Why a Familiar Factor Can Mislead
Independence is tested by comparing a joint distribution with the product of its marginals. The comparison can look deceptively simple: a common constant may appear in every probability, or an algebraic factor may appear on both sides of an equation. Cancelling without checking what the factor represents can hide a support restriction or leave important cases untested.
A related trap is to find a point where the joint probability or density equals the product of the marginals and treat that as proof. Independence requires a full factorization, not a successful check at one or several selected points. One counterexample within the support is enough to disprove independence.
What Cancellation Does—and Does Not—Show
A nonzero common factor can be cancelled from an equality. But cancellation only simplifies that equality where the factor is nonzero; it does not establish that an equality holds everywhere it needs to. In an independence test, that means a support indicator cannot be discarded just because there is also a constant factor.
For example, a joint density might have the form \(c\,g(x)h(y)I_S(x,y)\), where \(c\) is a normalizing constant and \(I_S\) equals 1 on the joint support \(S\) and 0 elsewhere. The constant \(c\) is not the whole story: the support factor \(I_S\) may depend jointly on \(x\) and \(y\). If \(S\) is not a product of an \(x\)-set and a \(y\)-set, the support restriction can prevent the density from factoring into a function of \(x\) times a function of \(y\).
A useful way to disprove independence is to find a point or cell where both marginal probabilities or densities are positive but the joint probability or density is zero. Their product is then positive, so factorization fails. For continuous variables, the disagreement must occur on a set of positive area (or, more generally, positive measure) to contradict equality almost everywhere; a single isolated point is not enough.
Worked Example: A Shared Normalizing Constant Does Not Remove a Support Restriction
A pair of discrete variables has joint probabilities \(1/3\) at \((0,0),(0,1),(1,0)\), and probability 0 at \((1,1)\). Determine whether \(X\) and \(Y\) are independent. A student argues that the common factor \(1/3\) can be cancelled, leaving no obstacle to independence.
Method. Compute the marginals and compare the joint probability with their product. The support restriction—the missing \((1,1)\) cell—must remain in the comparison.
The probabilities sum to \(1/3+1/3+1/3+0=1\), so this is a valid joint distribution. The marginal probabilities are
and, by summing the columns,
At \((X,Y)=(1,1)\), the joint probability is 0, while the product of the marginals is
Since \(0\ne 1/9\), the factorization fails and \(X,Y\) are not independent. Cancelling the common numerical factor \(1/3\) in the three nonzero cells would not make the missing cell independent: its joint probability is still zero, while both marginals there are positive. Common mistake avoided: treating a common constant as if it removed the support restriction.
Why the Full Factorization Is the Test
The factorization criterion follows directly from the definition of independence. For discrete variables, independence means that the events \(X=x\) and \(Y=y\) are independent for every pair \(x,y\), so their intersection must have probability equal to the product of their probabilities. For continuous variables, the corresponding statement is that the joint density is the product of the marginal densities almost everywhere. These are full-distribution conditions, not conditions to check at a convenient selection of points.
A point check can still be useful for disproving independence: one genuine mismatch is enough. But finding a match at one point proves only that factorization holds at that point. In continuous models, even several matching points cannot establish equality almost everywhere.
Worked Example: A Density Matches at One Point but Not Across the Support
Suppose \((X,Y)\) has joint density \(f_{X,Y}(x,y)=2\) when \(0<y<x<1\), and 0 elsewhere. Determine whether \(X\) and \(Y\) are independent. A proposed check finds a point where the joint density equals the product of the marginals.
Method. First verify the density, then find both marginals and compare at more than one point. The support restriction \(0<y<x<1\) is part of the model.
The density is normalized because
For \(0<x<1\), integrating over \(0<y<x\) gives
For \(0<y<1\), the allowed \(x\)-values run from \(y\) to 1, so
At the interior point \((x,y)=(3/4,1/3)\), the joint density is 2, and the product of the marginals is
This match does not prove independence. At the interior point \((x,y)=(1/2,1/4)\), the joint density is again 2, but
The densities disagree on a neighborhood of this point, so they do not agree almost everywhere. Thus \(X,Y\) are not independent. The triangular support is another warning: for example, points with \(x<y\) are outside the support even though both marginal densities can be positive there. Common mistake avoided: treating a match at a selected point as proof of a global factorization.
Handling Factors in Discrete Calculations
A shared denominator in a probability table is merely a convenient way to write the cell probabilities. It does not automatically cancel in the independence equation, because the marginal probabilities are sums of cells and their product has two marginal factors. A safe procedure is to calculate the marginals explicitly, then compare the joint probability and the product using the same scale.
