Why a Sum Requires Joint Information
For a single random variable, its characteristic function records its distribution through the expectations of \(e^{itX}\). For a sum \(X+Y\), the corresponding expression is \(e^{it(X+Y)}\). This depends on the two variables together: knowing how \(X\) behaves and how \(Y\) behaves separately does not, in general, tell us how often their values occur together.
The joint characteristic function keeps track of this joint behavior. Its values on a particular line—the diagonal where both arguments equal \(t\)—give exactly the characteristic function of the sum. This identity needs no assumption that \(X\) and \(Y\) are independent.
These expectations always exist: the complex exponential in each definition has absolute value \(1\), so its real and imaginary parts are bounded. In particular, no moment assumptions are needed. The joint characteristic function contains the marginal transforms as special cases:
The Diagonal Restriction
The key identity follows by substituting the same argument into both coordinates of the joint characteristic function.
Proof. Since \(X+Y\) is a measurable random variable, its characteristic function is defined. For every outcome, \(e^{it(X+Y)}=e^{i(tX+tY)}\). Taking expectations on both sides gives $$ \phi_{X+Y}(t) =E\!\left[e^{it(X+Y)}\right] =E\!\left[e^{i(tX+tY)}\right] =\phi_{X,Y}(t,t). $$ Each expectation exists because the integrands have absolute value \(1\). \(\square\)
The theorem gives a practical recipe: obtain or specify the joint transform, then evaluate it at \((t,t)\). The diagonal is important because both summands have the same coefficient in \(X+Y\). For a weighted sum \(aX+bY\), the same definition gives \(\phi_{aX+bY}(t)=\phi_{X,Y}(at,bt)\); here, however, the focus is the ordinary sum.
Worked Discrete Examples
Worked Example: Identical Bernoulli Marginals, Different Sums
Let \(X\) and \(Y\) each have the Bernoulli distribution with success probability \(1/2\). Compare two possible joint distributions: first \(Y=X\), and then \(Y=1-X\). Both constructions have the same marginal distribution for each variable.
For either marginal, $$ \phi_X(t)=\phi_Y(t) =\tfrac12 e^{i0t}+\tfrac12 e^{it} =\frac{1+e^{it}}{2}. $$ In the first construction, \((X,Y)\) equals \((0,0)\) or \((1,1)\), each with probability \(1/2\). Thus $$ \phi_{X,Y}(s,t)=\tfrac12+\tfrac12 e^{i(s+t)}, \qquad \phi_{X+Y}(t)=\phi_{X,Y}(t,t) =\tfrac12+\tfrac12 e^{i2t}. $$ Indeed, \(X+Y\) is \(0\) or \(2\), each with probability \(1/2\).
In the second construction, the pair is \((0,1)\) or \((1,0)\), each with probability \(1/2\). Its joint characteristic function is $$ \phi_{X,Y}(s,t)=\tfrac12 e^{it}+\tfrac12 e^{is}. $$ On the diagonal this becomes \(\phi_{X+Y}(t)=e^{it}\), the characteristic function of the constant \(1\), since now \(X+Y=1\) with probability \(1\). The marginal characteristic functions are unchanged, but the sum distributions are different. The joint transform distinguishes the two dependence structures; the marginal transforms alone do not.
Worked Example: A Finite Joint Table and Its Sum
Suppose \((X,Y)\) has the following probabilities: \(P(X=0,Y=0)=1/4\), \(P(X=0,Y=1)=1/4\), and \(P(X=1,Y=1)=1/2\). These probabilities are nonnegative and sum to \(1\), so they specify a joint distribution.
Adding the contribution from each possible pair gives the joint characteristic function $$ \phi_{X,Y}(s,t) =\tfrac14+\tfrac14 e^{it}+\tfrac12 e^{i(s+t)}. $$ Evaluating on the diagonal yields $$ \phi_{X+Y}(t) =\tfrac14+\tfrac14 e^{it}+\tfrac12 e^{i2t}. $$ Thus \(X+Y\) equals \(0,1,2\) with probabilities \(1/4,1/4,1/2\), respectively. The probabilities can also be checked directly: only \((0,0)\) gives sum \(0\), only \((0,1)\) gives sum \(1\), and only \((1,1)\) gives sum \(2\).
