or higher-order relationships
• Threshold effects (indicator functions)
• Mixtures where two regimes cancel linearly

The right fix depends on your goal. If you need full independence, you must test or justify it structurally. If you only need linear decorrelation, covariance is appropriate, but you should say that explicitly.

The right tool depends on what you need to control

Different tasks demand different notions.

| Goal | Notion that matches | Typical tool |

|—|—|—|

| Control linear prediction | Uncorrelatedness | covariance, least squares |

| Control mean dependence | $\mathbb{E}[Y\mid X]$ constant | conditional expectation |

| Control event factoring | Independence | product $\sigma$-algebras |

| Control tail interaction | Weak dependence | mixing, coupling, concentration |

The counterexample forces you to pick your notion rather than rely on a vague word like “unrelated.”

A short takeaway that is actually correct

The result is not “correlation is useless.” The correct takeaway is:

• Correlation is a measurement of linear alignment in $L^2$.
• Independence is a measurement of factorization of information.

When you move from one to the other without a theorem, you are smuggling in a model class.

If you remember only one thing, let it be this: a zero covariance can be caused by symmetry, not by separation. The counterexample $Y=X^2$ with symmetric $X$ is small enough to hold in your head, yet rich enough to calibrate your reasoning whenever probability starts to feel too intuitive.

Probability is unforgiving in a helpful way: a statement is either true under stated assumptions or it is not. Many mistakes come from skipping the assumptions that make a familiar identity valid. The fastest way to become reliable is to learn the standard failure modes and the correct replacement moves.

This article collects frequent mistakes that appear in coursework, research reading, and applied work. Each item is paired with a way to diagnose and repair the reasoning.

Mistake: treating probability zero as impossibility

A classic confusion is to read “$\mathbb{P}(A)=0$” as “$A$ cannot happen.” In many continuous models, single outcomes have probability zero and still occur.

Example: if $X\sim \mathrm{Unif}[0,1]$, then $\mathbb{P}(X=1/2)=0$, yet $X$ takes a single value in $[0,1]$ every time you sample it.

How to avoid it

Use the correct distinction:

• $\mathbb{P}(A)=0$ means $A$ is negligible with respect to the measure, not logically impossible.
• In measure-theoretic probability, “almost surely” means “outside a null set,” not “always.”

When a proof uses “almost surely,” track whether the null set can depend on parameters. If you need a statement that holds for every parameter value, you often need a uniform argument, not a pointwise one.

Mistake: confusing uncorrelatedness with independence

Covariance zero is not independence. The counterexample $Y=X^2$ with symmetric $X$ shows that dependence can be nonlinear and still have zero covariance.

How to avoid it

Replace the vague idea “unrelated” with a target notion:

• If you need event factorization, you need independence.
• If you need only linear prediction, uncorrelatedness may be enough.
• If you need stability under conditioning, examine $\mathbb{E}[Y\mid X]$.

A good habit is to write the definition you are using, even if only in your scratch work.

Mistake: using Bayes’ rule without checking denominators

Bayes’ rule is

but it requires $\mathbb{P}(B)>0$. In continuous settings, conditioning on events of probability zero cannot be handled by this formula.

How to avoid it

When $\mathbb{P}(B)=0$ is lurking, switch viewpoints:

• Use conditional densities with respect \to a dominating measure.
• Use regular conditional probabilities $\mathbb{P}(A\mid \mathcal{G})$ for a $\sigma$-algebra $\mathcal{G}$.
• In Euclidean settings, use the Radon–Nikodym derivative viewpoint: conditional density is a derivative of measures, not a ratio of point masses.

If you are conditioning on a value $X=x$ for a continuous $X$, treat it as conditioning on the $\sigma$-algebra generated by $X$, not the singleton event.

Mistake: swapping limits and expectations without justification

A pervasive mistake is to assert

without conditions. Sometimes it is true, but the conditions matter.

