Convolution Checks for Support and Normalization

Why Check a Convolution?

A convolution calculation can look algebraically plausible and still describe no probability distribution. A missing interval of integration can leave out probability; an incorrect range can include impossible values; and a density that integrates to one still does not describe the whole law if the sum also has an unaccounted-for atom. Checking support, nonnegativity, total probability, and boundary masses is therefore part of deriving a distribution, not an optional final step.

The checks apply to discrete, continuous, and mixed random variables. The common starting point is that \(X\) and \(Y\) are independent. If they are dependent, the marginal distributions alone do not determine the law of \(X+Y\), as established in the earlier material on why dependence invalidates convolution.

Support Determines the Limits

If \(X\) and \(Y\) have densities \(f_X\) and \(f_Y\), the density of their sum is $$ f_{X+Y}(z)=\int_{-\infty}^{\infty} f_X(x)f_Y(z-x)\,dx. $$ The integrand can be nonzero only when \(x\) is in the support of \(X\) and \(z-x\) is in the support of \(Y\). These two conditions determine the actual integration limits. If the supports are intervals \([a,b]\) and \([c,d]\), respectively, they require $$ \max(a,z-d)\leq x\leq \min(b,z-c). $$ An empty interval means the density is zero at that \(z\). The sum can only lie between \(a+c\) and \(b+d\), though it need not have positive density at every point in that range if the component supports have gaps.

For discrete variables, the corresponding rule is to sum only over values that are possible for both components: $$ P(X+Y=z)=\sum_x P(X=x)P(Y=z-x). $$ For mixed variables, it is safest to begin with the measure formula. A density convolution alone may fail to record atoms; the examples below show how to check for them.

Proof. Let \(S\) be the closure of \(S_X+S_Y\). If \(z\notin S\), some open interval \(I\) containing \(z\) is disjoint from \(S_X+S_Y\). With probability one, \(X\in S_X\) and \(Y\in S_Y\), so \(X+Y\notin I\) with probability one. Thus \(z\) is not in the support of the sum.

Now let \(z\in S\), and let \(I\) be any open interval containing \(z\). Since \(z\) is in the closure, there are \(x\in S_X\) and \(y\in S_Y\) with \(x+y\in I\). Because addition is continuous and \(I\) is open, there are open intervals \(U\) around \(x\) and \(V\) around \(y\) such that \(u+v\in I\) for every \(u\in U\) and \(v\in V\). The definition of support gives \(P(X\in U)>0\) and \(P(Y\in V)>0\). Independence then gives $$ P(X+Y\in I)\geq P(X\in U,Y\in V) =P(X\in U)P(Y\in V)>0. $$ Every interval around \(z\) therefore has positive probability, so \(z\) is in the support of the sum. This proves the claim. \(\square\)

Normalization and Boundary Masses

A valid density must be nonnegative and integrate to one. A valid probability mass function must be nonnegative and sum to one. Convolution preserves both properties under independence, but checking them in a calculation can expose a wrong formula or omitted part of the support.

Proof. Since densities are nonnegative, \(f_X(x)f_Y(z-x)\geq0\), and its integral in \(x\) is nonnegative. Tonelli’s theorem applies to the nonnegative integrand, so the integral over \(z\) can be evaluated by changing the order of integration. For every fixed \(x\), use \(u=z-x\): $$ \int_{-\infty}^{\infty} f_{X+Y}(z)\,dz =\int_{-\infty}^{\infty} f_X(x) \left(\int_{-\infty}^{\infty} f_Y(z-x)\,dz\right)dx =\int_{-\infty}^{\infty} f_X(x)\,dx=1. $$ Thus the convolution is nonnegative and integrates to one. \(\square\)

This proof depends on integrating over the entire real line, with the component densities set to zero outside their supports. In a piecewise calculation, the displayed intervals must cover all possible values of the sum. If a piece is omitted, the resulting expression need not integrate to one.

A support endpoint is not necessarily an atom. For example, a continuous variable may have support \([a,b]\) while \(P(X=a)=P(X=b)=0\). More generally, if \(X\geq a\) and \(Y\geq b\) almost surely, then the sum can equal \(a+b\) only when \(X=a\) and \(Y=b\). Independence gives $$ P(X+Y=a+b)=P(X=a)P(Y=b). $$ So a boundary atom occurs precisely when both component variables have atoms at their respective lower endpoints. The corresponding statement holds for upper endpoints when the variables are bounded above. For distributions whose support endpoint is only an infimum or supremum, the endpoint may not be attainable at all.

A Practical Diagnostic

When reviewing a convolution, use the following checks in order. They separate errors about where probability can occur from errors about how much probability the formula assigns.

For instance, keeping only the first branch \(f_Z(z)=z\) in the uniform-sum example would omit the interval \(1<z\leq2\). The retained formula would integrate to \(\int_0^1 z\,dz=1/2\), not one, immediately exposing the missing probability. Conversely, simply stretching that branch across \([0,2]\) would give a nonzero density where the convolution geometry requires a different overlap length. Support and normalization checks complement each other: neither replaces the derivation.

Common Mistake

• Assuming support endpoints must be atoms. A student sees support \([0,2]\) and assigns positive probability to \(Z=0\) or \(Z=2\). The correction is to distinguish neighborhoods from points: support means every neighborhood has positive probability, while an atom requires \(P(Z=z)>0\). For independent sums, check the component endpoint masses directly.
• Checking only that a formula integrates to one. A nonnegative function with total integral one can still put probability on impossible values and omit other parts. Verify the support and the limits before accepting normalization.
• Ignoring gaps in component supports. Adding only the smallest and largest possible values gives an outer range, not always the full support. The mixed example has sums in \([0,1]\cup[2,3]\), with no probability in between.