Why Dependence Invalidates Convolution

Why the Joint Law Matters

The distribution of $S=X+Y$ depends on how $X$ and $Y$ occur together, not only on their separate distributions. Convolution provides the law of a sum when the variables are independent: independence makes the joint distribution the product of the marginal distributions. If they are dependent, that factorization is unavailable, and multiplying the marginal laws can give the wrong answer.

The central calculation is always based on the joint law. For a particular sum, dependence does not guarantee that a convolution answer will differ from the truth; two different joint laws can sometimes give the same sum distribution. The point is that the marginals alone do not, in general, determine the distribution of a dependent sum.

Definition: Random variables $X$ and $Y$ are independent if $P(X\in A,\ Y\in B)=P(X\in A)P(Y\in B)$ for every pair of Borel sets $A,B\subseteq\mathbb R$. Their joint law records probabilities for pairs $(X,Y)$; their marginal laws record probabilities for $X$ and $Y$ separately. The convolution of two marginal laws is the distribution of $X+Y$ when the joint law is their product.

The Sum Is Determined by the Joint Law

Let $\mu$ be the joint probability law of $(X,Y)$. For every Borel set $C\subseteq\mathbb R$, the event $X+Y\in C$ consists of the pairs $(x,y)$ whose sum belongs to $C$. Thus the general law of the sum is obtained by measuring that set under the joint law—not by multiplying the marginals.

Theorem: For any real-valued random variables $X,Y$ and any Borel set $C$, $$ P(X+Y\in C)=\int_{\mathbb R^2}\mathbf{1}_{C}(x+y)\,\mu(dx,dy), $$ where $\mu$ is the joint law of $(X,Y)$ and $\mathbf{1}_{C}$ is 1 when its argument belongs to $C$, and 0 otherwise. If $X$ and $Y$ are independent, then $\mu=\mu_X\otimes\mu_Y$, so the same probability is $$ \int_{\mathbb R}\int_{\mathbb R}\mathbf{1}_{C}(x+y)\,\mu_Y(dy)\,\mu_X(dx). $$

Proof. By definition, $\mu$ assigns to a Borel set $D\subseteq\mathbb R^2$ the probability $P((X,Y)\in D)$. The set of pairs whose sum belongs to $C$ is the Borel set $D=\{(x,y):x+y\in C\}$, since addition is continuous. Therefore,

P(X+Y\in C) =P((X,Y)\in D) =\int_{\mathbb R^2}\mathbf{1}_{D}(x,y)\,\mu(dx,dy) =\int_{\mathbb R^2}\mathbf{1}_{C}(x+y)\,\mu(dx,dy).

When $X$ and $Y$ are independent, their joint law is the product measure $\mu_X\otimes\mu_Y$, which gives the displayed iterated integral. The indicator is bounded and measurable, so the product-measure integral is well-defined. $\square$

This result gives the discrete calculation directly. If $X$ and $Y$ have countable ranges, then for any possible sum value $s$,

P(X+Y=s)=\sum_x P(X=x,\ Y=s-x).

Only under independence may each joint probability in this sum be replaced by $P(X=x)P(Y=s-x)$. That replacement is precisely the step that produces the discrete convolution formula.

Worked Example: Perfectly Dependent Bernoulli Variables

Let $X$ be Bernoulli with $P(X=0)=P(X=1)=1/2$, and set $Y=X$. Find the law of $S=X+Y$, then compare it with the convolution of the two marginal laws.

Since $Y=X$, the joint law assigns probability $1/2$ to $(0,0)$, probability $1/2$ to $(1,1)$, and zero to the other pairs. The sum is $S=2X$, so

P(S=0)=\frac12,\qquad P(S=1)=0,\qquad P(S=2)=\frac12.

Both marginals are Bernoulli with success probability $1/2$. If we multiply their marginal probabilities as though the variables were independent, the resulting convolution gives

P_{\mathrm{conv}}(S=0)=\frac12\cdot\frac12=\frac14,\qquad P_{\mathrm{conv}}(S=1)=\frac12\cdot\frac12+\frac12\cdot\frac12=\frac12,\qquad P_{\mathrm{conv}}(S=2)=\frac12\cdot\frac12=\frac14.

These are not the actual probabilities. In particular, the dependent variables are always equal, so their sum cannot be 1. The marginals are correct; the error is treating their joint law as a product.

Continuous Convolution Also Uses Independence

When a pair has a joint density, the joint-law calculation can be written as an integral. For a general dependent pair with joint density $f_{X,Y}$, a density of $S=X+Y$ is

f_S(s)=\int_{\mathbb R} f_{X,Y}(x,s-x)\,dx

for almost every $s$. To see the formula, transform coordinates from $(x,y)$ to $(x,s)$, where $s=x+y$. The inverse transformation is $(x,s)\mapsto(x,s-x)$, whose Jacobian determinant has absolute value 1. Integrating the transformed joint density over $x$ gives the marginal density of $s$.

Independence supplies the additional factorization $f_{X,Y}(x,y)=f_X(x)f_Y(y)$ almost everywhere. Substitution then gives the familiar continuous convolution formula. Without independence, the joint density must remain in the integral.

Theorem: If $X$ and $Y$ are independent and have densities $f_X$ and $f_Y$, then $S=X+Y$ has a density given almost everywhere by $$ f_S(s)=\int_{\mathbb R} f_X(x)f_Y(s-x)\,dx. $$ For a dependent pair with joint density $f_{X,Y}$, the corresponding formula is $f_S(s)=\int_{\mathbb R}f_{X,Y}(x,s-x)\,dx$ almost everywhere.