How to avoid it

Know the three standard theorems and what they require:

• Monotone Convergence Theorem: if $0\le X_n\le X_{n+1}$ and $X_n\to X$ pointwise, then $\mathbb{E}[X_n]\to \mathbb{E}[X]$.
• Dominated Convergence Theorem: if $X_n\to X$ a.s. and $|X_n|\le Y$ with $\mathbb{E}[Y]<\infty$, then $\mathbb{E}[X_n]\to \mathbb{E}[X]$.
• Fatou’s Lemma: $\mathbb{E}[\liminf X_n]\le \liminf \mathbb{E}[X_n]$.

When you see a limit-inside-expectation step, identify which theorem is being used and where the domination or monotonicity comes from.

Mistake: confusing convergence modes

“Converges” can mean at least four different things:

• almost sure convergence
• convergence in probability
• convergence in $L^p$
• convergence in distribution

These implications are not symmetric. For example, convergence in probability implies convergence in distribution, but not conversely in general.

How to avoid it

Carry a small implication map in your head:

• $L^p\Rightarrow$ in probability $\Rightarrow$ in distribution.
• almost sure $\Rightarrow$ in probability.
• in distribution plus uniform integrability can upgrade to convergence of expectations.

When a result states “$X_n$ converges,” check which mode is meant. In many papers, “$X_n\Rightarrow X$” is distributional convergence, and expectations may not converge without extra hypotheses.

Mistake: assuming measurability instead of proving it

Many probabilistic statements are not well-formed until measurability is established.

Examples:

• $\sup_{t\in[0,1]} X_t$ is not automatically measurable unless $t\mapsto X_t(\omega)$ has suitable regularity.
• A stopping time $\tau$ must satisfy $\{\tau\le t\}\in \mathcal{F}_t$ for every $t$.

How to avoid it

Use standard measurable constructions:

• Replace a supremum over an uncountable set by a supremum over a countable dense set when paths are \right-continuous.
• In stochastic process arguments, explicitly state the path regularity you are using: càdlàg, continuous, or progressively measurable.

A practical rule: if the argument uses “take the first time,” you are probably proving a stopping-time property, and measurability is part of the work.

Mistake: treating “independent increments” as “independent values”

For a process $(X_t)$, independent increments means that $X_{t_1}-X_{s_1}$ is independent of $X_{t_2}-X_{s_2}$ when the intervals do not overlap. It does not imply that the random variables $X_{t_1}$ and $X_{t_2}$ are independent.

How to avoid it

Write the dependence explicitly:

so $X_{t_2}$ contains $X_{t_1}$. Independence can only apply to the increment, not to the cumulative value.

Mistake: forgetting conditioning changes distributions

People often compute with “the same distribution” after conditioning, as if conditioning were a harmless annotation. Conditioning is a transformation of measure.

How to avoid it

When you condition on $\mathcal{G}$, you are working with the random probability measure $\mathbb{P}(\cdot\mid\mathcal{G})$. Re-check independence assumptions under the conditioned measure. Many “nice” symmetries break when information is revealed.

A useful principle is to compute conditional expectations first and then average:

If you cannot compute a probability directly, compute it conditionally and then integrate.

Mistake: applying concentration bounds outside their scope

Tail inequalities like Hoeffding, Bernstein, or Azuma are powerful, but each has hypotheses: boundedness, sub-exponential tails, martingale differences, or conditional variance control. Using them blindly gives nonsense.

How to avoid it

Before quoting a bound, check:

• Are the variables bounded almost surely?
• Are they independent, or at least a martingale difference sequence?
• Is the variance finite, or is a conditional variance bound available?

If hypotheses fail, look for a theorem designed for heavy-tailed settings, truncation, or robust estimators.

Mistake: treating a heuristic density as a proof

In continuous probability, one often writes down a density and manipulates it as if it were a function, but not every distribution has a density, and not every operation is valid without integrability.

How to avoid it

When you use a density $f$, identify the dominating measure and the meaning:

• $f = d\mu/d\lambda$ is a Radon–Nikodym derivative.
• Identities about $f$ are identities about the measure $\mu$.

If you are unsure whether a density exists, either prove absolute continuity or reformulate in terms of distribution functions and measures.