Worked Example: A Common Denominator Is Not an Independence Test
The joint probability table for two indicators \(X,Y\), each taking values 0 and 1, is
| \(Y=0\) | \(Y=1\) | |
|---|---|---|
| \(X=0\) | \(2/8\) | \(2/8\) |
| \(X=1\) | \(1/8\) | \(3/8\) |
Determine whether \(X\) and \(Y\) are independent. The entries share denominator 8, which a student claims can be cancelled to establish independence.
Method. Sum the rows and columns to obtain the marginals, then compare one joint cell with the corresponding product. The table sums to \((2+2+1+3)/8=1\).
The row sums give
The column sums give
At \((0,0)\), the joint probability is \(2/8=1/4\), whereas
Since \(1/4=4/16\ne3/16\), the variables are not independent. Cancelling a denominator from the entries alone does not compare them with the products of the marginal sums. Common mistake avoided: cancelling the table’s shared denominator without accounting for how it appears in each marginal and in their product.
A Reliable Independence Check
Use a deliberate sequence rather than informal cancellation. In a discrete problem, write down the support and sum the joint probabilities to obtain each marginal. In a continuous problem, identify the support and integrate the joint density over the correct bounds to obtain each marginal. Then compare the joint distribution with the product of those marginals.
Include zero-probability cells or regions where the joint law vanishes. Do not replace a restricted support by a rectangle unless the model permits it.
Sum over the other variable in a discrete model, or integrate over its allowed values in a continuous model. Respect the support when choosing bounds.
Test the joint probability or density against the product of the marginal probabilities or densities. Keep any support indicator in the calculation.
A mismatch disproves independence. A match at one point does not prove it; verification requires factorization throughout the discrete support or almost everywhere for densities.
Common Traps and Why They Matter
- Dropping a support indicator. A density may look like a product on the region where it is positive, but a restriction such as \(y<x\) can prevent factorization over the full plane.
- Equating one successful check with proof. A joint density can match the product of its marginals at a particular point and fail elsewhere.
- Cancelling a factor that is zero. Dividing by a zero factor is not valid. An equation after cancellation gives no information about cases where the factor vanishes; those cases must be checked separately if they matter to the distribution.
- Misusing a common denominator. Marginal probabilities are sums, and the product of two marginals has to be calculated before comparing it with a joint cell.
These checks matter because independence is a claim about the entire joint law, not about how a formula looks after a convenient simplification. In exam problems, preserving the support and checking the actual joint-marginal equality is often more reliable than attempting to recognize independence by inspection. This method is aligned with the SOA Exam P (Probability) syllabus.
Check Your Understanding
For each problem, keep the support in view and check the full factorization required for the conclusion.
- A joint table assigns probability \(1/4\) to \((0,0),(0,1),(1,0)\) and 0 to \((1,1)\). Is this a valid probability table, and are \(X,Y\) independent? (Answer: It is valid; no, since \(P(X=1,Y=1)=0\) but \(P(X=1)P(Y=1)=1/16\).)
- A joint density is 1 on \(0<x<1,\ 0<y<1\), except it is zero on \(0<x<1/2,\ 0<y<1/2\), and zero elsewhere. Is this a normalized density, and can \(X,Y\) be independent? (Answer: It is not normalized: its integral is \(1-1/4=3/4\). Multiplying the positive part by \(4/3\) normalizes it, and the normalized variables are not independent because the joint density is zero on a positive-area region where both marginals are positive.)
- Let \(f_{X,Y}(x,y)=2\) for \(0<x<y<1\), and 0 elsewhere. Find the marginal densities and decide whether \(X,Y\) are independent. (Answer: \(f_X(x)=2(1-x)\) and \(f_Y(y)=2y\) on \((0,1)\); they are not independent because the support is triangular and the joint density is zero on a positive-area region where both marginals are positive.)
- A table has cell probabilities \(3/10,1/10,2/10,4/10\), in row order \((0,0),(0,1),(1,0),(1,1)\). Test independence by comparing the first cell with its marginal product. (Answer: No; \(P(X=0)=4/10\), \(P(Y=0)=5/10\), and \(P(X=0,Y=0)=3/10\ne(4/10)(5/10)=2/10\).)
- On the unit square, let \(f_{X,Y}(x,y)=x+y\). The density matches the product of its marginal densities at \((1/2,1/2)\). Are \(X,Y\) independent? (Answer: No; \(f_X(x)=x+1/2\) and \(f_Y(y)=y+1/2\). At \((1/4,1/4)\), the joint density is \(1/2\), while the marginal product is \((3/4)(3/4)=9/16\).)