Validity Diagnostics for Joint PMFs and Densities

What Makes a Joint Representation Valid?

A formula can look like a joint probability model and still fail a basic diagnostic. It might assign positive mass to a forbidden lattice point, take negative values, or have total mass different from one. Even a candidate with correctly normalized marginals can fail if the joint values themselves are negative. The checks in this tutorial separate these issues and show what each check can—and cannot—establish.

A joint PMF describes a pair of discrete random variables, while a joint density describes an absolutely continuous pair. Their validity conditions have the same basic shape, but the sums and integrals, and the treatment of exceptional points, differ.

Definition: A candidate joint PMF \(p(x,y)\) on a specified countable support \(S\) is valid if \(p(x,y)\geq 0\) for every \((x,y)\in S\) and \(\sum_{(x,y)\in S}p(x,y)=1\). It assigns probability zero to pairs outside \(S\). A candidate joint density \(f(x,y)\) on \(\mathbb R^2\) is valid if it is measurable, nonnegative almost everywhere with respect to area, and \(\int_{\mathbb R^2} f(x,y)\,dx\,dy=1\). If a support region \(D\) is specified, the density must be zero almost everywhere outside \(D\).

For a PMF, a negative value at even one allowed point is a failure. For a density, changing values on a set of area zero does not change the distribution, so nonnegativity and support conditions are stated almost everywhere. The normalization sum or integral must be well-defined; nonnegative functions can be summed or integrated in any order, including when first checking that the total is finite.

Necessary Checks and What They Establish

The three primary checks are support, nonnegativity, and total mass. Each answers a different question. Support asks where the model may put probability. Nonnegativity rules out assigning negative probability locally. Normalization ensures that the probability of the entire sample space is one. Passing all three checks is sufficient for validity under the definitions above.

Marginals and selected event probabilities provide useful additional diagnostics. For a valid discrete joint law, the marginal PMFs are obtained by summing over the other coordinate; for a valid density, marginal densities are obtained by integrating over the other coordinate. These operations should produce nonnegative functions with total mass one. But plausible marginals do not repair a negative joint value, as the first worked example demonstrates.

Worked Example: A Joint Table With Valid-Looking Marginals

On \(S=\{(0,0),(0,1),(1,0),(1,1)\}\), consider the candidate values \(p(0,0)=0.4\), \(p(0,1)=0.4\), \(p(1,0)=0.4\), and \(p(1,1)=-0.2\). Check the total, marginals, and validity.

The total is \(0.4+0.4+0.4-0.2=1\). The row sums, which would be the proposed marginal values for \(X\), are \(p_X(0)=0.4+0.4=0.8\) and \(p_X(1)=0.4-0.2=0.2\). The column sums are likewise \(p_Y(0)=0.8\) and \(p_Y(1)=0.2\). Each proposed marginal is nonnegative and sums to one.

Nevertheless, the joint candidate is not a PMF: \(p(1,1)=-0.2<0\). In particular, it assigns negative probability to the event \(\{X=1,Y=1\}\). The normalization and marginal checks do not override this failure.

For a continuous candidate, a negative value at an isolated point does not by itself invalidate a density, because a single point has area zero. A negative value throughout a region of positive area does invalidate it. In either case, the decisive condition is nonnegativity almost everywhere, not nonnegativity at every point.

Marginals as a Diagnostic

For a joint PMF \(p\) on a countable set, the marginal candidates are \(p_X(x)=\sum_y p(x,y)\) and \(p_Y(y)=\sum_x p(x,y)\), with sums restricted to the joint support. For a joint density \(f\), the corresponding formulas are \(f_X(x)=\int_{\mathbb R}f(x,y)\,dy\) and \(f_Y(y)=\int_{\mathbb R}f(x,y)\,dx\), wherever the integrals exist. The marginal formula must respect the joint support: values outside its permitted range contribute nothing.

Theorem: If \(p\) is a valid joint PMF, its marginal sums are nonnegative and each marginal sums to one. If \(f\) is a valid joint density, its marginal integrals are nonnegative almost everywhere and each marginal integrates to one.