A small checklist that saves real time

When you read or write a probabilistic argument, pause at the following steps:

• A limit is moved through an expectation, probability, or integral.
• A conditional probability is computed with a denominator that might be zero.
• A statement is claimed “almost surely” and then used uniformly in a parameter.
• Independence is inferred from a weak statistic like covariance.
• A supremum or hitting time is introduced without a measurability statement.

These are not pedantic concerns; they are exactly where otherwise-correct ideas fail.

What replaces the mistakes: a disciplined core

Probability becomes far easier when you treat it as measure theory with extra language:

• Events live in $\sigma$-algebras.
• Random variables are measurable maps.
• Expectation is integration.
• Conditioning is a projection in $L^1$ (and in $L^2$ it is literally an orthogonal projection).
• Convergence is not one concept; it is a family of concepts with named theorems relating them.

If you train yourself to ask “which theorem justifies this step,” you will start to see probability as a sequence of controlled reductions. That is the difference between intuition-driven manipulation and research-grade reliability.

Mistake: using Fubini and Tonelli without checking integrability

Interchanging integrals and expectations is often correct, but not automatic. A common error is to write

without verifying the conditions.

How to avoid it

Remember the division of labor:

• Tonelli’s theorem applies to nonnegative integrands. If $g\ge 0$, you may swap integrals and expectations freely, and both sides may be infinite.
• Fubini’s theorem applies to absolutely integrable integrands. If $\mathbb{E}[\int |g|]<\infty$, then the swap is justified and yields finite values.

In probability arguments, the hidden risk is a sign-changing integrand with an infinite positive and infinite negative part. When you see a swap, either reduce \to a nonnegative function or produce an absolute integrability bound.

Mistake: assuming a martingale has independent increments

Martingales generalize “fair game” behavior, but they do not typically have independent increments. The defining property is conditional mean zero:

Nothing here says increments are independent of the past, only that they have mean zero given the past.

How to avoid it

When you use a martingale inequality or an optional stopping theorem, check what is actually required:

• Do you need bounded increments, or bounded conditional variances?
• Do you need uniform integrability, or a bounded stopping time?
• Are you using the filtration $(\mathcal{F}_t)$ consistently, so that the stopping time is measurable with respect to the right information?

Treat “martingale” as “conditional expectation structure,” not as “independence structure.”

Probability lives in a world of measures on large spaces. Computing lives in a world of finite objects: floating-point numbers, arrays, finite graphs, finite random seeds. Almost every computational workflow in probability is a discretization, even when you do not call it that.

The interesting question is not whether discretization changes things. It always changes things. The question is: which probabilistic structures are stable under the kinds of discretization we actually use. When you know the stable structures, you can design algorithms that preserve the features you care about and you can state error claims that match the mathematics.

This article is a guide to what tends to survive and what tends to break when probability is made finite.

First principle: pick the topology you are approximating in

Two probability measures can be “close” in many inequivalent ways. Computation forces you to decide what “close” means, even if you never say it aloud.

Common choices include:

• Weak convergence: $\mu_n\Rightarrow \mu$ means $\int f\,d\mu_n\to \int f\,d\mu$ for every bounded continuous $f$. This is the default for approximating distributions on $\mathbb{R}^d$.
• Total variation: $\|\mu-\nu\|_{TV}$ controls sup error over all events. This is stronger and often too strong for continuous approximation unless densities are very close.
• Wasserstein metrics: these control errors in expectations of Lipschitz functions and are natural when geometry matters.

A discretization that is excellent in weak convergence might be terrible in total variation. Many computational confusions are just metric confusions.

Discretizing a distribution: bins, quadrature, and what stays true

Suppose $X$ is a real-valued random variable with law $\mu$. A simple discretization is binning: replace $X$ by $X^{(h)}$ that takes values on a grid of step size $h$ by rounding, truncation, or projection.

What survives automatically:

• Nonnegativity and normalization: probabilities remain probabilities if you renormalize correctly.
• Support constraints: if you project onto a grid inside a bounded interval, you preserve bounded support.
• Weak features: many smooth expectations $\mathbb{E}[f(X)]$ are approximated well when $f$ is regular at the scale $h$.

