Common Errors With Probability and Independence

Four Checks Before Using Probability or Independence

Probability calculations can look valid while relying on an event that has no defined probability, a map that is not a random variable, or an independence condition that is weaker than the conclusion requires. A related mistake is to treat zero covariance as proof of independence. These are not merely matters of terminology: each error can change the answer to a calculation.

Unless specified otherwise, let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, so \(\mathbb{P}(\Omega)=1\). No completeness assumption is made. An event is a set in \(\mathcal{F}\), and a real-valued random variable is an \(\mathcal{F}/\mathcal{B}_{\mathbb{R}}\)-measurable function \(X:\Omega\to\mathbb{R}\). Events \(A\) and \(B\) are independent when \(\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)\). A family of events is mutually independent when this product identity holds for every finite subfamily, not just every pair.

Probability-Zero Is Not the Same as Impossible

An event \(N\in\mathcal{F}\) is a null event if \(\mathbb{P}(N)=0\). This does not mean \(N=\varnothing\). It means that the event has probability zero. For example, on \(\Omega=[0,1]\) with the Borel sigma-algebra and Lebesgue probability measure, the event \(N=\{1/2\}\) is nonempty and has probability zero.

For events \(A,B\in\mathcal{F}\), the elementary conditional-probability formula is

The condition \(\mathbb{P}(A)>0\) is essential: if \(\mathbb{P}(A)=0\), the displayed ratio is undefined. In particular, for \(A=\{1/2\}\) in the example above, both \(\mathbb{P}(A)\) and \(\mathbb{P}(A\cap[0,1/2])\) are zero. The expression \(0/0\) does not define a conditional probability. This event-based formula alone gives no value for conditioning on \(A\).

There is a different useful fact about null events: every null event is independent of every event. Indeed, if \(\mathbb{P}(N)=0\) and \(B\in\mathcal{F}\), then \(N\cap B\subseteq N\), so \(\mathbb{P}(N\cap B)=0=\mathbb{P}(N)\mathbb{P}(B)\). Independence here does not turn conditioning on \(N\) into a defined ratio.

Proof. The symmetric difference assumption implies that \(A\setminus A'\) and \(A'\setminus A\) are null. Thus the parts of \(A\) and \(A'\) not shared by the other have probability zero, giving \(\mathbb{P}(A)=\mathbb{P}(A')\). Also, \((A\cap B)\triangle(A'\cap B)\subseteq A\triangle A'\), so \(\mathbb{P}(A\cap B)=\mathbb{P}(A'\cap B)\). Therefore the two independence identities have the same left side and the same right side. This proves both directions. \(\square\)

Verify Measurability Before Calling a Map a Random Variable

For a real-valued map \(X:\Omega\to\mathbb{R}\) to be a random variable, every inverse image \(X^{-1}(C)\) of a Borel set \(C\in\mathcal{B}_{\mathbb{R}}\) must belong to \(\mathcal{F}\). Assigning real values to every outcome is not enough. Without measurability, expressions such as \(\mathbb{P}(X\in C)\) need not be probabilities of events in the given space.

Proof. If \(E\in\mathcal{F}\), then for each Borel set \(C\subseteq\mathbb{R}\), the inverse image \(\chi_E^{-1}(C)\) is one of \(\varnothing,\Omega,E,E^c\), according to whether \(0\) and \(1\) belong to \(C\). All four sets are in \(\mathcal{F}\), so \(\chi_E\) is measurable. Conversely, if \(\chi_E\) is measurable, then \(E=\chi_E^{-1}(\{1\})\). Since \(\{1\}\) is Borel, measurability gives \(E\in\mathcal{F}\). \(\square\)

Pairwise Independence Does Not Guarantee Mutual Independence

For three events, pairwise independence checks the three pairs. Mutual independence requires the product identity for every subfamily, including the subfamily of all three. The extra condition cannot in general be inferred from the pair checks.

Common mistake. A wrong calculation is: “Each pair is independent, so the probability of all three events is \(\mathbb{P}(A)\mathbb{P}(B)\mathbb{P}(C)=1/8\).” The correction is to test mutual independence directly for the three-event intersection. In this example that intersection is empty, so its probability is \(0\), not \(1/8\). Pairwise independence supplies no missing triple-intersection condition.

Zero Covariance Is Not a Test for Independence

For finite-valued real random variables \(X,Y\), define \(\mathbb{E}[X]\) by summing each value times its probability. Their covariance is \(\operatorname{Cov}(X,Y)=\mathbb{E}[XY]-\mathbb{E}[X]\mathbb{E}[Y]\). They are uncorrelated if this covariance is zero. Independence is a stronger condition: it requires the probabilities of all Borel-set events \(\{X\in C\}\) and \(\{Y\in D\}\) to factor. Zero covariance alone does not check those event identities.

Proof. Write the distinct values of \(X\) as \(x_1,\ldots,x_r\) and those of \(Y\) as \(y_1,\ldots,y_s\). Their level sets are measurable because the variables are random variables. Independence gives \(\mathbb{P}(X=x_i,Y=y_j)=\mathbb{P}(X=x_i)\mathbb{P}(Y=y_j)\) for every \(i,j\). Since the variables have finite ranges, all expectations below are finite sums, and

Subtracting \(\mathbb{E}[X]\mathbb{E}[Y]\) gives \(\operatorname{Cov}(X,Y)=0\). The argument also covers a variable that is constant: its range has one value, and the same finite sums apply. \(\square\)

The theorem establishes one direction under its stated finite-valued hypotheses. The worked example disproves the converse, even for bounded variables on a finite probability space. A reliable check is therefore to keep the definitions separate: measurability makes a map a random variable; mutual independence requires all finite-subfamily identities; and zero covariance is not a substitute for independence.