Proof. For a PMF, every term \(p(x,y)\) is nonnegative, so each sum \(p_X(x)\) is nonnegative. Since all terms are nonnegative, the order of summation can be interchanged:

\sum_x p_X(x) =\sum_x\sum_y p(x,y) =\sum_{(x,y)\in S}p(x,y) =1.

The same argument with the coordinates reversed gives \(\sum_y p_Y(y)=1\). For a density, nonnegativity of \(f\) almost everywhere implies nonnegativity of its integrals almost everywhere. Tonelli’s theorem applies to the nonnegative measurable function \(f\), so

\int_{\mathbb R} f_X(x)\,dx =\int_{\mathbb R}\left(\int_{\mathbb R}f(x,y)\,dy\right)dx =\int_{\mathbb R^2}f(x,y)\,dx\,dy =1.

Interchanging the coordinates gives \(\int_{\mathbb R}f_Y(y)\,dy=1\). Thus valid joint models have normalized marginals. This implication is a diagnostic consequence of validity, not a substitute for checking the joint function itself. \(\square\)

Worked Example: Checking a Density on a Triangular Support

Let \(D=\{(x,y):0<x<y<1\}\), and define \(f(x,y)=6x\) on \(D\), with \(f(x,y)=0\) outside \(D\). Verify validity, find both marginals, and calculate \(P(Y<1/2)\).

On \(D\), \(x>0\), so \(6x\geq0\); outside \(D\), the function is zero. Its total integral is

\int_0^1\int_0^y 6x\,dx\,dy =\int_0^1 3y^2\,dy =\left[y^3\right]_0^1 =1.

The function is measurable, nonnegative, and integrates to one, so it is a valid density. For fixed \(0<x<1\), the allowed values of \(y\) run from \(x\) to \(1\). Hence \(f_X(x)=\int_x^1 6x\,dy=6x(1-x)\) for \(0<x<1\), and it is zero otherwise. For \(0<y<1\), the allowed values of \(x\) run from \(0\) to \(y\), giving \(f_Y(y)=\int_0^y6x\,dx=3y^2\), and zero otherwise.

As checks, \(\int_0^1 6x(1-x)\,dx=[3x^2-2x^3]_0^1=1\) and \(\int_0^1 3y^2\,dy=1\). Finally, \(Y<1/2\) cuts the triangle to \(0<y<1/2\), \(0<x<y\), so

P(Y<1/2)=\int_0^{1/2}\int_0^y6x\,dx\,dy =\int_0^{1/2}3y^2\,dy =\left[y^3\right]_0^{1/2} =\frac18.

Support and Region-Probability Checks

A support restriction affects every calculation, not just the normalization step. On a triangular or constrained lattice support, summing or integrating over a rectangle without restricting to the support can include forbidden pairs. A reliable calculation either writes the support limits directly or includes an indicator of the support in the integrand.

Theorem: For any valid joint PMF or density and any measurable event \(A\) in the pair’s sample space, \(0\leq P((X,Y)\in A)\leq1\). In the discrete case the probability is the sum of \(p\) over \(A\); in the density case it is the integral of \(f\) over \(A\).

Proof. In the discrete case, each summand is nonnegative, so the sum over \(A\) is at least zero. Since \(A\) is a subset of the full support, that sum is at most \(\sum_{(x,y)\in S}p(x,y)=1\). For a density, nonnegativity gives \(\int_A f\geq0\), and \(A\subseteq\mathbb R^2\) gives \(\int_A f\leq\int_{\mathbb R^2} f=1\). Therefore both forms of the event probability lie between zero and one. \(\square\)

This theorem supplies a quick check on a selected region calculation: a result below zero or above one signals an error in the proposed model or in the calculation. Passing the bound alone is not enough to establish validity; an invalid joint function may still produce plausible values for some regions.

Worked Example: A Valid Constrained-Lattice PMF

Let \(S=\{(0,0),(0,1),(1,0)\}\), and assign probabilities \(1/6\), \(2/6\), and \(3/6\) to these points, respectively. Verify the PMF, find its marginals, and calculate \(P(X+Y=1)\).

All three assigned values are nonnegative, and their sum is \(1/6+2/6+3/6=1\). There are no other allowed points, so this is a valid PMF supported on \(S\); in particular, \((1,1)\) has probability zero.