What does not survive automatically:

• Tail behavior: truncation changes rare-event probabilities, sometimes drastically.
• Exact moments: binning can bias $\mathbb{E}[X]$ and $\mathrm{Var}(X)$ unless designed to preserve them.
• Independence: discretizing two variables separately can create or destroy dependence structure if they share fine-scale coupling.

A good computational design is explicit about which moments or functionals you want to preserve.

A useful table: what you can promise from a grid approximation

| Target statement | Typical sufficient condition |

|—|—|

| $\mu_h\Rightarrow \mu$ | mesh $h\to 0$ and tightness of $\mu_h$ |

| $|\mathbb{E}[f(X^{(h)})]-\mathbb{E}[f(X)]|\le \varepsilon$ | $f$ Lipschitz and $\mathbb{E}[|X^{(h)}-X|]$ small |

| total variation closeness | density approximation in $L^1$ with uniform control |

This is why weak convergence is the natural baseline: it matches what most finite approximations can support.

Monte Carlo: the central invariant is unbiasedness

In Monte Carlo, you approximate $\mathbb{E}[f(X)]$ by

with $X_k$ i.i.d. copies of $X$.

The most important structure that survives discretization here is unbiasedness:

Unbiasedness is not automatic; it is built into the sampling scheme. When you discretize a continuous distribution by an approximate sampler, you may lose unbiasedness and you should quantify the bias.

What else survives, under mild hypotheses:

• Law of large numbers: $\hat{I}_N\to \mathbb{E}[f(X)]$ almost surely if $\mathbb{E}[|f(X)|]<\infty$.
• Central limit scaling: fluctuations behave like $N^{-1/2}$ when $\mathrm{Var}(f(X))<\infty$.

What fails in heavy tails is not “Monte Carlo” but the assumptions that give finite variance or even finite mean. In computation, you handle this by truncation, importance sampling, or robust estimators, but then you must track the new bias-variance tradeoff explicitly.

Markov chains: preserving stationarity is a design choice

Many computational probability tasks use Markov chain Monte Carlo (MCMC). A Markov chain is defined by a transition kernel $P(x,dy)$. The foundational object is a stationary distribution $\pi$ satisfying

When you implement a chain on a computer, you often discretize:

• the state space (finite grid or finite truncation)
• the kernel (numerical evaluation of densities)
• the acceptance step (finite precision)

What can survive well:

• Stationarity by construction: if you enforce detailed balance, $\pi(x)P(x,y)=\pi(y)P(y,x)$, then $\pi$ is stationary in the exact arithmetic model.
• Ergodicity on the discretized space: irreducibility and aperiodicity can be checked on the finite graph.

What can break:

• Target correctness under truncation: truncating the state space replaces $\pi$ by a conditional or renormalized $\pi_h$. That may be fine, but it must be stated.
• Mixing rates: a discretized chain can mix much slower than the continuous ideal, especially in high dimension.
• Reversibility under floating point: detailed balance relies on exact ratios. Finite precision can produce systematic drift if accept/reject thresholds are mishandled.

A practical principle is: if stationarity is essential, build it into the kernel symbolically, not numerically.

Coupling: the right way to compare the ideal and the computed

A clean way to reason about discretization error is coupling. You construct $(X, X^{(h)})$ on the same probability space and bound $\mathbb{P}(X\ne X^{(h)})$ or $\mathbb{E}[|X-X^{(h)}|]$.

Why this survives computation:

• Coupling converts measure comparison into a comparison of random variables.
• Many discretizations are literally defined by transforming the same underlying randomness, such as rounding a continuous sample.

Once you have a coupling, you can bound errors in expectations:

for Lipschitz $f$. This is exactly the sort of statement computation can support.

Concentration survives if you preserve boundedness or variance control

Concentration inequalities are computationally friendly because they turn finite samples into explicit error bars. What discretization threatens is the hypothesis.