For comparison, the marginal probabilities are \(P(X=1)=1/2\) and \(P(Y=1)=3/4\). If one incorrectly treated these variables as independent, the resulting sum probabilities would be \(1/8,1/2,3/8\) at \(0,1,2\). These differ from the actual probabilities, illustrating why the joint distribution—not merely the two marginal distributions—is needed.
A Continuous Example With Uniform Marginals
Dependence can change a sum just as decisively for continuous variables. The following example uses a uniform random variable and a deterministic function of it. The two variables have identical continuous marginal distributions, but their sum behaves very differently from what the marginals alone would reveal.
Worked Example: Uniform Variables With a Constant Sum
Let \(U\) be uniform on \((0,1)\), and set \(X=U\) and \(Y=1-U\). Reflection about \(1/2\) preserves the uniform distribution, so both \(X\) and \(Y\) are uniform on \((0,1)\). For \(t\ne 0\), direct integration gives $$ \phi_X(t)=\int_0^1 e^{itu}\,du =\frac{e^{it}-1}{it}, \qquad \phi_Y(t)=\int_0^1 e^{it(1-u)}\,du =\frac{e^{it}-1}{it}. $$ At \(t=0\), both characteristic functions equal \(1\), as follows directly from their definition.
For the joint transform, integrate over the values \(X=u\) and \(Y=1-u\): $$ \phi_{X,Y}(s,t) =\int_0^1 e^{i(su+t(1-u))}\,du. $$ The sum is \(X+Y=1\) for every outcome, so the diagonal restriction gives $$ \phi_{X+Y}(t)=\phi_{X,Y}(t,t) =\int_0^1 e^{it}\,du =e^{it}. $$ This is the characteristic function of the constant \(1\). Thus two uniform marginals are consistent with a sum that has no variation at all. The dependence between the variables is essential information.
What the Marginals Do—and Do Not—Determine
The marginal transforms reveal the distributions of \(X\) and \(Y\) separately, but they only give the joint transform along the two coordinate axes. To find the sum transform, we need its values along the diagonal. Knowing the axes does not generally determine the diagonal, as the Bernoulli and uniform examples demonstrate.
If \(X\) and \(Y\) are independent, the joint transform factors as \(\phi_{X,Y}(s,t)=\phi_X(s)\phi_Y(t)\). In that special case, the diagonal identity yields \(\phi_{X+Y}(t)=\phi_X(t)\phi_Y(t)\). Without independence, using that product is not justified: the joint transform need not factor, and the sum’s distribution can change while both marginals stay fixed.
This distinction matters whenever the variables are linked by a shared subject, environment, or mechanism. Repeated measurements on the same person, paired outcomes, and components subject to a common source of variation can be dependent. For such pairs, the correct starting point is their joint law or joint characteristic function, not a product of separately computed marginal transforms.
Common Mistake
A reliable check is to ask what information has been supplied. If it is only the two marginal distributions, a sum distribution may not be determined. If the joint distribution or joint characteristic function is known, restrict the latter to \((t,t)\), then use uniqueness of characteristic functions to identify the distribution when needed.
Check Your Understanding
Use the diagonal identity and the examples above to answer the following questions.
- Write the joint characteristic function of \((X,Y)\) and express \(\phi_{X+Y}(t)\) in terms of it.
- Why does the diagonal identity require neither independence nor finite moments?
- In the Bernoulli example, what are the possible values of \(X+Y\) when \(Y=X\), and when \(Y=1-X\)?
- For the finite joint table, explain why the coefficient of \(e^{i2t}\) in the sum characteristic function is \(1/2\).
- What additional condition justifies replacing the joint-transform calculation with \(\phi_X(t)\phi_Y(t)\)?