Summing over \(y\) gives \(p_X(0)=1/6+2/6=1/2\) and \(p_X(1)=3/6=1/2\). Summing over \(x\) gives \(p_Y(0)=1/6+3/6=2/3\) and \(p_Y(1)=2/6=1/3\). Both marginal lists are nonnegative and sum to one.

The event \(X+Y=1\) consists of \((0,1)\) and \((1,0)\), so \(P(X+Y=1)=2/6+3/6=5/6\). The forbidden pair \((1,1)\) is not included in this sum.

When and How to Normalize

Sometimes a candidate is presented as an unnormalized function \(h\), with the instruction to find a constant that makes it a density or PMF. If \(h\geq0\) and its total \(Z\) is finite and strictly positive, then \(h/Z\) is nonnegative and has total one. But dividing an arbitrary signed function by its total is not automatically safe: the signs after division depend on both the sign of the total and the signs of the function.

Key takeaway: Let \(h\) be integrable, or summable on a discrete support, with finite nonzero total \(Z\). The normalized function \(h/Z\) has total one, but it is a valid PMF or density only if \(h/Z\) is nonnegative on the allowed points, or almost everywhere in the continuous case. If \(Z>0\), division preserves signs; if \(Z<0\), division reverses signs. If \(Z=0\), division by \(Z\) is undefined.

Worked Example: Normalizing a Signed Function

On the unit square \(D=(0,1)^2\), let \(h(x,y)=-1\), and set \(h=0\) outside \(D\). Its integral is \(Z=\int_D -1\,dx\,dy=-1\). Dividing by this negative total gives \(h/Z=1\) on \(D\), zero outside, which is a valid density because it is nonnegative and integrates to one.

This example shows why the statement “a negative region can never be fixed by dividing by the total” is too broad. Here the total is negative, so division reverses the sign everywhere. By contrast, if \(Z>0\) and \(h\) is negative on a set of positive area, \(h/Z\) remains negative there and cannot be a density. More generally, the actual condition to check after normalization is that \(h/Z\geq0\) almost everywhere. In the discrete case, use the same sign check at every allowed point.

Common Mistakes and a Reliable Diagnostic Order

A candidate’s total mass is often the first calculation, but it should not be treated as the only one. The sequence below keeps the checks distinct and makes it harder to overlook a support or sign error.

Identify the allowed support.
Write down which lattice points or regions are permitted; require a PMF to assign zero outside its support and a density to be zero almost everywhere outside its specified region.

Check signs.
For a PMF, inspect every allowed point. For a density, verify nonnegativity almost everywhere on the support.

Check total mass.
Sum over the discrete support or integrate over the continuous support, using limits that respect its constraints.

Check consequences.
Derive marginals and probabilities of selected regions; confirm that marginal totals and event probabilities are plausible and within their required bounds.

Mistake: treating total one as proof of validity. A signed function can integrate to one while being negative on part of its support. Check nonnegativity separately.
Mistake: trusting normalized marginals as proof of a valid joint law. The negative table above has normalized marginal sums, yet one joint entry is negative. Check the joint values themselves.
Mistake: integrating over a larger region than the support. The density \(6x\) in the triangular example is only used where \(0<x<y<1\). Its marginal limits come from that constraint.
Mistake: treating a density value as a point probability. A density is integrated over a region to obtain probability; a value at one point is not the probability of that point.

These diagnostics matter because later calculations inherit every defect in the joint representation. A mistaken support or negative value can distort marginals, event probabilities, and any summary derived from them. Establish validity first; then use the representation to calculate.

Check Your Understanding

For each question, distinguish checks that establish validity from checks that only provide supporting diagnostics.

A candidate PMF has total one but assigns \(-0.1\) to one allowed point. Is it valid? Which check fails?
A density candidate is negative at one isolated point and nonnegative elsewhere. Does that alone invalidate it? State the relevant condition.
For a joint density supported on \(0<x<y<1\), which limits should be used to integrate out \(x\) when finding the marginal density of \(Y\)?
Can normalized, nonnegative marginal PMFs guarantee that all entries of the proposed joint PMF are nonnegative? Explain using the table example.
A signed function \(h\) has total \(Z<0\). What sign condition on \(h\) is needed so that \(h/Z\) is nonnegative?