If your discretization preserves:

• boundedness of $f(X)$, or
• a usable variance proxy,

then bounds like Hoeffding or Bernstein remain meaningful. If discretization introduces rare large spikes, concentration becomes misleading.

A strong engineering habit is to monitor empirical tails and to design transforms that keep $f(X)$ within controlled ranges when possible.

Floating point: measure preservation can fail quietly

Floating point errors usually appear small, but they can break exact invariants:

• probabilities that should sum to one can drift
• nonnegativity can be violated by subtraction
• accept/reject rules can become biased near thresholds

Two simple safeguards preserve probabilistic structure:

• renormalize explicitly when building discrete distributions
• avoid subtractive cancellation in probability calculations by working in log-space for very small probabilities

These are not numerical tricks; they are measure-theoretic safeguards.

What you should say in a research-grade computational claim

A computational probability statement is strongest when it names:

• the ideal object (measure, kernel, expectation)
• the discretized object (grid, truncation, finite chain)
• the comparison notion (weak, total variation, Wasserstein, coupling)
• the error parameter (mesh size $h$, sample size $N$, truncation radius $R$)

Then you can say something precise, such as:

• “As $h\to 0$, $\mu_h\Rightarrow \mu$.”
• “For Lipschitz $f$, $|\mathbb{E}_{\mu_h}f-\mathbb{E}_{\mu}f|\le C h$.”
• “The computed Markov chain has stationary distribution $\pi_h$ and $\|\pi_h-\pi\|$ is controlled in Wasserstein distance by the truncation level.”

This is what “what survives discretization” really means: you isolate the invariants your algorithm respects and you measure the rest.

A compact takeaway

Discretization does not have a single effect on probability. Its effect depends on the structure you are trying to preserve.

• Weak distributional features survive many discretizations.
• Tail events and fine dependence structure are fragile.
• Stationarity and unbiasedness can survive, but only if you build them in.
• The most transparent error analysis comes from coupling and metric choice.

Once you think this way, computational probability stops being a collection of ad hoc approximations. It becomes a controlled translation between infinite objects and finite representations, with a named contract for what is preserved.

Time-stepping stochastic processes: preserving the right martingale structure

When you simulate a stochastic process, you typically replace continuous time by a time grid $t_k = k\Delta t$. Even when the underlying model is not written as a differential equation, the computational object is a recursion.

Two features matter more than they first appear:

• adaptedness: the next step must depend only on information available up to the current time
• conditional mean structure: martingale increments should have conditional mean zero when that is part of the model

A simple example is a random walk $S_{k+1}=S_k+\xi_{k+1}$ with $\mathbb{E}[\xi_{k+1}\mid \mathcal{F}_k]=0$. If you approximate $\xi_{k+1}$ by a discretized noise $\tilde{\xi}_{k+1}$, you should preserve the conditional mean zero property. Otherwise you introduce drift, and the long-run behavior changes qualitatively.

In time-stepping schemes for continuous models, the same principle appears as “weak order” versus “strong order”:

• Weak accuracy means expectations of test functions $\mathbb{E}[f(X_{t_k})]$ are approximated well.
• Strong accuracy means pathwise errors $\mathbb{E}[|X_{t_k}-\tilde{X}_{t_k}|]$ are small.

Most statistical tasks care about weak accuracy. Many path-dependent functionals (such as hitting probabilities) care about strong accuracy or at least careful control of rare-event distortion.

Rare events: discretization can destroy the quantity you care about

If your target is a rare probability $\mathbb{P}(X\in A)$ with very small mass, naive discretization can be disastrous:

• truncation can remove the rare region entirely
• binning can smear a sharp boundary
• finite sampling can produce zero hits, giving a misleading estimate

This is where importance sampling and splitting methods belong. The mathematical point is: you are changing the sampling measure on purpose and correcting by a likelihood ratio. The invariant you preserve is still unbiasedness, but now it is unbiasedness under a weighted estimator.

A reliable practice is to pair any rare-event computation with a diagnostic coupling or a variance estimate that confirms you are not just measuring the discretization artifact.