Proof. Under the coordinate transformation $(x,y)\mapsto(x,s)=(x,x+y)$, the inverse is $(x,s)\mapsto(x,s-x)$, and its Jacobian determinant is

\det\begin{pmatrix}1&0\\-1&1\end{pmatrix}=1.

The change-of-variables formula therefore gives the joint density of $(X,S)$ as $f_{X,Y}(x,s-x)$. Integrating out $x$ gives the density of $S$ for almost every $s$. For independent $X,Y$, their joint density is the product $f_X(x)f_Y(y)$ almost everywhere, yielding the stated convolution formula. These integrations are valid for the nonnegative density; Tonelli's theorem permits the order of integration to be exchanged. $\square$

Worked Example: A Continuous Pair with Complete Dependence

Let $X\sim\operatorname{Uniform}(0,1)$ and set $Y=X$. Compare the true density of $S=X+Y$ with the density obtained by convolving the marginals as if $X$ and $Y$ were independent.

Because $S=2X$, for $0<s<2$,

F_S(s)=P(2X\le s)=P(X\le s/2)=\frac{s}{2}.

The CDF is 0 for $s\le0$ and 1 for $s\ge2$. Differentiating on $(0,2)$ shows that the actual density is $f_S(s)=1/2$ there, and 0 outside that interval.

Each marginal density is $f(x)=1$ for $0<x<1$ and 0 elsewhere. Their independent convolution would be

\int_{\mathbb R} f(x)f(s-x)\,dx = \begin{cases} s, & 0<s<1,\\ 2-s, & 1\le s<2,\\ 0, & \text{otherwise}. \end{cases}

Indeed, the integrand is 1 precisely when both $x$ and $s-x$ lie in $(0,1)$; the length of the permissible interval for $x$ is $s$ for $0<s<1$, and $2-s$ for $1\le s<2$. At $s=1/2$, the actual density is $1/2$, whereas the independent convolution density is $1/2$ as well; at $s=1/4$, they are $1/2$ and $1/4$, respectively. Thus the densities differ, even though they happen to agree at some points.

This dependent pair lies on the diagonal $y=x$, a set of area zero in the plane. Its joint law has no density with respect to area, so it cannot be handled by substituting a joint density into the dependent-pair formula. The joint-law theorem still applies: the pair has a well-defined probability law even when it has no joint density.

Worked Directly from a Dependent Joint Law

For discrete variables, a joint probability table makes the required diagonal summation visible. Each probability for $X+Y=s$ is the sum of the joint probabilities at all table entries whose coordinates add to $s$.

Worked Example: Summing a Joint Probability Table

Suppose the joint probability mass function of $X$ and $Y$ is given below. Find the distribution of $S=X+Y$.

$X\backslash Y$	$Y=0$	$Y=1$	$Y=2$
$X=0$	0.10	0.20	0.10
$X=1$	0.30	0.10	0.20

First check that the entries sum to 1: $0.10+0.20+0.10+0.30+0.10+0.20=1.00$. Now group entries by their sums. The sum 0 occurs only at $(0,0)$; the sum 1 occurs at $(0,1)$ and $(1,0)$; the sum 2 occurs at $(0,2)$ and $(1,1)$; and the sum 3 occurs only at $(1,2)$. Therefore,

P(S=0)=0.10,\quad P(S=1)=0.20+0.30=0.50,\quad P(S=2)=0.10+0.10=0.20,\quad P(S=3)=0.20.

The probabilities sum to $0.10+0.50+0.20+0.20=1.00$, as required. To see why the marginals would not suffice here, the marginal probabilities are $P(X=0)=0.40$, $P(X=1)=0.60$, and $P(Y=0)=0.40$, $P(Y=1)=0.30$, $P(Y=2)=0.30$. Multiplying these marginals would instead produce sum probabilities $0.16, 0.36, 0.30, 0.18$ for sums 0 through 3. For example, the actual probability $P(X=0,Y=0)=0.10$ is not the product $0.40(0.40)=0.16$, confirming that the variables are dependent.

Common Mistake: Treating Marginals as a Joint Law

A frequent error is to see that $X$ and $Y$ have known marginal distributions and immediately multiply their probability masses or densities. That multiplication is not a neutral way of combining the marginals: it constructs the product joint law, which is the joint law for independent variables.

The correction is to ask what information about the pair is available. If independence is established, use the convolution formula. If a joint mass function or joint density is given, calculate from it. If only the marginals are known and dependence is possible, the law of the sum is generally not determined. Also, dependence alone does not prove that a particular convolution answer is wrong; it means the product-law justification is missing, so the answer must be checked against the actual joint law.

Key takeaway: A sum is governed by the joint distribution of its summands. Convolution replaces that joint law by a product of marginals, which is justified by independence—not by knowing the marginals alone.

Check Your Understanding

Use the joint-law formulas to answer the following.

For countable-valued $X,Y$, write the formula for $P(X+Y=s)$ without assuming independence. Which step changes under independence?
Let $X$ be Bernoulli with success probability $p$ and $Y=1-X$. What values can $X+Y$ take? Why need the marginals alone not establish this sum law?
If a dependent pair has a joint density, what density formula applies to $X+Y$? Identify the extra factorization needed to obtain convolution of the marginal densities.
In the joint-table example, which table entries contribute to $P(X+Y=2)$, and why are their probabilities added?
Does dependence necessarily imply that the convolution of the marginals gives a different sum distribution? Explain the distinction between a formula being unjustified and its coinciding with the correct answer.

\(X\backslash Y\)	\(Y=0\)	\(Y=1\)	\(Y=2\)
\(X=0\)	0.10	0.20	0.10
\(X=1\)	0.30	0.10	0.20